PHYS 232 - UNIVERSITY PHYSICS
II - Calculation of electric forces
between point charges
Question Bank - Set 8
Liberty University
Question 1
Question
Two point charges, Q1= 4.0µC and Q2= 6.0µC, are placed 10.0 cm apart in
air. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs since 1µC = 10−6C.
Q1= 4.0µC = 4.0×10−6C
Q2= 6.0µC = 6.0×10−6C
Step 2: Calculate the distance between the charges, r, in meters.
r= 10.0cm = 10.0×10−2m= 0.10 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law:
F=k|Q1Q2|
r2. where k= 8.99 ×109Nm2/C2is the Coulomb constant.
|F|=(8.99 ×109Nm2/C2)(|4.0×10−6C||6.0×10−6C|)
(0.10 m)2
=(8.99 ×109)(24 ×10−12)
0.01
=215.76 ×10−3
0.01
= 21.576 N
Therefore, the magnitude of the electric force between the charges is 21.576 N.
Question 2
Question
Three point charges are arranged on the corners of an equilateral triangle as
shown below. Charge 1 (q1= +3µC) is located at the top vertex, charge 2
(q2=−2µC) is at the bottom left vertex, and charge 3 (q3= +4µC) is at the
bottom right vertex. The side length of the triangle is 2 meters. Calculate the
net electric force acting on charge 2 due to charges 1 and 3.
q1= +3µC
q3= +4µC
q2=−2µC
Solution
Step 1: Calculate the distance between charges 2 and 1 (top vertex). The
distance between neighboring charges can be calculated using the side length of
the equilateral triangle. Since the triangle is equilateral, each side is equal to
2 meters. The distance between charge 2 and charge 1 is the diagonal of the
triangle, which can be calculated using Pythagorean theorem. Let d21 be the
distance between charge 2 and 1.
d21 =p22+ 22=√8 = 2√2meters
Step 2: Calculate the magnitude and direction of the electric force on charge
2 due to charge 1. The electric force between two charges can be calculated using
Coulomb’s law:
F12 =k|q1q2|
r2
12
where - kis Coulomb’s constant (8.99 ×109Nm2/C2), - q1and q2are the
magnitudes of the charges, and - r12 is the distance between the charges.
The direction is attractive if charges have opposite signs and repulsive if the
charges have the same sign.
F12 =(8.99 ×109)× |3×10−6|×|−2×10−6|
(2√2)2=26.97 ×10−15
8= 3.37×10−15 N
The force is attractive because q1and q2have opposite signs.
Step 3: Calculate the distance between charges 2 and 3 (bottom left and
right vertices). Since the triangle is equilateral, the distance between these two
charges is also equal to 2√2meters.
Step 4: Calculate the magnitude and direction of the electric force on charge
2 due to charge 3.
2
Question 3
Question
Three point charges are arranged as shown in the diagram:
+3q+2q+q
d d
If the charges are separated by a distance d(which is the same between all
the pairs of charges), what is the magnitude of the net electrostatic force acting
on the charge +2qdue to the other two charges?
Given: - Electric force between two point charges q1and q2:F=k|q1q2|
r2-
Coulomb’s constant k= 8.99 ×109N m2/C2
Solution
Step 1: Calculate the force F1,2between charge +3qand charge +2q:
F1,2=k|+ 3q·+2q|
d2
F1,2=k6q2
d2
Step 2: Calculate the force F2,3between charge +2qand charge +q:
F2,3=k|+ 2q·+q|
d2
F2,3=k2q2
d2
Step 3: Calculate the net force acting on charge +2q: The forces F1,2and
F2,3are in the same direction, so we add their magnitudes:
Fnet =F1,2+F2,3
Fnet =k6q2
d2+k2q2
d2
Fnet =k8q2
d2
Therefore, the magnitude of the net electrostatic force acting on the charge
+2qdue to the other two charges is 8kq2/d2.
3
Question 4
Question
Two point charges, q1and q2, are separated by a distance of 4 cm. If q1= 5 µC
and q2=−3µC, calculate the magnitude and direction of the electric force
between the charges.
Solution
Let’s start by using Coulomb’s Law to calculate the electric force between the
two charges.
Step 1: Identify the given information
• Charge q1= 5 µC
• Charge q2=−3µC
• Distance between the charges r= 4 cm = 0.04 m
Step 2: Calculate the electric force The magnitude of the electric force
between two charges is given by Coulomb’s Law:
F=k|q1q2|
r2
where - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the charges, and - ris
the distance between the charges.
Plugging in the values:
F= (8.99 ×109)|5×10−6× −3×10−6|
(0.04)2
Step 3: Calculate the magnitude of the electric force
F= 8.99 ×109×15 ×10−12
0.0016
F= 8.99 ×15
1.6×10−3
F≈8.43 ×10−2N
Step 4: Determine the direction of the electric force Since q1and q2
have opposite charges, the electric force is attractive and directed from q1to q2.
Therefore, the magnitude of the electric force between the charges is approx-
imately 8.43 ×10−2N, and it is directed from q1to q2.
4
Question 5
Question
A small charge of +3.0µC is placed 20 cm to the right of a larger charge of
−6.0µC. Calculate the magnitude and direction of the electric force that the
larger charge exerts on the smaller charge.
Solution
Step 1: Determine the distance between the charges in meters. Given that the
smaller charge is placed 20 cm to the right of the larger charge, the distance
between the charges is d= 20 cm = 0.20 m.
Step 2: Calculate the magnitude of the electric force using Coulomb’s Law.
The magnitude of the electric force Fbetween two point charges q1and q2
separated by a distance ris given by:
F=k
q1·q2
r2
where kis Coulomb’s constant, 8.99 ×109N m2/C2.
Plugging in the given values:
F= (8.99 ×109)
(3.0×10−6)·(−6.0×10−6)
(0.20)2
Step 3: Calculate the magnitude of the electric force.
F= (8.99 ×109)
−18 ×10−12
0.04
F= 8.99 ×109×4.5×10−10
F= 4.0455 N
Step 4: Determine the direction of the electric force. The electric force
between the two charges is attractive since they have opposite signs. Therefore,
the larger charge of −6.0µC exerts a force to the right on the smaller charge of
+3.0µC.
Thus, the magnitude of the electric force that the larger charge exerts on
the smaller charge is 4.0455 N to the right.
Question 6
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. Charge q1=−2nC is located at the top vertex, charge q2= 4
nC is located at the bottom left vertex, and charge q3=−3nC is located at the
5
bottom right vertex. Calculate the magnitude and direction of the net electric
force experienced by the charge at the top vertex.
q2= 4 nC
q1=−2nC q3=−3nC
Solution
Step 1: We start by determining the direction and magnitude of the electric
force between q1and q2. The electric force between two point charges is given
by Coulomb’s Law:
F=k· |q1|·|q2|
r2
The direction of the force is along the line connecting the charges, and the
force is repulsive if both charges have the same sign and attractive if they have
opposite signs.
Step 2: Calculate the distance rbetween q1and q2. Since the charges are
at the vertices of an equilateral triangle,
r=2
√3a
where ais the side length of the triangle.
Step 3: Calculate the magnitude of the electric force between q1and q2using
Coulomb’s Law:
F12 =k· |q1|·|q2|
r2
Step 4: Next, we determine the direction and magnitude of the electric force
between q1and q3. Similarly to Step 1, calculate the distance between q1and
q3and then the force between them.
Step 5: Finally, calculate the net electric force experienced by the charge at
the top vertex by considering the vector sum of the forces between q1and q2and
between q1and q3. Remember to consider both the magnitude and direction of
each force.
Question 7
Question
Two point charges, Q1=−3.0µC and Q2= 6.0µC, are placed 10.0 cm apart in
a vacuum. Calculate the magnitude of the electric force that Q1exerts on Q2.
Solution
Let’s first write down the formula for the electric force between two point
charges:
F=k|Q1Q2|
r2
6
where F= magnitude of the electric force, k= Coulomb’s constant (8.99 ×
109N m2/C2), Q1and Q2= magnitudes of the charges, and r= separation
distance between the charges.
Step 1: Convert the charges to Coulombs. Since 1µC= 10−6C, Q1=
−3.0µC=−3.0×10−6C and Q2= 6.0µC= 6.0×10−6C.
Step 2: Calculate the force using the formula.
F=(8.99 ×109)×|−3.0×10−6×6.0×10−6|
(0.10)2
F=8.99 ×109×18 ×10−12
0.01
F=161.82 ×10−3
0.01 = 16.182 N
So, the magnitude of the electric force that Q1exerts on Q2is 16.182 N.
Question 8
Question
Three point charges are arranged in a line as shown below:
+2 µC−5µC+3 µC
The charges are located at positions x=−5cm, x= 0 cm, and x= 5 cm
respectively. Determine the magnitude and direction of the net electrostatic
force acting on the +3 µC charge.
Solution
Step 1: Calculate the distance between each pair of charges. First, we will
determine the distances r1,r2, and r3between the charges +2 µC and −5µC,
−5µC and +3 µC, and +2 µC and +3 µC respectively.
r1= 5 cm −(−5cm) = 10 cm = 0.10 m
r2= 0 cm −(−5cm) = 5 cm = 0.05 m
r3= 5 cm −0cm = 5 cm = 0.05 m
Step 2: Calculate the magnitude of the force between each pair of charges.
The magnitude of the electrostatic force between two point charges q1and
q2separated by a distance ris given by Coulomb’s Law:
F=k|q1q2|
r2
Where k= 8.99 ×109N m2/C2is Coulomb’s constant.
7
For charges +2 µC and −5µC:
F1= 8.99 ×109|2×10−6C×(−5×10−6C)|
(0.10 m)2
F1= 8.99 ×109×1×10−11 N= 8.99 ×10−2N= 0.0899 N
For charges −5µC and +3 µC:
F2= 8.99 ×109| − 5×10−6C×3×10−6C|
(0.05 m)2
F2= 8.99 ×109×1.5×10−11 N= 1.35 ×10−2N= 0.0135 N
For charges +2 µC and +3 µC:
F3= 8.99 ×109|2×10−6C×3×10−6C|
(0.05 m)2
F3= 8.99
Question 9
Question
Three point charges are placed at the corners of an equilateral triangle of side
d. The charges are +q,−2q, and +3q. What is the net force on the charge +q
due to the other two charges? Assume the charges are placed at the vertices of
the triangle and the sides are of equal length.
Solution
Let’s denote the charge +qas Q1, the charge −2qas Q2, and the charge +3qas
Q3. We will first find the force on Q1due to Q2and then the force on Q1due
to Q3. Finally, we will find the net force by summing up the forces.
Step 1: Calculate the force on Q1due to Q2. The electric force between
two charges Q1and Q2is given by Coulomb’s law:
F12 =k|Q1Q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2) and ris the distance
between the charges. The distance between Q1and Q2in this case, considering
the equilateral triangle, is d. So, the force F12 is given by:
F12 =k|q·(−2q)|
d2=k|q2·2|
d2=2kq2
d2
8
Step 2: Calculate the force on Q1due to Q3. Following the same steps as
above, we find the force F13 on Q1due to Q3:
F13 =k|q·3q|
d2=k|q2·3|
d2=3kq2
d2
Step 3: Calculate the net force on Q1. The net force is the vector sum of
F12 and F13, where the direction of F12 is opposite to F13 (since Q2is negative
and Q3is positive).
Fnet =F13 −F12 =3kq2
d2−2kq2
d2=kq2
d2
Therefore, the net force on the charge +qdue to the other two charges is
kq2
d2directed towards the charge +3q.
Question 10
Question
Three point charges are arranged along the x-axis as follows: Q1=−2µC at
the origin, Q2= 4 µC at x= 3 m, and Q3=−3µC at x= 5 m. Calculate the
net force on Q2due to Q1and Q3.
Solution
Step 1: Determine the force on Q2due to Q1. The force F12 on Q2due to Q1
can be calculated using the electric force equation:
F12 =k|Q1Q2|
r2
12
where: - k= 8.99 ×109Nm2/C2is the Coulomb constant, - Q1=−2×10−6C,
-Q2= 4 ×10−6C, - r12 = 3 mis the distance between Q1and Q2.
Plugging in the values:
F12 =8.99 ×109×|−2×10−6×4×10−6|
(3)2
F12 =8.99 ×109×8×10−12
9= 7.99 ×10−3N
Step 2: Determine the force on Q2due to Q3. The force F23 on Q2due to
Q3can be calculated using the electric force equation:
F23 =k|Q2Q3|
r2
23
9
where: - Q2= 4 ×10−6C, - Q3=−3×10−6C, - r23 = 2 mis the distance
between Q2and Q3.
Plugging in the values:
F23 =8.99 ×109× |4×10−6× −3×10−6|
(2)2
F23 =8.99 ×109×12 ×10−12
4= 26.97 ×10−3= 2.697 ×10−2N
Step 3: Calculate the net force on Q2. The net force on Q2is the vector
sum of forces F12 and F23.
Fnet =F12 +F23 = 7.99 ×10−3+ 2.697 ×10−2
Fnet ≈3.49 ×10−2N
Therefore, the net force on Q2due to Q1and Q3is approximately 3.49 ×
10−2Nin the positive x-direction.
Question 12
Question
Three point charges are arranged in the vertices of an equilateral triangle of side
length a= 2 m as shown in the figure below. Charge q1=−3µC is at the top
vertex, charge q2= 5 µC is at the bottom left vertex, and charge q3=−4µC is
at the bottom right vertex. Calculate the net electric force acting on q3.
Solution
Step 1: Calculate the electric force acting on charge q3due to charges q1and
q2separately.
The electric force between two charges q1and q3is given by Coulomb’s Law:
F13 =k|q1||q3|
r2
13
where kis Coulomb’s constant, r13 is the distance between charges q1and q3,
and |q1|= 3 ×10−6C, |q3|= 4 ×10−6C.
Given that the side length of the equilateral triangle is a= 2 m, the distance
r13 between charges q1and q3is:
r13 =a
Plugging the given values into Coulomb’s Law, we get:
F13 =9×109×3×10−6×4×10−6
(2)2
10
F13 =108
4×10−9×103
F13 = 27 ×10−3N
So, the electric force between charges q1and q3is 27 ×10−3N.
Using the same method, we can calculate the electric force between charges
q2and q3as:
F23 =9×109×5×10−6×4×10−6
(2)2
F23 =180
4×10−9×103
F23 = 45 ×10−3N
So, the electric force between charges q2and q3is 45 ×10−3N.
Step 2: Calculate the net electric force acting on charge q3.
To find the net force, we need to consider the direction of the forces. Since
the charges q1and q3have opposite signs, their forces will be in opposite direc-
tions. However, the forces between q2and q3will both be pointing towards q3.
Calculating the net force:
Fnet =F23 −F13
Fnet = 45 ×10−3−27 ×10−3
Fnet = 18 ×10−3
Question 13
Question
Three point charges are placed at the vertices of an equilateral triangle of side
2.0 cm. Two of the charges are +3.0 nC, and the third charge is +5.0 nC.
a) What is the net electric force exerted on the charge of +3.0 nC?
b) What is the net electric force exerted on the charge of +5.0 nC?
Solution
a) To find the net electric force on the +3.0 nC charge, we need to consider the
forces exerted by the other two charges.
Let’s label the charges at the vertices of the triangle: A (+3.0 nC), B (+3.0
nC), and C (+5.0 nC).
Step 1: Calculate the distance between charges A and B. Since they are
at the vertices of an equilateral triangle of side 2.0 cm, the distance between A
and B is 2.0cm.
11
Step 2: Calculate the magnitude of the electric force between charges A
and B using Coulomb’s law:
FAB =k·q1·q2
r2
where k= 8.988×109N m2/C2is the electrostatic constant, q1= +3.0×10−9C
is the charge of A, q2= +3.0×10−9C is the charge of B, and r= 0.02 m is the
distance between A and B.
FAB = (8.988 ×109)·(3.0×10−9)2
(0.02)2
FAB ≈1.343 ×10−3N
This force is repulsive since the charges have the same sign.
Step 3: Find the net electric force on charge A. The force
FAB acts along
the line joining charges A and B, and it has the direction away from charge B.
Since A and C are equidistant from A, the force
FAC also acts along the line
joining A and C, but in the opposite direction to that of
FAB .
The net force on charge A will be the vector sum of
FAB and
FAC .
Fnet =
FAB +
FAC
Since the forces are along the same line, we can add them as scalars.
Fnet = 1.343 ×10−3N+ 1.343 ×10−3N
Fnet ≈2.686 ×10−3N
Therefore, the net electric force on the +3.0 nC charge is approximately
2.686 ×10−3N away from the charge at C.
b) The net electric force on the +5.0 nC charge can be calculated in a similar
way considering the forces from charges A and B.
Question 14
Question
Two point charges, q1=−4µC and q2= 2 µC, are located 5 cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
To find the electric force between the charges, we can use Coulomb’s law, which
states that the magnitude of the electric force between two point charges is given
by:
F=k·|q1·q2|
r2
12
where kis the Coulomb’s constant (8.9875 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 1: Convert the distance between charges from cm to meters:
r= 5 cm = 0.05 m
Step 2: Substitute the given values into Coulomb’s law to find the magni-
tude of the electric force:
F= (8.9875 ×109)·| − 4×10−6·2×10−6|
(0.05)2
Step 3: Calculate the magnitude of the electric force:
F= (8.9875 ×109)·8×10−12
0.0025
F= (8.9875 ×109)·3.2×10−9
F= 2.876 ×10−7N
Therefore, the magnitude of the electric force between the charges is 2.876 ×
10−7N.
Question 15
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length a. Charge q1is at the top vertex, charge q2is at the bottom left
vertex, and charge q3is at the bottom right vertex. Calculate the magnitude of
the net electric force on charge q1.
Solution
Step 1: Calculate the electric force between q1and q2. The distance between q1
and q2is a
√3(half of a side length of an equilateral triangle). The magnitude of
the electric force between two charges q1and q2is given by Coulomb’s law:
F12 =k|q1q2|
r2
12
where - kis Coulomb’s constant, k= 8.99 ×109N m2/C2, - r12 =a
√3is the dis-
tance between q1and q2, - q1=q2=q, as the charges are the same magnitude.
Thus,
F12 =(8.99 ×109N m2/C2)(q2)
a
√32
13
Step 2: Calculate the electric force between q1and q3. The distance between
q1and q3is 2a
√3(the full diagonal length of the equilateral triangle). Similarly,
applying Coulomb’s law:
F13 =(8.99 ×109N m2/C2)(q2)
2a
√32
Step 3: Find the net electric force on q1. The net electric force on charge
q1is in the upward direction due to the forces from q2and q3. The vertical
components of F12 and F13 add up to give the net force on q1. To find the net
force, subtract the force due to q3from the force due to q2.
Fnet = 2F12 sin(30◦)−F13
Solving this expression using the values from Steps 1 and 2 will give the
magnitude of the net electric force on charge q1.
Question 16
Question
Three point charges are placed at the corners of an equilateral triangle with side
length d. The charges are +q,+2q, and −3q. Calculate the magnitude of the
net electrostatic force on the +2qcharge due to the other two charges.
Solution
To find the net electrostatic force on the +2qcharge, we need to calculate
the individual forces on it due to the other two charges and then add them
vectorially.
Step 1: Calculate the force on the +2qcharge due to the +qcharge
The magnitude of the force between two charges q1and q2separated by a
distance ris given by Coulomb’s law:
F=k|q1||q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2). The distance between
the +2qand +qcharges in the triangle can be calculated using the height of
the equilateral triangle. The height of an equilateral triangle with side length d
is √3
2d. So, the distance between the +2qand +qcharges is d. Therefore, the
magnitude of the force on the +2qcharge due to the +qcharge is:
F1=k|2q||q|
d2
Step 2: Calculate the force on the +2qcharge due to the −3qcharge
Using the same reasoning as in Step 1, the distance between the +2qand −3q
14
charges is also d. Therefore, the magnitude of the force on the +2qcharge due
to the −3qcharge is:
F2=k|2q||3q|
d2
Step 3: Determine the net force on the +2qcharge The net force on
the +2qcharge is the vector sum of F1and F2. The forces F1and F2act in
opposite directions along the same line passing through the +2qcharge, so we
just need to subtract their magnitudes:
Fnet =F1−F2=k|2q||q|
d2−k|2q||3q|
d2
Solving this expression will give us the magnitude of the net electrostatic
force on the +2qcharge.
Question 17
Question
Three point charges are arranged in a line. Charge q1=−4.0µC is located at
the origin, charge q2= 6.0µC is located at x= 3.0m, and charge q3=−2.0µC
is located at x= 5.0m. Calculate the net electrostatic force on charge q2due
to charges q1and q3.
Solution
To find the net electrostatic force on charge q2, we first need to calculate the
forces acting on charge q2due to charges q1and q3separately, then sum these
forces vectorially.
Step 1: Calculate force on charge q2due to charge q1
The magnitude of the force on charge q2due to charge q1is given by
Coulomb’s law:
F1 on 2 =k|q1q2|
r2
12
where - kis the electrostatic constant, 8.99 ×109N m2/C2, - q1=−4.0µC=
−4.0×10−6C, - q2= 6.0µC= 6.0×10−6C, - r12 = 3.0m.
Plugging in these values, we get:
F1 on 2 = (8.99 ×109)| − 4.0×10−6×6.0×10−6|
(3.0)2
F1 on 2 = (8.99 ×109)24 ×10−12
9= 23.97 N (repulsive)
So, the force on charge q2due to charge q1is 23.97 N in the positive direction.
Step 2: Calculate force on charge q2due to charge q3
15
Similarly, the magnitude of the force on charge q2due to charge q3is given
by:
F3 on 2 =k|q3q2|
r2
32
where - q3=−2.0µC=−2.0×10−6C, - r32 = 2.0m.
Plugging in these values, we get:
F3 on 2 = (8.99 ×109)| − 2.0×10−6×6.0×10−6|
(2.0)2
F3 on 2 = (8.99 ×109)12 ×10−12
4= 26.97 N (attractive)
Therefore, the force on charge q2due to charge q3is 26.97 N in the negative
direction.
Step 3: Calculate the net force on charge q2
The net force exerted on charge q2is the vector sum of the forces due to
charges q1and q3:
Net force on q2=F1 on 2 +F3 on 2 = 23.97 N+ (−26.97 N)
Question 18
Question
Two point charges, q1= +4.0µC and q2=−2.0µC, are initially separated by
a distance of 10.0cm. If the charges are brought 20.0cm closer together, what
is the change in the magnitude of the electric force between them?
Solution
Step 1: Calculate the initial force between the charges using Coulomb’s law:
The electric force between two point charges can be calculated using Coulomb’s
law:
F=k|q1·q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant.
Substitute the given values:
F1= (8.99 ×109)|4.0×10−6·(−2.0×10−6)|
(0.10)2
Calculating F1gives:
F1≈5.758 N
Step 2: Calculate the final force when the charges are 20.0cm closer together:
After the charges are brought 20.0cm closer together, the new separation dis-
tance is r′= 10.0cm −20.0cm =−10.0cm.
16
Calculate the final force F2using the new separation distance r′:
F2= (8.99 ×109)|4.0×10−6·(−2.0×10−6)|
(0.10 −0.20)2
Calculating F2gives:
F2≈23.032 N
Step 3: Calculate the change in magnitude of the electric force: The change
in the magnitude of the electric force is given by:
∆F=|F2−F1|
Substitute the calculated values to find ∆F:
∆F=|23.032 −5.758|
∆F≈17.274 N
Therefore, the change in the magnitude of the electric force between the
charges is approximately 17.274 N.
Question 19
Question
Three point charges are located on the x-axis: a charge of +3 nC at x = 0, a
charge of -2 nC at x = 60 cm, and a charge of +5 nC at x = 100 cm. Calculate
the net electric force on the +5 nC charge.
Solution
Step 1: Convert all distances to meters. Given: q1= +3 nC = 3 ×10−9C at
x= 0 m, q2=−2nC =−2×10−9C at x= 60 cm = 0.6m, q3= +5 nC =
5×10−9C at x= 100 cm = 1.0m.
Step 2: Calculate the net electric force on the +5 nC charge. The net force
on the +5 nC charge is the sum of the forces due to the other charges. The
force between two charges is given by Coulomb’s Law:
F=k|q1q2|
r2,
where k= 8.99 ×109N m2/C2is the electrostatic constant, q1and q2are the
magnitudes of the two charges, and ris the separation between the charges.
The force between q1and q3is:
F13 =k|q1q3|
r2
13
.
17
The force between q2and q3is:
F23 =k|q2q3|
r2
23
.
The net force on the +5 nC charge due to q1and q2is:
Fnet =F13 +F23.
Substitute the given values and solve for the net force.
Question 20
Question
Two point charges, +3.0µC and −6.0µC, are placed 10.0cm apart in air. Calcu-
late the magnitude and direction of the electric force experienced by the positive
charge due to the negative charge.
Solution
Step 1: Convert the charges to coulombs. Given that 1µC= 10−6C, the first
charge +3.0µC becomes +3.0×10−6C and the second charge −6.0µC becomes
−6.0×10−6C.
Step 2: Calculate the distance between the charges in meters. The distance
between the charges is 10.0cm = 0.10 m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
The magnitude of the electric force Fbetween two point charges is given by
Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Substitute the given values into the formula:
F=(8.99 ×109)× |3.0×10−6×6.0×10−6|
(0.10)2
Step 4: Calculate the magnitude of the electric force.
F=(8.99 ×109)×18 ×10−12
0.01
Step 5: Simplify the expression.
F= 161.82 N
18
Step 6: Determine the direction of the electric force. The electric force is
attractive since the charges are of opposite sign. Therefore, the electric force
experienced by the positive charge is directed towards the negative charge.
Therefore, the magnitude of the electric force experienced by the positive
charge due to the negative charge is 161.82 N and it is directed towards the
negative charge.
Question 21
Question
Two point charges, q1=−6µC and q2= 4 µC, are placed 10 cm apart in air.
Calculate the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−6µC =−6×10−6C
q2= 4 µC = 4 ×10−6C
Step 2: Calculate the electric force. The electric force between two point
charges can be calculated using Coulomb’s Law:
F=k· |q1·q2|
r2
where Fis the magnitude of the electric force, kis the Coulomb constant (8.99×
109Nm2/C2), q1and q2are the magnitudes of the charges, and ris the distance
between the charges.
Plugging in the values:
F=(8.99 ×109Nm2/C2)· |(−6×10−6C)·(4 ×10−6C)|
(0.1m)2
Step 3: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(6 ×10−6)·(4 ×10−6)
0.01
F=215.76
0.01
F= 21576 N
Therefore, the magnitude of the electric force between the charges is 21576
N.
Question 22
Question
Two point charges of +4.0 µC and -6.0 µC are placed 0.10 m apart. Calculate
the magnitude of the electric force between them.
19
Solution
Step 1: Convert the charges to SI units. Given: q1= +4.0µC= 4.0×10−6C
q2=−6.0µC=−6.0×10−6C
Step 2: Calculate the distance between the charges. Given: r= 0.10 m
Step 3: Calculate the magnitude of the electric force between the charges
using Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the Coulomb constant with a value of 8.99 ×109N·m2/C2.
Substitute the given values into the formula:
F=(8.99 ×109)·(4.0×10−6)·(6.0×10−6)
(0.10)2
Step 4: Calculate the magnitude of the electric force.
F=35.96 ×103
0.01 = 3.596 ×106N
Therefore, the magnitude of the electric force between the charges is 3.596 ×
106N.
Question 23
Question
Two point charges with magnitudes q1= 3.0µC and q2=−5.0µC are placed
20.0 cm apart in a vacuum. Calculate the magnitude and direction of the electric
force that each charge exerts on the other.
Solution
To calculate the magnitude and direction of the electric force between the two
charges, we will use Coulomb’s law.
Step 1: Calculate the electric force F12 that charge q2exerts on charge q1.
Coulomb’s law is given by:
F12 =k|q1q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the two charges.
Plugging in the values, we get:
F12 = (8.99 ×109)(3.0×10−6)(5.0×10−6)
(0.20)2
F12 = 2.248 N
20
The direction of the force is from charge q2to charge q1.
Step 2: Calculate the electric force F21 that charge q1exerts on charge q2.
Since the forces act along the line joining the charges, the magnitude of F21 is
the same as that of F12.
Therefore, the magnitude of the electric force that each charge exerts on the
other is 2.248 N, and the direction of the force from charge q1to charge q2.
Question 24
Question
Two point charges, q1=−3.0µC and q2= 4.0µC, are placed 6.0 cm apart.
Calculate the magnitude and direction of the electric force exerted on q1.
Solution
Step 1: Convert the given charges to their equivalent in coulombs. Using the
conversion factor 1µC= 10−6C, we have: q1=−3.0µC=−3.0×10−6C, and
q2= 4.0µC= 4.0×10−6C.
Step 2: Given data: q1=−3.0×10−6Cq2= 4.0×10−6Cr= 6.0cm =
0.06 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k·
q1·q2
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging the values into the formula gives:
F= 8.99 ×109·3.0×10−6·4.0×10−6
(0.06)2
F= 8.99 ×109·12 ×10−12
0.0036
F= 8.99 ×109·0.000003333
F= 29.97 N
Step 4: Determine the direction of the electric force. Since q1is negative and
q2is positive, the electric force exerted on q1is attractive, pulling q1towards
q2. Therefore, the direction of the electric force is towards q2.
Question 25
Question
Two point charges, q1= 4.0µC and q2=−6.0µC, are placed 12.0cm apart in
air. Calculate the magnitude of the electric force between these charges.
21
Solution
Step 1: Let’s first convert the charges to Coulombs by using the relationship
1µC= 10−6C. Thus, q1= 4.0×10−6C and q2=−6.0×10−6C.
Step 2: Next, let’s calculate the distance between the charges in meters.
This can be done by converting 12.0cm to meters: 12.0cm = 0.12 m.
Step 3: Now, we can use Coulomb’s law to find the magnitude of the electric
force between the two charges. Coulomb’s law states that the magnitude of the
electric force between two point charges is given by:
F=k
q1q2
r2
,
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 4: Substituting the given values into Coulomb’s law, we have:
F= (8.99 ×109N m2/C2)|4.0×10−6C· −6.0×10−6C|
(0.12 m)2
Step 5: Calculating the magnitude of the electric force:
F= (8.99 ×109)(2.4×10−11)/(0.0144) N
F≈1.495 ×10−3N
Therefore, the magnitude of the electric force between the charges q1=
4.0µC and q2=−6.0µC, separated by a distance of 12.0cm, is approximately
1.495 ×10−3N.
22
q1
q2q3
23
Question 2
Question
Three point charges are arranged on the corners of an equilateral triangle as
shown below. Charge 1 (q1= +3µC) is located at the top vertex, charge 2
(q2=−2µC) is at the bottom left vertex, and charge 3 (q3= +4µC) is at the
bottom right vertex. The side length of the triangle is 2 meters. Calculate the
net electric force acting on charge 2 due to charges 1 and 3.
q1= +3µC
q3= +4µC
q2=−2µC
Solution
Step 1: Calculate the distance between charges 2 and 1 (top vertex). The
distance between neighboring charges can be calculated using the side length of
the equilateral triangle. Since the triangle is equilateral, each side is equal to
2 meters. The distance between charge 2 and charge 1 is the diagonal of the
triangle, which can be calculated using Pythagorean theorem. Let d21 be the
distance between charge 2 and 1.
d21 =p22+ 22=√8 = 2√2meters
Step 2: Calculate the magnitude and direction of the electric force on charge
2 due to charge 1. The electric force between two charges can be calculated using
Coulomb’s law:
F12 =k|q1q2|
r2
12
where - kis Coulomb’s constant (8.99 ×109Nm2/C2), - q1and q2are the
magnitudes of the charges, and - r12 is the distance between the charges.
The direction is attractive if charges have opposite signs and repulsive if the
charges have the same sign.
F12 =(8.99 ×109)× |3×10−6|×|−2×10−6|
(2√2)2=26.97 ×10−15
8= 3.37×10−15 N
The force is attractive because q1and q2have opposite signs.
Step 3: Calculate the distance between charges 2 and 3 (bottom left and
right vertices). Since the triangle is equilateral, the distance between these two
charges is also equal to 2√2meters.
Step 4: Calculate the magnitude and direction of the electric force on charge
2 due to charge 3.
2
Question 3
Question
Three point charges are arranged as shown in the diagram:
+3q+2q+q
d d
If the charges are separated by a distance d(which is the same between all
the pairs of charges), what is the magnitude of the net electrostatic force acting
on the charge +2qdue to the other two charges?
Given: - Electric force between two point charges q1and q2:F=k|q1q2|
r2-
Coulomb’s constant k= 8.99 ×109N m2/C2
Solution
Step 1: Calculate the force F1,2between charge +3qand charge +2q:
F1,2=k|+ 3q·+2q|
d2
F1,2=k6q2
d2
Step 2: Calculate the force F2,3between charge +2qand charge +q:
F2,3=k|+ 2q·+q|
d2
F2,3=k2q2
d2
Step 3: Calculate the net force acting on charge +2q: The forces F1,2and
F2,3are in the same direction, so we add their magnitudes:
Fnet =F1,2+F2,3
Fnet =k6q2
d2+k2q2
d2
Fnet =k8q2
d2
Therefore, the magnitude of the net electrostatic force acting on the charge
+2qdue to the other two charges is 8kq2/d2.
3
Question 4
Question
Two point charges, q1and q2, are separated by a distance of 4 cm. If q1= 5 µC
and q2=−3µC, calculate the magnitude and direction of the electric force
between the charges.
Solution
Let’s start by using Coulomb’s Law to calculate the electric force between the
two charges.
Step 1: Identify the given information
• Charge q1= 5 µC
• Charge q2=−3µC
• Distance between the charges r= 4 cm = 0.04 m
Step 2: Calculate the electric force The magnitude of the electric force
between two charges is given by Coulomb’s Law:
F=k|q1q2|
r2
where - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the charges, and - ris
the distance between the charges.
Plugging in the values:
F= (8.99 ×109)|5×10−6× −3×10−6|
(0.04)2
Step 3: Calculate the magnitude of the electric force
F= 8.99 ×109×15 ×10−12
0.0016
F= 8.99 ×15
1.6×10−3
F≈8.43 ×10−2N
Step 4: Determine the direction of the electric force Since q1and q2
have opposite charges, the electric force is attractive and directed from q1to q2.
Therefore, the magnitude of the electric force between the charges is approx-
imately 8.43 ×10−2N, and it is directed from q1to q2.
4
Question 5
Question
A small charge of +3.0µC is placed 20 cm to the right of a larger charge of
−6.0µC. Calculate the magnitude and direction of the electric force that the
larger charge exerts on the smaller charge.
Solution
Step 1: Determine the distance between the charges in meters. Given that the
smaller charge is placed 20 cm to the right of the larger charge, the distance
between the charges is d= 20 cm = 0.20 m.
Step 2: Calculate the magnitude of the electric force using Coulomb’s Law.
The magnitude of the electric force Fbetween two point charges q1and q2
separated by a distance ris given by:
F=k
q1·q2
r2
where kis Coulomb’s constant, 8.99 ×109N m2/C2.
Plugging in the given values:
F= (8.99 ×109)
(3.0×10−6)·(−6.0×10−6)
(0.20)2
Step 3: Calculate the magnitude of the electric force.
F= (8.99 ×109)
−18 ×10−12
0.04
F= 8.99 ×109×4.5×10−10
F= 4.0455 N
Step 4: Determine the direction of the electric force. The electric force
between the two charges is attractive since they have opposite signs. Therefore,
the larger charge of −6.0µC exerts a force to the right on the smaller charge of
+3.0µC.
Thus, the magnitude of the electric force that the larger charge exerts on
the smaller charge is 4.0455 N to the right.
Question 6
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. Charge q1=−2nC is located at the top vertex, charge q2= 4
nC is located at the bottom left vertex, and charge q3=−3nC is located at the
5
bottom right vertex. Calculate the magnitude and direction of the net electric
force experienced by the charge at the top vertex.
q2= 4 nC
q1=−2nC q3=−3nC
Solution
Step 1: We start by determining the direction and magnitude of the electric
force between q1and q2. The electric force between two point charges is given
by Coulomb’s Law:
F=k· |q1|·|q2|
r2
The direction of the force is along the line connecting the charges, and the
force is repulsive if both charges have the same sign and attractive if they have
opposite signs.
Step 2: Calculate the distance rbetween q1and q2. Since the charges are
at the vertices of an equilateral triangle,
r=2
√3a
where ais the side length of the triangle.
Step 3: Calculate the magnitude of the electric force between q1and q2using
Coulomb’s Law:
F12 =k· |q1|·|q2|
r2
Step 4: Next, we determine the direction and magnitude of the electric force
between q1and q3. Similarly to Step 1, calculate the distance between q1and
q3and then the force between them.
Step 5: Finally, calculate the net electric force experienced by the charge at
the top vertex by considering the vector sum of the forces between q1and q2and
between q1and q3. Remember to consider both the magnitude and direction of
each force.
Question 7
Question
Two point charges, Q1=−3.0µC and Q2= 6.0µC, are placed 10.0 cm apart in
a vacuum. Calculate the magnitude of the electric force that Q1exerts on Q2.
Solution
Let’s first write down the formula for the electric force between two point
charges:
F=k|Q1Q2|
r2
6
where F= magnitude of the electric force, k= Coulomb’s constant (8.99 ×
109N m2/C2), Q1and Q2= magnitudes of the charges, and r= separation
distance between the charges.
Step 1: Convert the charges to Coulombs. Since 1µC= 10−6C, Q1=
−3.0µC=−3.0×10−6C and Q2= 6.0µC= 6.0×10−6C.
Step 2: Calculate the force using the formula.
F=(8.99 ×109)×|−3.0×10−6×6.0×10−6|
(0.10)2
F=8.99 ×109×18 ×10−12
0.01
F=161.82 ×10−3
0.01 = 16.182 N
So, the magnitude of the electric force that Q1exerts on Q2is 16.182 N.
Question 8
Question
Three point charges are arranged in a line as shown below:
+2 µC−5µC+3 µC
The charges are located at positions x=−5cm, x= 0 cm, and x= 5 cm
respectively. Determine the magnitude and direction of the net electrostatic
force acting on the +3 µC charge.
Solution
Step 1: Calculate the distance between each pair of charges. First, we will
determine the distances r1,r2, and r3between the charges +2 µC and −5µC,
−5µC and +3 µC, and +2 µC and +3 µC respectively.
r1= 5 cm −(−5cm) = 10 cm = 0.10 m
r2= 0 cm −(−5cm) = 5 cm = 0.05 m
r3= 5 cm −0cm = 5 cm = 0.05 m
Step 2: Calculate the magnitude of the force between each pair of charges.
The magnitude of the electrostatic force between two point charges q1and
q2separated by a distance ris given by Coulomb’s Law:
F=k|q1q2|
r2
Where k= 8.99 ×109N m2/C2is Coulomb’s constant.
7
For charges +2 µC and −5µC:
F1= 8.99 ×109|2×10−6C×(−5×10−6C)|
(0.10 m)2
F1= 8.99 ×109×1×10−11 N= 8.99 ×10−2N= 0.0899 N
For charges −5µC and +3 µC:
F2= 8.99 ×109| − 5×10−6C×3×10−6C|
(0.05 m)2
F2= 8.99 ×109×1.5×10−11 N= 1.35 ×10−2N= 0.0135 N
For charges +2 µC and +3 µC:
F3= 8.99 ×109|2×10−6C×3×10−6C|
(0.05 m)2
F3= 8.99
Question 9
Question
Three point charges are placed at the corners of an equilateral triangle of side
d. The charges are +q,−2q, and +3q. What is the net force on the charge +q
due to the other two charges? Assume the charges are placed at the vertices of
the triangle and the sides are of equal length.
Solution
Let’s denote the charge +qas Q1, the charge −2qas Q2, and the charge +3qas
Q3. We will first find the force on Q1due to Q2and then the force on Q1due
to Q3. Finally, we will find the net force by summing up the forces.
Step 1: Calculate the force on Q1due to Q2. The electric force between
two charges Q1and Q2is given by Coulomb’s law:
F12 =k|Q1Q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2) and ris the distance
between the charges. The distance between Q1and Q2in this case, considering
the equilateral triangle, is d. So, the force F12 is given by:
F12 =k|q·(−2q)|
d2=k|q2·2|
d2=2kq2
d2
8
Step 2: Calculate the force on Q1due to Q3. Following the same steps as
above, we find the force F13 on Q1due to Q3:
F13 =k|q·3q|
d2=k|q2·3|
d2=3kq2
d2
Step 3: Calculate the net force on Q1. The net force is the vector sum of
F12 and F13, where the direction of F12 is opposite to F13 (since Q2is negative
and Q3is positive).
Fnet =F13 −F12 =3kq2
d2−2kq2
d2=kq2
d2
Therefore, the net force on the charge +qdue to the other two charges is
kq2
d2directed towards the charge +3q.
Question 10
Question
Three point charges are arranged along the x-axis as follows: Q1=−2µC at
the origin, Q2= 4 µC at x= 3 m, and Q3=−3µC at x= 5 m. Calculate the
net force on Q2due to Q1and Q3.
Solution
Step 1: Determine the force on Q2due to Q1. The force F12 on Q2due to Q1
can be calculated using the electric force equation:
F12 =k|Q1Q2|
r2
12
where: - k= 8.99 ×109Nm2/C2is the Coulomb constant, - Q1=−2×10−6C,
-Q2= 4 ×10−6C, - r12 = 3 mis the distance between Q1and Q2.
Plugging in the values:
F12 =8.99 ×109×|−2×10−6×4×10−6|
(3)2
F12 =8.99 ×109×8×10−12
9= 7.99 ×10−3N
Step 2: Determine the force on Q2due to Q3. The force F23 on Q2due to
Q3can be calculated using the electric force equation:
F23 =k|Q2Q3|
r2
23
9
where: - Q2= 4 ×10−6C, - Q3=−3×10−6C, - r23 = 2 mis the distance
between Q2and Q3.
Plugging in the values:
F23 =8.99 ×109× |4×10−6× −3×10−6|
(2)2
F23 =8.99 ×109×12 ×10−12
4= 26.97 ×10−3= 2.697 ×10−2N
Step 3: Calculate the net force on Q2. The net force on Q2is the vector
sum of forces F12 and F23.
Fnet =F12 +F23 = 7.99 ×10−3+ 2.697 ×10−2
Fnet ≈3.49 ×10−2N
Therefore, the net force on Q2due to Q1and Q3is approximately 3.49 ×
10−2Nin the positive x-direction.
Question 12
Question
Three point charges are arranged in the vertices of an equilateral triangle of side
length a= 2 m as shown in the figure below. Charge q1=−3µC is at the top
vertex, charge q2= 5 µC is at the bottom left vertex, and charge q3=−4µC is
at the bottom right vertex. Calculate the net electric force acting on q3.
Solution
Step 1: Calculate the electric force acting on charge q3due to charges q1and
q2separately.
The electric force between two charges q1and q3is given by Coulomb’s Law:
F13 =k|q1||q3|
r2
13
where kis Coulomb’s constant, r13 is the distance between charges q1and q3,
and |q1|= 3 ×10−6C, |q3|= 4 ×10−6C.
Given that the side length of the equilateral triangle is a= 2 m, the distance
r13 between charges q1and q3is:
r13 =a
Plugging the given values into Coulomb’s Law, we get:
F13 =9×109×3×10−6×4×10−6
(2)2
10
F13 =108
4×10−9×103
F13 = 27 ×10−3N
So, the electric force between charges q1and q3is 27 ×10−3N.
Using the same method, we can calculate the electric force between charges
q2and q3as:
F23 =9×109×5×10−6×4×10−6
(2)2
F23 =180
4×10−9×103
F23 = 45 ×10−3N
So, the electric force between charges q2and q3is 45 ×10−3N.
Step 2: Calculate the net electric force acting on charge q3.
To find the net force, we need to consider the direction of the forces. Since
the charges q1and q3have opposite signs, their forces will be in opposite direc-
tions. However, the forces between q2and q3will both be pointing towards q3.
Calculating the net force:
Fnet =F23 −F13
Fnet = 45 ×10−3−27 ×10−3
Fnet = 18 ×10−3
Question 13
Question
Three point charges are placed at the vertices of an equilateral triangle of side
2.0 cm. Two of the charges are +3.0 nC, and the third charge is +5.0 nC.
a) What is the net electric force exerted on the charge of +3.0 nC?
b) What is the net electric force exerted on the charge of +5.0 nC?
Solution
a) To find the net electric force on the +3.0 nC charge, we need to consider the
forces exerted by the other two charges.
Let’s label the charges at the vertices of the triangle: A (+3.0 nC), B (+3.0
nC), and C (+5.0 nC).
Step 1: Calculate the distance between charges A and B. Since they are
at the vertices of an equilateral triangle of side 2.0 cm, the distance between A
and B is 2.0cm.
11
Step 2: Calculate the magnitude of the electric force between charges A
and B using Coulomb’s law:
FAB =k·q1·q2
r2
where k= 8.988×109N m2/C2is the electrostatic constant, q1= +3.0×10−9C
is the charge of A, q2= +3.0×10−9C is the charge of B, and r= 0.02 m is the
distance between A and B.
FAB = (8.988 ×109)·(3.0×10−9)2
(0.02)2
FAB ≈1.343 ×10−3N
This force is repulsive since the charges have the same sign.
Step 3: Find the net electric force on charge A. The force
FAB acts along
the line joining charges A and B, and it has the direction away from charge B.
Since A and C are equidistant from A, the force
FAC also acts along the line
joining A and C, but in the opposite direction to that of
FAB .
The net force on charge A will be the vector sum of
FAB and
FAC .
Fnet =
FAB +
FAC
Since the forces are along the same line, we can add them as scalars.
Fnet = 1.343 ×10−3N+ 1.343 ×10−3N
Fnet ≈2.686 ×10−3N
Therefore, the net electric force on the +3.0 nC charge is approximately
2.686 ×10−3N away from the charge at C.
b) The net electric force on the +5.0 nC charge can be calculated in a similar
way considering the forces from charges A and B.
Question 14
Question
Two point charges, q1=−4µC and q2= 2 µC, are located 5 cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
To find the electric force between the charges, we can use Coulomb’s law, which
states that the magnitude of the electric force between two point charges is given
by:
F=k·|q1·q2|
r2
12
where kis the Coulomb’s constant (8.9875 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 1: Convert the distance between charges from cm to meters:
r= 5 cm = 0.05 m
Step 2: Substitute the given values into Coulomb’s law to find the magni-
tude of the electric force:
F= (8.9875 ×109)·| − 4×10−6·2×10−6|
(0.05)2
Step 3: Calculate the magnitude of the electric force:
F= (8.9875 ×109)·8×10−12
0.0025
F= (8.9875 ×109)·3.2×10−9
F= 2.876 ×10−7N
Therefore, the magnitude of the electric force between the charges is 2.876 ×
10−7N.
Question 15
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length a. Charge q1is at the top vertex, charge q2is at the bottom left
vertex, and charge q3is at the bottom right vertex. Calculate the magnitude of
the net electric force on charge q1.
Solution
Step 1: Calculate the electric force between q1and q2. The distance between q1
and q2is a
√3(half of a side length of an equilateral triangle). The magnitude of
the electric force between two charges q1and q2is given by Coulomb’s law:
F12 =k|q1q2|
r2
12
where - kis Coulomb’s constant, k= 8.99 ×109N m2/C2, - r12 =a
√3is the dis-
tance between q1and q2, - q1=q2=q, as the charges are the same magnitude.
Thus,
F12 =(8.99 ×109N m2/C2)(q2)
a
√32
13
Step 2: Calculate the electric force between q1and q3. The distance between
q1and q3is 2a
√3(the full diagonal length of the equilateral triangle). Similarly,
applying Coulomb’s law:
F13 =(8.99 ×109N m2/C2)(q2)
2a
√32
Step 3: Find the net electric force on q1. The net electric force on charge
q1is in the upward direction due to the forces from q2and q3. The vertical
components of F12 and F13 add up to give the net force on q1. To find the net
force, subtract the force due to q3from the force due to q2.
Fnet = 2F12 sin(30◦)−F13
Solving this expression using the values from Steps 1 and 2 will give the
magnitude of the net electric force on charge q1.
Question 16
Question
Three point charges are placed at the corners of an equilateral triangle with side
length d. The charges are +q,+2q, and −3q. Calculate the magnitude of the
net electrostatic force on the +2qcharge due to the other two charges.
Solution
To find the net electrostatic force on the +2qcharge, we need to calculate
the individual forces on it due to the other two charges and then add them
vectorially.
Step 1: Calculate the force on the +2qcharge due to the +qcharge
The magnitude of the force between two charges q1and q2separated by a
distance ris given by Coulomb’s law:
F=k|q1||q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2). The distance between
the +2qand +qcharges in the triangle can be calculated using the height of
the equilateral triangle. The height of an equilateral triangle with side length d
is √3
2d. So, the distance between the +2qand +qcharges is d. Therefore, the
magnitude of the force on the +2qcharge due to the +qcharge is:
F1=k|2q||q|
d2
Step 2: Calculate the force on the +2qcharge due to the −3qcharge
Using the same reasoning as in Step 1, the distance between the +2qand −3q
14
charges is also d. Therefore, the magnitude of the force on the +2qcharge due
to the −3qcharge is:
F2=k|2q||3q|
d2
Step 3: Determine the net force on the +2qcharge The net force on
the +2qcharge is the vector sum of F1and F2. The forces F1and F2act in
opposite directions along the same line passing through the +2qcharge, so we
just need to subtract their magnitudes:
Fnet =F1−F2=k|2q||q|
d2−k|2q||3q|
d2
Solving this expression will give us the magnitude of the net electrostatic
force on the +2qcharge.
Question 17
Question
Three point charges are arranged in a line. Charge q1=−4.0µC is located at
the origin, charge q2= 6.0µC is located at x= 3.0m, and charge q3=−2.0µC
is located at x= 5.0m. Calculate the net electrostatic force on charge q2due
to charges q1and q3.
Solution
To find the net electrostatic force on charge q2, we first need to calculate the
forces acting on charge q2due to charges q1and q3separately, then sum these
forces vectorially.
Step 1: Calculate force on charge q2due to charge q1
The magnitude of the force on charge q2due to charge q1is given by
Coulomb’s law:
F1 on 2 =k|q1q2|
r2
12
where - kis the electrostatic constant, 8.99 ×109N m2/C2, - q1=−4.0µC=
−4.0×10−6C, - q2= 6.0µC= 6.0×10−6C, - r12 = 3.0m.
Plugging in these values, we get:
F1 on 2 = (8.99 ×109)| − 4.0×10−6×6.0×10−6|
(3.0)2
F1 on 2 = (8.99 ×109)24 ×10−12
9= 23.97 N (repulsive)
So, the force on charge q2due to charge q1is 23.97 N in the positive direction.
Step 2: Calculate force on charge q2due to charge q3
15
Similarly, the magnitude of the force on charge q2due to charge q3is given
by:
F3 on 2 =k|q3q2|
r2
32
where - q3=−2.0µC=−2.0×10−6C, - r32 = 2.0m.
Plugging in these values, we get:
F3 on 2 = (8.99 ×109)| − 2.0×10−6×6.0×10−6|
(2.0)2
F3 on 2 = (8.99 ×109)12 ×10−12
4= 26.97 N (attractive)
Therefore, the force on charge q2due to charge q3is 26.97 N in the negative
direction.
Step 3: Calculate the net force on charge q2
The net force exerted on charge q2is the vector sum of the forces due to
charges q1and q3:
Net force on q2=F1 on 2 +F3 on 2 = 23.97 N+ (−26.97 N)
Question 18
Question
Two point charges, q1= +4.0µC and q2=−2.0µC, are initially separated by
a distance of 10.0cm. If the charges are brought 20.0cm closer together, what
is the change in the magnitude of the electric force between them?
Solution
Step 1: Calculate the initial force between the charges using Coulomb’s law:
The electric force between two point charges can be calculated using Coulomb’s
law:
F=k|q1·q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant.
Substitute the given values:
F1= (8.99 ×109)|4.0×10−6·(−2.0×10−6)|
(0.10)2
Calculating F1gives:
F1≈5.758 N
Step 2: Calculate the final force when the charges are 20.0cm closer together:
After the charges are brought 20.0cm closer together, the new separation dis-
tance is r′= 10.0cm −20.0cm =−10.0cm.
16
Calculate the final force F2using the new separation distance r′:
F2= (8.99 ×109)|4.0×10−6·(−2.0×10−6)|
(0.10 −0.20)2
Calculating F2gives:
F2≈23.032 N
Step 3: Calculate the change in magnitude of the electric force: The change
in the magnitude of the electric force is given by:
∆F=|F2−F1|
Substitute the calculated values to find ∆F:
∆F=|23.032 −5.758|
∆F≈17.274 N
Therefore, the change in the magnitude of the electric force between the
charges is approximately 17.274 N.
Question 19
Question
Three point charges are located on the x-axis: a charge of +3 nC at x = 0, a
charge of -2 nC at x = 60 cm, and a charge of +5 nC at x = 100 cm. Calculate
the net electric force on the +5 nC charge.
Solution
Step 1: Convert all distances to meters. Given: q1= +3 nC = 3 ×10−9C at
x= 0 m, q2=−2nC =−2×10−9C at x= 60 cm = 0.6m, q3= +5 nC =
5×10−9C at x= 100 cm = 1.0m.
Step 2: Calculate the net electric force on the +5 nC charge. The net force
on the +5 nC charge is the sum of the forces due to the other charges. The
force between two charges is given by Coulomb’s Law:
F=k|q1q2|
r2,
where k= 8.99 ×109N m2/C2is the electrostatic constant, q1and q2are the
magnitudes of the two charges, and ris the separation between the charges.
The force between q1and q3is:
F13 =k|q1q3|
r2
13
.
17
The force between q2and q3is:
F23 =k|q2q3|
r2
23
.
The net force on the +5 nC charge due to q1and q2is:
Fnet =F13 +F23.
Substitute the given values and solve for the net force.
Question 20
Question
Two point charges, +3.0µC and −6.0µC, are placed 10.0cm apart in air. Calcu-
late the magnitude and direction of the electric force experienced by the positive
charge due to the negative charge.
Solution
Step 1: Convert the charges to coulombs. Given that 1µC= 10−6C, the first
charge +3.0µC becomes +3.0×10−6C and the second charge −6.0µC becomes
−6.0×10−6C.
Step 2: Calculate the distance between the charges in meters. The distance
between the charges is 10.0cm = 0.10 m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
The magnitude of the electric force Fbetween two point charges is given by
Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Substitute the given values into the formula:
F=(8.99 ×109)× |3.0×10−6×6.0×10−6|
(0.10)2
Step 4: Calculate the magnitude of the electric force.
F=(8.99 ×109)×18 ×10−12
0.01
Step 5: Simplify the expression.
F= 161.82 N
18
Step 6: Determine the direction of the electric force. The electric force is
attractive since the charges are of opposite sign. Therefore, the electric force
experienced by the positive charge is directed towards the negative charge.
Therefore, the magnitude of the electric force experienced by the positive
charge due to the negative charge is 161.82 N and it is directed towards the
negative charge.
Question 21
Question
Two point charges, q1=−6µC and q2= 4 µC, are placed 10 cm apart in air.
Calculate the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−6µC =−6×10−6C
q2= 4 µC = 4 ×10−6C
Step 2: Calculate the electric force. The electric force between two point
charges can be calculated using Coulomb’s Law:
F=k· |q1·q2|
r2
where Fis the magnitude of the electric force, kis the Coulomb constant (8.99×
109Nm2/C2), q1and q2are the magnitudes of the charges, and ris the distance
between the charges.
Plugging in the values:
F=(8.99 ×109Nm2/C2)· |(−6×10−6C)·(4 ×10−6C)|
(0.1m)2
Step 3: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(6 ×10−6)·(4 ×10−6)
0.01
F=215.76
0.01
F= 21576 N
Therefore, the magnitude of the electric force between the charges is 21576
N.
Question 22
Question
Two point charges of +4.0 µC and -6.0 µC are placed 0.10 m apart. Calculate
the magnitude of the electric force between them.
19
Solution
Step 1: Convert the charges to SI units. Given: q1= +4.0µC= 4.0×10−6C
q2=−6.0µC=−6.0×10−6C
Step 2: Calculate the distance between the charges. Given: r= 0.10 m
Step 3: Calculate the magnitude of the electric force between the charges
using Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the Coulomb constant with a value of 8.99 ×109N·m2/C2.
Substitute the given values into the formula:
F=(8.99 ×109)·(4.0×10−6)·(6.0×10−6)
(0.10)2
Step 4: Calculate the magnitude of the electric force.
F=35.96 ×103
0.01 = 3.596 ×106N
Therefore, the magnitude of the electric force between the charges is 3.596 ×
106N.
Question 23
Question
Two point charges with magnitudes q1= 3.0µC and q2=−5.0µC are placed
20.0 cm apart in a vacuum. Calculate the magnitude and direction of the electric
force that each charge exerts on the other.
Solution
To calculate the magnitude and direction of the electric force between the two
charges, we will use Coulomb’s law.
Step 1: Calculate the electric force F12 that charge q2exerts on charge q1.
Coulomb’s law is given by:
F12 =k|q1q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the two charges.
Plugging in the values, we get:
F12 = (8.99 ×109)(3.0×10−6)(5.0×10−6)
(0.20)2
F12 = 2.248 N
20
The direction of the force is from charge q2to charge q1.
Step 2: Calculate the electric force F21 that charge q1exerts on charge q2.
Since the forces act along the line joining the charges, the magnitude of F21 is
the same as that of F12.
Therefore, the magnitude of the electric force that each charge exerts on the
other is 2.248 N, and the direction of the force from charge q1to charge q2.
Question 24
Question
Two point charges, q1=−3.0µC and q2= 4.0µC, are placed 6.0 cm apart.
Calculate the magnitude and direction of the electric force exerted on q1.
Solution
Step 1: Convert the given charges to their equivalent in coulombs. Using the
conversion factor 1µC= 10−6C, we have: q1=−3.0µC=−3.0×10−6C, and
q2= 4.0µC= 4.0×10−6C.
Step 2: Given data: q1=−3.0×10−6Cq2= 4.0×10−6Cr= 6.0cm =
0.06 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k·
q1·q2
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging the values into the formula gives:
F= 8.99 ×109·3.0×10−6·4.0×10−6
(0.06)2
F= 8.99 ×109·12 ×10−12
0.0036
F= 8.99 ×109·0.000003333
F= 29.97 N
Step 4: Determine the direction of the electric force. Since q1is negative and
q2is positive, the electric force exerted on q1is attractive, pulling q1towards
q2. Therefore, the direction of the electric force is towards q2.
Question 25
Question
Two point charges, q1= 4.0µC and q2=−6.0µC, are placed 12.0cm apart in
air. Calculate the magnitude of the electric force between these charges.
21
Solution
Step 1: Let’s first convert the charges to Coulombs by using the relationship
1µC= 10−6C. Thus, q1= 4.0×10−6C and q2=−6.0×10−6C.
Step 2: Next, let’s calculate the distance between the charges in meters.
This can be done by converting 12.0cm to meters: 12.0cm = 0.12 m.
Step 3: Now, we can use Coulomb’s law to find the magnitude of the electric
force between the two charges. Coulomb’s law states that the magnitude of the
electric force between two point charges is given by:
F=k
q1q2
r2
,
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 4: Substituting the given values into Coulomb’s law, we have:
F= (8.99 ×109N m2/C2)|4.0×10−6C· −6.0×10−6C|
(0.12 m)2
Step 5: Calculating the magnitude of the electric force:
F= (8.99 ×109)(2.4×10−11)/(0.0144) N
F≈1.495 ×10−3N
Therefore, the magnitude of the electric force between the charges q1=
4.0µC and q2=−6.0µC, separated by a distance of 12.0cm, is approximately
1.495 ×10−3N.
22
q1
q2q3
23
Question 2
Question
Three point charges are arranged on the corners of an equilateral triangle as
shown below. Charge 1 (q1= +3µC) is located at the top vertex, charge 2
(q2=−2µC) is at the bottom left vertex, and charge 3 (q3= +4µC) is at the
bottom right vertex. The side length of the triangle is 2 meters. Calculate the
net electric force acting on charge 2 due to charges 1 and 3.
q1= +3µC
q3= +4µC
q2=−2µC
Solution
Step 1: Calculate the distance between charges 2 and 1 (top vertex). The
distance between neighboring charges can be calculated using the side length of
the equilateral triangle. Since the triangle is equilateral, each side is equal to
2 meters. The distance between charge 2 and charge 1 is the diagonal of the
triangle, which can be calculated using Pythagorean theorem. Let d21 be the
distance between charge 2 and 1.
d21 =p22+ 22=√8 = 2√2meters
Step 2: Calculate the magnitude and direction of the electric force on charge
2 due to charge 1. The electric force between two charges can be calculated using
Coulomb’s law:
F12 =k|q1q2|
r2
12
where - kis Coulomb’s constant (8.99 ×109Nm2/C2), - q1and q2are the
magnitudes of the charges, and - r12 is the distance between the charges.
The direction is attractive if charges have opposite signs and repulsive if the
charges have the same sign.
F12 =(8.99 ×109)× |3×10−6|×|−2×10−6|
(2√2)2=26.97 ×10−15
8= 3.37×10−15 N
The force is attractive because q1and q2have opposite signs.
Step 3: Calculate the distance between charges 2 and 3 (bottom left and
right vertices). Since the triangle is equilateral, the distance between these two
charges is also equal to 2√2meters.
Step 4: Calculate the magnitude and direction of the electric force on charge
2 due to charge 3.
2
Question 3
Question
Three point charges are arranged as shown in the diagram:
+3q+2q+q
d d
If the charges are separated by a distance d(which is the same between all
the pairs of charges), what is the magnitude of the net electrostatic force acting
on the charge +2qdue to the other two charges?
Given: - Electric force between two point charges q1and q2:F=k|q1q2|
r2-
Coulomb’s constant k= 8.99 ×109N m2/C2
Solution
Step 1: Calculate the force F1,2between charge +3qand charge +2q:
F1,2=k|+ 3q·+2q|
d2
F1,2=k6q2
d2
Step 2: Calculate the force F2,3between charge +2qand charge +q:
F2,3=k|+ 2q·+q|
d2
F2,3=k2q2
d2
Step 3: Calculate the net force acting on charge +2q: The forces F1,2and
F2,3are in the same direction, so we add their magnitudes:
Fnet =F1,2+F2,3
Fnet =k6q2
d2+k2q2
d2
Fnet =k8q2
d2
Therefore, the magnitude of the net electrostatic force acting on the charge
+2qdue to the other two charges is 8kq2/d2.
3
Question 4
Question
Two point charges, q1and q2, are separated by a distance of 4 cm. If q1= 5 µC
and q2=−3µC, calculate the magnitude and direction of the electric force
between the charges.
Solution
Let’s start by using Coulomb’s Law to calculate the electric force between the
two charges.
Step 1: Identify the given information
• Charge q1= 5 µC
• Charge q2=−3µC
• Distance between the charges r= 4 cm = 0.04 m
Step 2: Calculate the electric force The magnitude of the electric force
between two charges is given by Coulomb’s Law:
F=k|q1q2|
r2
where - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the charges, and - ris
the distance between the charges.
Plugging in the values:
F= (8.99 ×109)|5×10−6× −3×10−6|
(0.04)2
Step 3: Calculate the magnitude of the electric force
F= 8.99 ×109×15 ×10−12
0.0016
F= 8.99 ×15
1.6×10−3
F≈8.43 ×10−2N
Step 4: Determine the direction of the electric force Since q1and q2
have opposite charges, the electric force is attractive and directed from q1to q2.
Therefore, the magnitude of the electric force between the charges is approx-
imately 8.43 ×10−2N, and it is directed from q1to q2.
4
Question 5
Question
A small charge of +3.0µC is placed 20 cm to the right of a larger charge of
−6.0µC. Calculate the magnitude and direction of the electric force that the
larger charge exerts on the smaller charge.
Solution
Step 1: Determine the distance between the charges in meters. Given that the
smaller charge is placed 20 cm to the right of the larger charge, the distance
between the charges is d= 20 cm = 0.20 m.
Step 2: Calculate the magnitude of the electric force using Coulomb’s Law.
The magnitude of the electric force Fbetween two point charges q1and q2
separated by a distance ris given by:
F=k
q1·q2
r2
where kis Coulomb’s constant, 8.99 ×109N m2/C2.
Plugging in the given values:
F= (8.99 ×109)
(3.0×10−6)·(−6.0×10−6)
(0.20)2
Step 3: Calculate the magnitude of the electric force.
F= (8.99 ×109)
−18 ×10−12
0.04
F= 8.99 ×109×4.5×10−10
F= 4.0455 N
Step 4: Determine the direction of the electric force. The electric force
between the two charges is attractive since they have opposite signs. Therefore,
the larger charge of −6.0µC exerts a force to the right on the smaller charge of
+3.0µC.
Thus, the magnitude of the electric force that the larger charge exerts on
the smaller charge is 4.0455 N to the right.
Question 6
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. Charge q1=−2nC is located at the top vertex, charge q2= 4
nC is located at the bottom left vertex, and charge q3=−3nC is located at the
5
bottom right vertex. Calculate the magnitude and direction of the net electric
force experienced by the charge at the top vertex.
q2= 4 nC
q1=−2nC q3=−3nC
Solution
Step 1: We start by determining the direction and magnitude of the electric
force between q1and q2. The electric force between two point charges is given
by Coulomb’s Law:
F=k· |q1|·|q2|
r2
The direction of the force is along the line connecting the charges, and the
force is repulsive if both charges have the same sign and attractive if they have
opposite signs.
Step 2: Calculate the distance rbetween q1and q2. Since the charges are
at the vertices of an equilateral triangle,
r=2
√3a
where ais the side length of the triangle.
Step 3: Calculate the magnitude of the electric force between q1and q2using
Coulomb’s Law:
F12 =k· |q1|·|q2|
r2
Step 4: Next, we determine the direction and magnitude of the electric force
between q1and q3. Similarly to Step 1, calculate the distance between q1and
q3and then the force between them.
Step 5: Finally, calculate the net electric force experienced by the charge at
the top vertex by considering the vector sum of the forces between q1and q2and
between q1and q3. Remember to consider both the magnitude and direction of
each force.
Question 7
Question
Two point charges, Q1=−3.0µC and Q2= 6.0µC, are placed 10.0 cm apart in
a vacuum. Calculate the magnitude of the electric force that Q1exerts on Q2.
Solution
Let’s first write down the formula for the electric force between two point
charges:
F=k|Q1Q2|
r2
6
where F= magnitude of the electric force, k= Coulomb’s constant (8.99 ×
109N m2/C2), Q1and Q2= magnitudes of the charges, and r= separation
distance between the charges.
Step 1: Convert the charges to Coulombs. Since 1µC= 10−6C, Q1=
−3.0µC=−3.0×10−6C and Q2= 6.0µC= 6.0×10−6C.
Step 2: Calculate the force using the formula.
F=(8.99 ×109)×|−3.0×10−6×6.0×10−6|
(0.10)2
F=8.99 ×109×18 ×10−12
0.01
F=161.82 ×10−3
0.01 = 16.182 N
So, the magnitude of the electric force that Q1exerts on Q2is 16.182 N.
Question 8
Question
Three point charges are arranged in a line as shown below:
+2 µC−5µC+3 µC
The charges are located at positions x=−5cm, x= 0 cm, and x= 5 cm
respectively. Determine the magnitude and direction of the net electrostatic
force acting on the +3 µC charge.
Solution
Step 1: Calculate the distance between each pair of charges. First, we will
determine the distances r1,r2, and r3between the charges +2 µC and −5µC,
−5µC and +3 µC, and +2 µC and +3 µC respectively.
r1= 5 cm −(−5cm) = 10 cm = 0.10 m
r2= 0 cm −(−5cm) = 5 cm = 0.05 m
r3= 5 cm −0cm = 5 cm = 0.05 m
Step 2: Calculate the magnitude of the force between each pair of charges.
The magnitude of the electrostatic force between two point charges q1and
q2separated by a distance ris given by Coulomb’s Law:
F=k|q1q2|
r2
Where k= 8.99 ×109N m2/C2is Coulomb’s constant.
7
For charges +2 µC and −5µC:
F1= 8.99 ×109|2×10−6C×(−5×10−6C)|
(0.10 m)2
F1= 8.99 ×109×1×10−11 N= 8.99 ×10−2N= 0.0899 N
For charges −5µC and +3 µC:
F2= 8.99 ×109| − 5×10−6C×3×10−6C|
(0.05 m)2
F2= 8.99 ×109×1.5×10−11 N= 1.35 ×10−2N= 0.0135 N
For charges +2 µC and +3 µC:
F3= 8.99 ×109|2×10−6C×3×10−6C|
(0.05 m)2
F3= 8.99
Question 9
Question
Three point charges are placed at the corners of an equilateral triangle of side
d. The charges are +q,−2q, and +3q. What is the net force on the charge +q
due to the other two charges? Assume the charges are placed at the vertices of
the triangle and the sides are of equal length.
Solution
Let’s denote the charge +qas Q1, the charge −2qas Q2, and the charge +3qas
Q3. We will first find the force on Q1due to Q2and then the force on Q1due
to Q3. Finally, we will find the net force by summing up the forces.
Step 1: Calculate the force on Q1due to Q2. The electric force between
two charges Q1and Q2is given by Coulomb’s law:
F12 =k|Q1Q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2) and ris the distance
between the charges. The distance between Q1and Q2in this case, considering
the equilateral triangle, is d. So, the force F12 is given by:
F12 =k|q·(−2q)|
d2=k|q2·2|
d2=2kq2
d2
8
Step 2: Calculate the force on Q1due to Q3. Following the same steps as
above, we find the force F13 on Q1due to Q3:
F13 =k|q·3q|
d2=k|q2·3|
d2=3kq2
d2
Step 3: Calculate the net force on Q1. The net force is the vector sum of
F12 and F13, where the direction of F12 is opposite to F13 (since Q2is negative
and Q3is positive).
Fnet =F13 −F12 =3kq2
d2−2kq2
d2=kq2
d2
Therefore, the net force on the charge +qdue to the other two charges is
kq2
d2directed towards the charge +3q.
Question 10
Question
Three point charges are arranged along the x-axis as follows: Q1=−2µC at
the origin, Q2= 4 µC at x= 3 m, and Q3=−3µC at x= 5 m. Calculate the
net force on Q2due to Q1and Q3.
Solution
Step 1: Determine the force on Q2due to Q1. The force F12 on Q2due to Q1
can be calculated using the electric force equation:
F12 =k|Q1Q2|
r2
12
where: - k= 8.99 ×109Nm2/C2is the Coulomb constant, - Q1=−2×10−6C,
-Q2= 4 ×10−6C, - r12 = 3 mis the distance between Q1and Q2.
Plugging in the values:
F12 =8.99 ×109×|−2×10−6×4×10−6|
(3)2
F12 =8.99 ×109×8×10−12
9= 7.99 ×10−3N
Step 2: Determine the force on Q2due to Q3. The force F23 on Q2due to
Q3can be calculated using the electric force equation:
F23 =k|Q2Q3|
r2
23
9
where: - Q2= 4 ×10−6C, - Q3=−3×10−6C, - r23 = 2 mis the distance
between Q2and Q3.
Plugging in the values:
F23 =8.99 ×109× |4×10−6× −3×10−6|
(2)2
F23 =8.99 ×109×12 ×10−12
4= 26.97 ×10−3= 2.697 ×10−2N
Step 3: Calculate the net force on Q2. The net force on Q2is the vector
sum of forces F12 and F23.
Fnet =F12 +F23 = 7.99 ×10−3+ 2.697 ×10−2
Fnet ≈3.49 ×10−2N
Therefore, the net force on Q2due to Q1and Q3is approximately 3.49 ×
10−2Nin the positive x-direction.
Question 12
Question
Three point charges are arranged in the vertices of an equilateral triangle of side
length a= 2 m as shown in the figure below. Charge q1=−3µC is at the top
vertex, charge q2= 5 µC is at the bottom left vertex, and charge q3=−4µC is
at the bottom right vertex. Calculate the net electric force acting on q3.
Solution
Step 1: Calculate the electric force acting on charge q3due to charges q1and
q2separately.
The electric force between two charges q1and q3is given by Coulomb’s Law:
F13 =k|q1||q3|
r2
13
where kis Coulomb’s constant, r13 is the distance between charges q1and q3,
and |q1|= 3 ×10−6C, |q3|= 4 ×10−6C.
Given that the side length of the equilateral triangle is a= 2 m, the distance
r13 between charges q1and q3is:
r13 =a
Plugging the given values into Coulomb’s Law, we get:
F13 =9×109×3×10−6×4×10−6
(2)2
10
F13 =108
4×10−9×103
F13 = 27 ×10−3N
So, the electric force between charges q1and q3is 27 ×10−3N.
Using the same method, we can calculate the electric force between charges
q2and q3as:
F23 =9×109×5×10−6×4×10−6
(2)2
F23 =180
4×10−9×103
F23 = 45 ×10−3N
So, the electric force between charges q2and q3is 45 ×10−3N.
Step 2: Calculate the net electric force acting on charge q3.
To find the net force, we need to consider the direction of the forces. Since
the charges q1and q3have opposite signs, their forces will be in opposite direc-
tions. However, the forces between q2and q3will both be pointing towards q3.
Calculating the net force:
Fnet =F23 −F13
Fnet = 45 ×10−3−27 ×10−3
Fnet = 18 ×10−3
Question 13
Question
Three point charges are placed at the vertices of an equilateral triangle of side
2.0 cm. Two of the charges are +3.0 nC, and the third charge is +5.0 nC.
a) What is the net electric force exerted on the charge of +3.0 nC?
b) What is the net electric force exerted on the charge of +5.0 nC?
Solution
a) To find the net electric force on the +3.0 nC charge, we need to consider the
forces exerted by the other two charges.
Let’s label the charges at the vertices of the triangle: A (+3.0 nC), B (+3.0
nC), and C (+5.0 nC).
Step 1: Calculate the distance between charges A and B. Since they are
at the vertices of an equilateral triangle of side 2.0 cm, the distance between A
and B is 2.0cm.
11
Step 2: Calculate the magnitude of the electric force between charges A
and B using Coulomb’s law:
FAB =k·q1·q2
r2
where k= 8.988×109N m2/C2is the electrostatic constant, q1= +3.0×10−9C
is the charge of A, q2= +3.0×10−9C is the charge of B, and r= 0.02 m is the
distance between A and B.
FAB = (8.988 ×109)·(3.0×10−9)2
(0.02)2
FAB ≈1.343 ×10−3N
This force is repulsive since the charges have the same sign.
Step 3: Find the net electric force on charge A. The force
FAB acts along
the line joining charges A and B, and it has the direction away from charge B.
Since A and C are equidistant from A, the force
FAC also acts along the line
joining A and C, but in the opposite direction to that of
FAB .
The net force on charge A will be the vector sum of
FAB and
FAC .
Fnet =
FAB +
FAC
Since the forces are along the same line, we can add them as scalars.
Fnet = 1.343 ×10−3N+ 1.343 ×10−3N
Fnet ≈2.686 ×10−3N
Therefore, the net electric force on the +3.0 nC charge is approximately
2.686 ×10−3N away from the charge at C.
b) The net electric force on the +5.0 nC charge can be calculated in a similar
way considering the forces from charges A and B.
Question 14
Question
Two point charges, q1=−4µC and q2= 2 µC, are located 5 cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
To find the electric force between the charges, we can use Coulomb’s law, which
states that the magnitude of the electric force between two point charges is given
by:
F=k·|q1·q2|
r2
12
where kis the Coulomb’s constant (8.9875 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 1: Convert the distance between charges from cm to meters:
r= 5 cm = 0.05 m
Step 2: Substitute the given values into Coulomb’s law to find the magni-
tude of the electric force:
F= (8.9875 ×109)·| − 4×10−6·2×10−6|
(0.05)2
Step 3: Calculate the magnitude of the electric force:
F= (8.9875 ×109)·8×10−12
0.0025
F= (8.9875 ×109)·3.2×10−9
F= 2.876 ×10−7N
Therefore, the magnitude of the electric force between the charges is 2.876 ×
10−7N.
Question 15
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length a. Charge q1is at the top vertex, charge q2is at the bottom left
vertex, and charge q3is at the bottom right vertex. Calculate the magnitude of
the net electric force on charge q1.
Solution
Step 1: Calculate the electric force between q1and q2. The distance between q1
and q2is a
√3(half of a side length of an equilateral triangle). The magnitude of
the electric force between two charges q1and q2is given by Coulomb’s law:
F12 =k|q1q2|
r2
12
where - kis Coulomb’s constant, k= 8.99 ×109N m2/C2, - r12 =a
√3is the dis-
tance between q1and q2, - q1=q2=q, as the charges are the same magnitude.
Thus,
F12 =(8.99 ×109N m2/C2)(q2)
a
√32
13
Step 2: Calculate the electric force between q1and q3. The distance between
q1and q3is 2a
√3(the full diagonal length of the equilateral triangle). Similarly,
applying Coulomb’s law:
F13 =(8.99 ×109N m2/C2)(q2)
2a
√32
Step 3: Find the net electric force on q1. The net electric force on charge
q1is in the upward direction due to the forces from q2and q3. The vertical
components of F12 and F13 add up to give the net force on q1. To find the net
force, subtract the force due to q3from the force due to q2.
Fnet = 2F12 sin(30◦)−F13
Solving this expression using the values from Steps 1 and 2 will give the
magnitude of the net electric force on charge q1.
Question 16
Question
Three point charges are placed at the corners of an equilateral triangle with side
length d. The charges are +q,+2q, and −3q. Calculate the magnitude of the
net electrostatic force on the +2qcharge due to the other two charges.
Solution
To find the net electrostatic force on the +2qcharge, we need to calculate
the individual forces on it due to the other two charges and then add them
vectorially.
Step 1: Calculate the force on the +2qcharge due to the +qcharge
The magnitude of the force between two charges q1and q2separated by a
distance ris given by Coulomb’s law:
F=k|q1||q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2). The distance between
the +2qand +qcharges in the triangle can be calculated using the height of
the equilateral triangle. The height of an equilateral triangle with side length d
is √3
2d. So, the distance between the +2qand +qcharges is d. Therefore, the
magnitude of the force on the +2qcharge due to the +qcharge is:
F1=k|2q||q|
d2
Step 2: Calculate the force on the +2qcharge due to the −3qcharge
Using the same reasoning as in Step 1, the distance between the +2qand −3q
14
charges is also d. Therefore, the magnitude of the force on the +2qcharge due
to the −3qcharge is:
F2=k|2q||3q|
d2
Step 3: Determine the net force on the +2qcharge The net force on
the +2qcharge is the vector sum of F1and F2. The forces F1and F2act in
opposite directions along the same line passing through the +2qcharge, so we
just need to subtract their magnitudes:
Fnet =F1−F2=k|2q||q|
d2−k|2q||3q|
d2
Solving this expression will give us the magnitude of the net electrostatic
force on the +2qcharge.
Question 17
Question
Three point charges are arranged in a line. Charge q1=−4.0µC is located at
the origin, charge q2= 6.0µC is located at x= 3.0m, and charge q3=−2.0µC
is located at x= 5.0m. Calculate the net electrostatic force on charge q2due
to charges q1and q3.
Solution
To find the net electrostatic force on charge q2, we first need to calculate the
forces acting on charge q2due to charges q1and q3separately, then sum these
forces vectorially.
Step 1: Calculate force on charge q2due to charge q1
The magnitude of the force on charge q2due to charge q1is given by
Coulomb’s law:
F1 on 2 =k|q1q2|
r2
12
where - kis the electrostatic constant, 8.99 ×109N m2/C2, - q1=−4.0µC=
−4.0×10−6C, - q2= 6.0µC= 6.0×10−6C, - r12 = 3.0m.
Plugging in these values, we get:
F1 on 2 = (8.99 ×109)| − 4.0×10−6×6.0×10−6|
(3.0)2
F1 on 2 = (8.99 ×109)24 ×10−12
9= 23.97 N (repulsive)
So, the force on charge q2due to charge q1is 23.97 N in the positive direction.
Step 2: Calculate force on charge q2due to charge q3
15
Similarly, the magnitude of the force on charge q2due to charge q3is given
by:
F3 on 2 =k|q3q2|
r2
32
where - q3=−2.0µC=−2.0×10−6C, - r32 = 2.0m.
Plugging in these values, we get:
F3 on 2 = (8.99 ×109)| − 2.0×10−6×6.0×10−6|
(2.0)2
F3 on 2 = (8.99 ×109)12 ×10−12
4= 26.97 N (attractive)
Therefore, the force on charge q2due to charge q3is 26.97 N in the negative
direction.
Step 3: Calculate the net force on charge q2
The net force exerted on charge q2is the vector sum of the forces due to
charges q1and q3:
Net force on q2=F1 on 2 +F3 on 2 = 23.97 N+ (−26.97 N)
Question 18
Question
Two point charges, q1= +4.0µC and q2=−2.0µC, are initially separated by
a distance of 10.0cm. If the charges are brought 20.0cm closer together, what
is the change in the magnitude of the electric force between them?
Solution
Step 1: Calculate the initial force between the charges using Coulomb’s law:
The electric force between two point charges can be calculated using Coulomb’s
law:
F=k|q1·q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant.
Substitute the given values:
F1= (8.99 ×109)|4.0×10−6·(−2.0×10−6)|
(0.10)2
Calculating F1gives:
F1≈5.758 N
Step 2: Calculate the final force when the charges are 20.0cm closer together:
After the charges are brought 20.0cm closer together, the new separation dis-
tance is r′= 10.0cm −20.0cm =−10.0cm.
16
Calculate the final force F2using the new separation distance r′:
F2= (8.99 ×109)|4.0×10−6·(−2.0×10−6)|
(0.10 −0.20)2
Calculating F2gives:
F2≈23.032 N
Step 3: Calculate the change in magnitude of the electric force: The change
in the magnitude of the electric force is given by:
∆F=|F2−F1|
Substitute the calculated values to find ∆F:
∆F=|23.032 −5.758|
∆F≈17.274 N
Therefore, the change in the magnitude of the electric force between the
charges is approximately 17.274 N.
Question 19
Question
Three point charges are located on the x-axis: a charge of +3 nC at x = 0, a
charge of -2 nC at x = 60 cm, and a charge of +5 nC at x = 100 cm. Calculate
the net electric force on the +5 nC charge.
Solution
Step 1: Convert all distances to meters. Given: q1= +3 nC = 3 ×10−9C at
x= 0 m, q2=−2nC =−2×10−9C at x= 60 cm = 0.6m, q3= +5 nC =
5×10−9C at x= 100 cm = 1.0m.
Step 2: Calculate the net electric force on the +5 nC charge. The net force
on the +5 nC charge is the sum of the forces due to the other charges. The
force between two charges is given by Coulomb’s Law:
F=k|q1q2|
r2,
where k= 8.99 ×109N m2/C2is the electrostatic constant, q1and q2are the
magnitudes of the two charges, and ris the separation between the charges.
The force between q1and q3is:
F13 =k|q1q3|
r2
13
.
17
The force between q2and q3is:
F23 =k|q2q3|
r2
23
.
The net force on the +5 nC charge due to q1and q2is:
Fnet =F13 +F23.
Substitute the given values and solve for the net force.
Question 20
Question
Two point charges, +3.0µC and −6.0µC, are placed 10.0cm apart in air. Calcu-
late the magnitude and direction of the electric force experienced by the positive
charge due to the negative charge.
Solution
Step 1: Convert the charges to coulombs. Given that 1µC= 10−6C, the first
charge +3.0µC becomes +3.0×10−6C and the second charge −6.0µC becomes
−6.0×10−6C.
Step 2: Calculate the distance between the charges in meters. The distance
between the charges is 10.0cm = 0.10 m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
The magnitude of the electric force Fbetween two point charges is given by
Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Substitute the given values into the formula:
F=(8.99 ×109)× |3.0×10−6×6.0×10−6|
(0.10)2
Step 4: Calculate the magnitude of the electric force.
F=(8.99 ×109)×18 ×10−12
0.01
Step 5: Simplify the expression.
F= 161.82 N
18
Step 6: Determine the direction of the electric force. The electric force is
attractive since the charges are of opposite sign. Therefore, the electric force
experienced by the positive charge is directed towards the negative charge.
Therefore, the magnitude of the electric force experienced by the positive
charge due to the negative charge is 161.82 N and it is directed towards the
negative charge.
Question 21
Question
Two point charges, q1=−6µC and q2= 4 µC, are placed 10 cm apart in air.
Calculate the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−6µC =−6×10−6C
q2= 4 µC = 4 ×10−6C
Step 2: Calculate the electric force. The electric force between two point
charges can be calculated using Coulomb’s Law:
F=k· |q1·q2|
r2
where Fis the magnitude of the electric force, kis the Coulomb constant (8.99×
109Nm2/C2), q1and q2are the magnitudes of the charges, and ris the distance
between the charges.
Plugging in the values:
F=(8.99 ×109Nm2/C2)· |(−6×10−6C)·(4 ×10−6C)|
(0.1m)2
Step 3: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(6 ×10−6)·(4 ×10−6)
0.01
F=215.76
0.01
F= 21576 N
Therefore, the magnitude of the electric force between the charges is 21576
N.
Question 22
Question
Two point charges of +4.0 µC and -6.0 µC are placed 0.10 m apart. Calculate
the magnitude of the electric force between them.
19
Solution
Step 1: Convert the charges to SI units. Given: q1= +4.0µC= 4.0×10−6C
q2=−6.0µC=−6.0×10−6C
Step 2: Calculate the distance between the charges. Given: r= 0.10 m
Step 3: Calculate the magnitude of the electric force between the charges
using Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the Coulomb constant with a value of 8.99 ×109N·m2/C2.
Substitute the given values into the formula:
F=(8.99 ×109)·(4.0×10−6)·(6.0×10−6)
(0.10)2
Step 4: Calculate the magnitude of the electric force.
F=35.96 ×103
0.01 = 3.596 ×106N
Therefore, the magnitude of the electric force between the charges is 3.596 ×
106N.
Question 23
Question
Two point charges with magnitudes q1= 3.0µC and q2=−5.0µC are placed
20.0 cm apart in a vacuum. Calculate the magnitude and direction of the electric
force that each charge exerts on the other.
Solution
To calculate the magnitude and direction of the electric force between the two
charges, we will use Coulomb’s law.
Step 1: Calculate the electric force F12 that charge q2exerts on charge q1.
Coulomb’s law is given by:
F12 =k|q1q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the two charges.
Plugging in the values, we get:
F12 = (8.99 ×109)(3.0×10−6)(5.0×10−6)
(0.20)2
F12 = 2.248 N
20
The direction of the force is from charge q2to charge q1.
Step 2: Calculate the electric force F21 that charge q1exerts on charge q2.
Since the forces act along the line joining the charges, the magnitude of F21 is
the same as that of F12.
Therefore, the magnitude of the electric force that each charge exerts on the
other is 2.248 N, and the direction of the force from charge q1to charge q2.
Question 24
Question
Two point charges, q1=−3.0µC and q2= 4.0µC, are placed 6.0 cm apart.
Calculate the magnitude and direction of the electric force exerted on q1.
Solution
Step 1: Convert the given charges to their equivalent in coulombs. Using the
conversion factor 1µC= 10−6C, we have: q1=−3.0µC=−3.0×10−6C, and
q2= 4.0µC= 4.0×10−6C.
Step 2: Given data: q1=−3.0×10−6Cq2= 4.0×10−6Cr= 6.0cm =
0.06 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k·
q1·q2
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging the values into the formula gives:
F= 8.99 ×109·3.0×10−6·4.0×10−6
(0.06)2
F= 8.99 ×109·12 ×10−12
0.0036
F= 8.99 ×109·0.000003333
F= 29.97 N
Step 4: Determine the direction of the electric force. Since q1is negative and
q2is positive, the electric force exerted on q1is attractive, pulling q1towards
q2. Therefore, the direction of the electric force is towards q2.
Question 25
Question
Two point charges, q1= 4.0µC and q2=−6.0µC, are placed 12.0cm apart in
air. Calculate the magnitude of the electric force between these charges.
21
Solution
Step 1: Let’s first convert the charges to Coulombs by using the relationship
1µC= 10−6C. Thus, q1= 4.0×10−6C and q2=−6.0×10−6C.
Step 2: Next, let’s calculate the distance between the charges in meters.
This can be done by converting 12.0cm to meters: 12.0cm = 0.12 m.
Step 3: Now, we can use Coulomb’s law to find the magnitude of the electric
force between the two charges. Coulomb’s law states that the magnitude of the
electric force between two point charges is given by:
F=k
q1q2
r2
,
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 4: Substituting the given values into Coulomb’s law, we have:
F= (8.99 ×109N m2/C2)|4.0×10−6C· −6.0×10−6C|
(0.12 m)2
Step 5: Calculating the magnitude of the electric force:
F= (8.99 ×109)(2.4×10−11)/(0.0144) N
F≈1.495 ×10−3N
Therefore, the magnitude of the electric force between the charges q1=
4.0µC and q2=−6.0µC, separated by a distance of 12.0cm, is approximately
1.495 ×10−3N.
22
q1
q2q3
23
Question 2
Question
Three point charges are arranged on the corners of an equilateral triangle as
shown below. Charge 1 (q1= +3µC) is located at the top vertex, charge 2
(q2=−2µC) is at the bottom left vertex, and charge 3 (q3= +4µC) is at the
bottom right vertex. The side length of the triangle is 2 meters. Calculate the
net electric force acting on charge 2 due to charges 1 and 3.
q1= +3µC
q3= +4µC
q2=−2µC
Solution
Step 1: Calculate the distance between charges 2 and 1 (top vertex). The
distance between neighboring charges can be calculated using the side length of
the equilateral triangle. Since the triangle is equilateral, each side is equal to
2 meters. The distance between charge 2 and charge 1 is the diagonal of the
triangle, which can be calculated using Pythagorean theorem. Let d21 be the
distance between charge 2 and 1.
d21 =p22+ 22=√8 = 2√2meters
Step 2: Calculate the magnitude and direction of the electric force on charge
2 due to charge 1. The electric force between two charges can be calculated using
Coulomb’s law:
F12 =k|q1q2|
r2
12
where - kis Coulomb’s constant (8.99 ×109Nm2/C2), - q1and q2are the
magnitudes of the charges, and - r12 is the distance between the charges.
The direction is attractive if charges have opposite signs and repulsive if the
charges have the same sign.
F12 =(8.99 ×109)× |3×10−6|×|−2×10−6|
(2√2)2=26.97 ×10−15
8= 3.37×10−15 N
The force is attractive because q1and q2have opposite signs.
Step 3: Calculate the distance between charges 2 and 3 (bottom left and
right vertices). Since the triangle is equilateral, the distance between these two
charges is also equal to 2√2meters.
Step 4: Calculate the magnitude and direction of the electric force on charge
2 due to charge 3.
2
Question 3
Question
Three point charges are arranged as shown in the diagram:
+3q+2q+q
d d
If the charges are separated by a distance d(which is the same between all
the pairs of charges), what is the magnitude of the net electrostatic force acting
on the charge +2qdue to the other two charges?
Given: - Electric force between two point charges q1and q2:F=k|q1q2|
r2-
Coulomb’s constant k= 8.99 ×109N m2/C2
Solution
Step 1: Calculate the force F1,2between charge +3qand charge +2q:
F1,2=k|+ 3q·+2q|
d2
F1,2=k6q2
d2
Step 2: Calculate the force F2,3between charge +2qand charge +q:
F2,3=k|+ 2q·+q|
d2
F2,3=k2q2
d2
Step 3: Calculate the net force acting on charge +2q: The forces F1,2and
F2,3are in the same direction, so we add their magnitudes:
Fnet =F1,2+F2,3
Fnet =k6q2
d2+k2q2
d2
Fnet =k8q2
d2
Therefore, the magnitude of the net electrostatic force acting on the charge
+2qdue to the other two charges is 8kq2/d2.
3
Question 4
Question
Two point charges, q1and q2, are separated by a distance of 4 cm. If q1= 5 µC
and q2=−3µC, calculate the magnitude and direction of the electric force
between the charges.
Solution
Let’s start by using Coulomb’s Law to calculate the electric force between the
two charges.
Step 1: Identify the given information
• Charge q1= 5 µC
• Charge q2=−3µC
• Distance between the charges r= 4 cm = 0.04 m
Step 2: Calculate the electric force The magnitude of the electric force
between two charges is given by Coulomb’s Law:
F=k|q1q2|
r2
where - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the charges, and - ris
the distance between the charges.
Plugging in the values:
F= (8.99 ×109)|5×10−6× −3×10−6|
(0.04)2
Step 3: Calculate the magnitude of the electric force
F= 8.99 ×109×15 ×10−12
0.0016
F= 8.99 ×15
1.6×10−3
F≈8.43 ×10−2N
Step 4: Determine the direction of the electric force Since q1and q2
have opposite charges, the electric force is attractive and directed from q1to q2.
Therefore, the magnitude of the electric force between the charges is approx-
imately 8.43 ×10−2N, and it is directed from q1to q2.
4
Question 5
Question
A small charge of +3.0µC is placed 20 cm to the right of a larger charge of
−6.0µC. Calculate the magnitude and direction of the electric force that the
larger charge exerts on the smaller charge.
Solution
Step 1: Determine the distance between the charges in meters. Given that the
smaller charge is placed 20 cm to the right of the larger charge, the distance
between the charges is d= 20 cm = 0.20 m.
Step 2: Calculate the magnitude of the electric force using Coulomb’s Law.
The magnitude of the electric force Fbetween two point charges q1and q2
separated by a distance ris given by:
F=k
q1·q2
r2
where kis Coulomb’s constant, 8.99 ×109N m2/C2.
Plugging in the given values:
F= (8.99 ×109)
(3.0×10−6)·(−6.0×10−6)
(0.20)2
Step 3: Calculate the magnitude of the electric force.
F= (8.99 ×109)
−18 ×10−12
0.04
F= 8.99 ×109×4.5×10−10
F= 4.0455 N
Step 4: Determine the direction of the electric force. The electric force
between the two charges is attractive since they have opposite signs. Therefore,
the larger charge of −6.0µC exerts a force to the right on the smaller charge of
+3.0µC.
Thus, the magnitude of the electric force that the larger charge exerts on
the smaller charge is 4.0455 N to the right.
Question 6
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. Charge q1=−2nC is located at the top vertex, charge q2= 4
nC is located at the bottom left vertex, and charge q3=−3nC is located at the
5
bottom right vertex. Calculate the magnitude and direction of the net electric
force experienced by the charge at the top vertex.
q2= 4 nC
q1=−2nC q3=−3nC
Solution
Step 1: We start by determining the direction and magnitude of the electric
force between q1and q2. The electric force between two point charges is given
by Coulomb’s Law:
F=k· |q1|·|q2|
r2
The direction of the force is along the line connecting the charges, and the
force is repulsive if both charges have the same sign and attractive if they have
opposite signs.
Step 2: Calculate the distance rbetween q1and q2. Since the charges are
at the vertices of an equilateral triangle,
r=2
√3a
where ais the side length of the triangle.
Step 3: Calculate the magnitude of the electric force between q1and q2using
Coulomb’s Law:
F12 =k· |q1|·|q2|
r2
Step 4: Next, we determine the direction and magnitude of the electric force
between q1and q3. Similarly to Step 1, calculate the distance between q1and
q3and then the force between them.
Step 5: Finally, calculate the net electric force experienced by the charge at
the top vertex by considering the vector sum of the forces between q1and q2and
between q1and q3. Remember to consider both the magnitude and direction of
each force.
Question 7
Question
Two point charges, Q1=−3.0µC and Q2= 6.0µC, are placed 10.0 cm apart in
a vacuum. Calculate the magnitude of the electric force that Q1exerts on Q2.
Solution
Let’s first write down the formula for the electric force between two point
charges:
F=k|Q1Q2|
r2
6
where F= magnitude of the electric force, k= Coulomb’s constant (8.99 ×
109N m2/C2), Q1and Q2= magnitudes of the charges, and r= separation
distance between the charges.
Step 1: Convert the charges to Coulombs. Since 1µC= 10−6C, Q1=
−3.0µC=−3.0×10−6C and Q2= 6.0µC= 6.0×10−6C.
Step 2: Calculate the force using the formula.
F=(8.99 ×109)×|−3.0×10−6×6.0×10−6|
(0.10)2
F=8.99 ×109×18 ×10−12
0.01
F=161.82 ×10−3
0.01 = 16.182 N
So, the magnitude of the electric force that Q1exerts on Q2is 16.182 N.
Question 8
Question
Three point charges are arranged in a line as shown below:
+2 µC−5µC+3 µC
The charges are located at positions x=−5cm, x= 0 cm, and x= 5 cm
respectively. Determine the magnitude and direction of the net electrostatic
force acting on the +3 µC charge.
Solution
Step 1: Calculate the distance between each pair of charges. First, we will
determine the distances r1,r2, and r3between the charges +2 µC and −5µC,
−5µC and +3 µC, and +2 µC and +3 µC respectively.
r1= 5 cm −(−5cm) = 10 cm = 0.10 m
r2= 0 cm −(−5cm) = 5 cm = 0.05 m
r3= 5 cm −0cm = 5 cm = 0.05 m
Step 2: Calculate the magnitude of the force between each pair of charges.
The magnitude of the electrostatic force between two point charges q1and
q2separated by a distance ris given by Coulomb’s Law:
F=k|q1q2|
r2
Where k= 8.99 ×109N m2/C2is Coulomb’s constant.
7
For charges +2 µC and −5µC:
F1= 8.99 ×109|2×10−6C×(−5×10−6C)|
(0.10 m)2
F1= 8.99 ×109×1×10−11 N= 8.99 ×10−2N= 0.0899 N
For charges −5µC and +3 µC:
F2= 8.99 ×109| − 5×10−6C×3×10−6C|
(0.05 m)2
F2= 8.99 ×109×1.5×10−11 N= 1.35 ×10−2N= 0.0135 N
For charges +2 µC and +3 µC:
F3= 8.99 ×109|2×10−6C×3×10−6C|
(0.05 m)2
F3= 8.99
Question 9
Question
Three point charges are placed at the corners of an equilateral triangle of side
d. The charges are +q,−2q, and +3q. What is the net force on the charge +q
due to the other two charges? Assume the charges are placed at the vertices of
the triangle and the sides are of equal length.
Solution
Let’s denote the charge +qas Q1, the charge −2qas Q2, and the charge +3qas
Q3. We will first find the force on Q1due to Q2and then the force on Q1due
to Q3. Finally, we will find the net force by summing up the forces.
Step 1: Calculate the force on Q1due to Q2. The electric force between
two charges Q1and Q2is given by Coulomb’s law:
F12 =k|Q1Q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2) and ris the distance
between the charges. The distance between Q1and Q2in this case, considering
the equilateral triangle, is d. So, the force F12 is given by:
F12 =k|q·(−2q)|
d2=k|q2·2|
d2=2kq2
d2
8
Step 2: Calculate the force on Q1due to Q3. Following the same steps as
above, we find the force F13 on Q1due to Q3:
F13 =k|q·3q|
d2=k|q2·3|
d2=3kq2
d2
Step 3: Calculate the net force on Q1. The net force is the vector sum of
F12 and F13, where the direction of F12 is opposite to F13 (since Q2is negative
and Q3is positive).
Fnet =F13 −F12 =3kq2
d2−2kq2
d2=kq2
d2
Therefore, the net force on the charge +qdue to the other two charges is
kq2
d2directed towards the charge +3q.
Question 10
Question
Three point charges are arranged along the x-axis as follows: Q1=−2µC at
the origin, Q2= 4 µC at x= 3 m, and Q3=−3µC at x= 5 m. Calculate the
net force on Q2due to Q1and Q3.
Solution
Step 1: Determine the force on Q2due to Q1. The force F12 on Q2due to Q1
can be calculated using the electric force equation:
F12 =k|Q1Q2|
r2
12
where: - k= 8.99 ×109Nm2/C2is the Coulomb constant, - Q1=−2×10−6C,
-Q2= 4 ×10−6C, - r12 = 3 mis the distance between Q1and Q2.
Plugging in the values:
F12 =8.99 ×109×|−2×10−6×4×10−6|
(3)2
F12 =8.99 ×109×8×10−12
9= 7.99 ×10−3N
Step 2: Determine the force on Q2due to Q3. The force F23 on Q2due to
Q3can be calculated using the electric force equation:
F23 =k|Q2Q3|
r2
23
9
where: - Q2= 4 ×10−6C, - Q3=−3×10−6C, - r23 = 2 mis the distance
between Q2and Q3.
Plugging in the values:
F23 =8.99 ×109× |4×10−6× −3×10−6|
(2)2
F23 =8.99 ×109×12 ×10−12
4= 26.97 ×10−3= 2.697 ×10−2N
Step 3: Calculate the net force on Q2. The net force on Q2is the vector
sum of forces F12 and F23.
Fnet =F12 +F23 = 7.99 ×10−3+ 2.697 ×10−2
Fnet ≈3.49 ×10−2N
Therefore, the net force on Q2due to Q1and Q3is approximately 3.49 ×
10−2Nin the positive x-direction.
Question 12
Question
Three point charges are arranged in the vertices of an equilateral triangle of side
length a= 2 m as shown in the figure below. Charge q1=−3µC is at the top
vertex, charge q2= 5 µC is at the bottom left vertex, and charge q3=−4µC is
at the bottom right vertex. Calculate the net electric force acting on q3.
Solution
Step 1: Calculate the electric force acting on charge q3due to charges q1and
q2separately.
The electric force between two charges q1and q3is given by Coulomb’s Law:
F13 =k|q1||q3|
r2
13
where kis Coulomb’s constant, r13 is the distance between charges q1and q3,
and |q1|= 3 ×10−6C, |q3|= 4 ×10−6C.
Given that the side length of the equilateral triangle is a= 2 m, the distance
r13 between charges q1and q3is:
r13 =a
Plugging the given values into Coulomb’s Law, we get:
F13 =9×109×3×10−6×4×10−6
(2)2
10
F13 =108
4×10−9×103
F13 = 27 ×10−3N
So, the electric force between charges q1and q3is 27 ×10−3N.
Using the same method, we can calculate the electric force between charges
q2and q3as:
F23 =9×109×5×10−6×4×10−6
(2)2
F23 =180
4×10−9×103
F23 = 45 ×10−3N
So, the electric force between charges q2and q3is 45 ×10−3N.
Step 2: Calculate the net electric force acting on charge q3.
To find the net force, we need to consider the direction of the forces. Since
the charges q1and q3have opposite signs, their forces will be in opposite direc-
tions. However, the forces between q2and q3will both be pointing towards q3.
Calculating the net force:
Fnet =F23 −F13
Fnet = 45 ×10−3−27 ×10−3
Fnet = 18 ×10−3
Question 13
Question
Three point charges are placed at the vertices of an equilateral triangle of side
2.0 cm. Two of the charges are +3.0 nC, and the third charge is +5.0 nC.
a) What is the net electric force exerted on the charge of +3.0 nC?
b) What is the net electric force exerted on the charge of +5.0 nC?
Solution
a) To find the net electric force on the +3.0 nC charge, we need to consider the
forces exerted by the other two charges.
Let’s label the charges at the vertices of the triangle: A (+3.0 nC), B (+3.0
nC), and C (+5.0 nC).
Step 1: Calculate the distance between charges A and B. Since they are
at the vertices of an equilateral triangle of side 2.0 cm, the distance between A
and B is 2.0cm.
11
Step 2: Calculate the magnitude of the electric force between charges A
and B using Coulomb’s law:
FAB =k·q1·q2
r2
where k= 8.988×109N m2/C2is the electrostatic constant, q1= +3.0×10−9C
is the charge of A, q2= +3.0×10−9C is the charge of B, and r= 0.02 m is the
distance between A and B.
FAB = (8.988 ×109)·(3.0×10−9)2
(0.02)2
FAB ≈1.343 ×10−3N
This force is repulsive since the charges have the same sign.
Step 3: Find the net electric force on charge A. The force
FAB acts along
the line joining charges A and B, and it has the direction away from charge B.
Since A and C are equidistant from A, the force
FAC also acts along the line
joining A and C, but in the opposite direction to that of
FAB .
The net force on charge A will be the vector sum of
FAB and
FAC .
Fnet =
FAB +
FAC
Since the forces are along the same line, we can add them as scalars.
Fnet = 1.343 ×10−3N+ 1.343 ×10−3N
Fnet ≈2.686 ×10−3N
Therefore, the net electric force on the +3.0 nC charge is approximately
2.686 ×10−3N away from the charge at C.
b) The net electric force on the +5.0 nC charge can be calculated in a similar
way considering the forces from charges A and B.
Question 14
Question
Two point charges, q1=−4µC and q2= 2 µC, are located 5 cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
To find the electric force between the charges, we can use Coulomb’s law, which
states that the magnitude of the electric force between two point charges is given
by:
F=k·|q1·q2|
r2
12
where kis the Coulomb’s constant (8.9875 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 1: Convert the distance between charges from cm to meters:
r= 5 cm = 0.05 m
Step 2: Substitute the given values into Coulomb’s law to find the magni-
tude of the electric force:
F= (8.9875 ×109)·| − 4×10−6·2×10−6|
(0.05)2
Step 3: Calculate the magnitude of the electric force:
F= (8.9875 ×109)·8×10−12
0.0025
F= (8.9875 ×109)·3.2×10−9
F= 2.876 ×10−7N
Therefore, the magnitude of the electric force between the charges is 2.876 ×
10−7N.
Question 15
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length a. Charge q1is at the top vertex, charge q2is at the bottom left
vertex, and charge q3is at the bottom right vertex. Calculate the magnitude of
the net electric force on charge q1.
Solution
Step 1: Calculate the electric force between q1and q2. The distance between q1
and q2is a
√3(half of a side length of an equilateral triangle). The magnitude of
the electric force between two charges q1and q2is given by Coulomb’s law:
F12 =k|q1q2|
r2
12
where - kis Coulomb’s constant, k= 8.99 ×109N m2/C2, - r12 =a
√3is the dis-
tance between q1and q2, - q1=q2=q, as the charges are the same magnitude.
Thus,
F12 =(8.99 ×109N m2/C2)(q2)
a
√32
13
Step 2: Calculate the electric force between q1and q3. The distance between
q1and q3is 2a
√3(the full diagonal length of the equilateral triangle). Similarly,
applying Coulomb’s law:
F13 =(8.99 ×109N m2/C2)(q2)
2a
√32
Step 3: Find the net electric force on q1. The net electric force on charge
q1is in the upward direction due to the forces from q2and q3. The vertical
components of F12 and F13 add up to give the net force on q1. To find the net
force, subtract the force due to q3from the force due to q2.
Fnet = 2F12 sin(30◦)−F13
Solving this expression using the values from Steps 1 and 2 will give the
magnitude of the net electric force on charge q1.
Question 16
Question
Three point charges are placed at the corners of an equilateral triangle with side
length d. The charges are +q,+2q, and −3q. Calculate the magnitude of the
net electrostatic force on the +2qcharge due to the other two charges.
Solution
To find the net electrostatic force on the +2qcharge, we need to calculate
the individual forces on it due to the other two charges and then add them
vectorially.
Step 1: Calculate the force on the +2qcharge due to the +qcharge
The magnitude of the force between two charges q1and q2separated by a
distance ris given by Coulomb’s law:
F=k|q1||q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2). The distance between
the +2qand +qcharges in the triangle can be calculated using the height of
the equilateral triangle. The height of an equilateral triangle with side length d
is √3
2d. So, the distance between the +2qand +qcharges is d. Therefore, the
magnitude of the force on the +2qcharge due to the +qcharge is:
F1=k|2q||q|
d2
Step 2: Calculate the force on the +2qcharge due to the −3qcharge
Using the same reasoning as in Step 1, the distance between the +2qand −3q
14
charges is also d. Therefore, the magnitude of the force on the +2qcharge due
to the −3qcharge is:
F2=k|2q||3q|
d2
Step 3: Determine the net force on the +2qcharge The net force on
the +2qcharge is the vector sum of F1and F2. The forces F1and F2act in
opposite directions along the same line passing through the +2qcharge, so we
just need to subtract their magnitudes:
Fnet =F1−F2=k|2q||q|
d2−k|2q||3q|
d2
Solving this expression will give us the magnitude of the net electrostatic
force on the +2qcharge.
Question 17
Question
Three point charges are arranged in a line. Charge q1=−4.0µC is located at
the origin, charge q2= 6.0µC is located at x= 3.0m, and charge q3=−2.0µC
is located at x= 5.0m. Calculate the net electrostatic force on charge q2due
to charges q1and q3.
Solution
To find the net electrostatic force on charge q2, we first need to calculate the
forces acting on charge q2due to charges q1and q3separately, then sum these
forces vectorially.
Step 1: Calculate force on charge q2due to charge q1
The magnitude of the force on charge q2due to charge q1is given by
Coulomb’s law:
F1 on 2 =k|q1q2|
r2
12
where - kis the electrostatic constant, 8.99 ×109N m2/C2, - q1=−4.0µC=
−4.0×10−6C, - q2= 6.0µC= 6.0×10−6C, - r12 = 3.0m.
Plugging in these values, we get:
F1 on 2 = (8.99 ×109)| − 4.0×10−6×6.0×10−6|
(3.0)2
F1 on 2 = (8.99 ×109)24 ×10−12
9= 23.97 N (repulsive)
So, the force on charge q2due to charge q1is 23.97 N in the positive direction.
Step 2: Calculate force on charge q2due to charge q3
15
Similarly, the magnitude of the force on charge q2due to charge q3is given
by:
F3 on 2 =k|q3q2|
r2
32
where - q3=−2.0µC=−2.0×10−6C, - r32 = 2.0m.
Plugging in these values, we get:
F3 on 2 = (8.99 ×109)| − 2.0×10−6×6.0×10−6|
(2.0)2
F3 on 2 = (8.99 ×109)12 ×10−12
4= 26.97 N (attractive)
Therefore, the force on charge q2due to charge q3is 26.97 N in the negative
direction.
Step 3: Calculate the net force on charge q2
The net force exerted on charge q2is the vector sum of the forces due to
charges q1and q3:
Net force on q2=F1 on 2 +F3 on 2 = 23.97 N+ (−26.97 N)
Question 18
Question
Two point charges, q1= +4.0µC and q2=−2.0µC, are initially separated by
a distance of 10.0cm. If the charges are brought 20.0cm closer together, what
is the change in the magnitude of the electric force between them?
Solution
Step 1: Calculate the initial force between the charges using Coulomb’s law:
The electric force between two point charges can be calculated using Coulomb’s
law:
F=k|q1·q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant.
Substitute the given values:
F1= (8.99 ×109)|4.0×10−6·(−2.0×10−6)|
(0.10)2
Calculating F1gives:
F1≈5.758 N
Step 2: Calculate the final force when the charges are 20.0cm closer together:
After the charges are brought 20.0cm closer together, the new separation dis-
tance is r′= 10.0cm −20.0cm =−10.0cm.
16
Calculate the final force F2using the new separation distance r′:
F2= (8.99 ×109)|4.0×10−6·(−2.0×10−6)|
(0.10 −0.20)2
Calculating F2gives:
F2≈23.032 N
Step 3: Calculate the change in magnitude of the electric force: The change
in the magnitude of the electric force is given by:
∆F=|F2−F1|
Substitute the calculated values to find ∆F:
∆F=|23.032 −5.758|
∆F≈17.274 N
Therefore, the change in the magnitude of the electric force between the
charges is approximately 17.274 N.
Question 19
Question
Three point charges are located on the x-axis: a charge of +3 nC at x = 0, a
charge of -2 nC at x = 60 cm, and a charge of +5 nC at x = 100 cm. Calculate
the net electric force on the +5 nC charge.
Solution
Step 1: Convert all distances to meters. Given: q1= +3 nC = 3 ×10−9C at
x= 0 m, q2=−2nC =−2×10−9C at x= 60 cm = 0.6m, q3= +5 nC =
5×10−9C at x= 100 cm = 1.0m.
Step 2: Calculate the net electric force on the +5 nC charge. The net force
on the +5 nC charge is the sum of the forces due to the other charges. The
force between two charges is given by Coulomb’s Law:
F=k|q1q2|
r2,
where k= 8.99 ×109N m2/C2is the electrostatic constant, q1and q2are the
magnitudes of the two charges, and ris the separation between the charges.
The force between q1and q3is:
F13 =k|q1q3|
r2
13
.
17
The force between q2and q3is:
F23 =k|q2q3|
r2
23
.
The net force on the +5 nC charge due to q1and q2is:
Fnet =F13 +F23.
Substitute the given values and solve for the net force.
Question 20
Question
Two point charges, +3.0µC and −6.0µC, are placed 10.0cm apart in air. Calcu-
late the magnitude and direction of the electric force experienced by the positive
charge due to the negative charge.
Solution
Step 1: Convert the charges to coulombs. Given that 1µC= 10−6C, the first
charge +3.0µC becomes +3.0×10−6C and the second charge −6.0µC becomes
−6.0×10−6C.
Step 2: Calculate the distance between the charges in meters. The distance
between the charges is 10.0cm = 0.10 m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
The magnitude of the electric force Fbetween two point charges is given by
Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Substitute the given values into the formula:
F=(8.99 ×109)× |3.0×10−6×6.0×10−6|
(0.10)2
Step 4: Calculate the magnitude of the electric force.
F=(8.99 ×109)×18 ×10−12
0.01
Step 5: Simplify the expression.
F= 161.82 N
18
Step 6: Determine the direction of the electric force. The electric force is
attractive since the charges are of opposite sign. Therefore, the electric force
experienced by the positive charge is directed towards the negative charge.
Therefore, the magnitude of the electric force experienced by the positive
charge due to the negative charge is 161.82 N and it is directed towards the
negative charge.
Question 21
Question
Two point charges, q1=−6µC and q2= 4 µC, are placed 10 cm apart in air.
Calculate the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−6µC =−6×10−6C
q2= 4 µC = 4 ×10−6C
Step 2: Calculate the electric force. The electric force between two point
charges can be calculated using Coulomb’s Law:
F=k· |q1·q2|
r2
where Fis the magnitude of the electric force, kis the Coulomb constant (8.99×
109Nm2/C2), q1and q2are the magnitudes of the charges, and ris the distance
between the charges.
Plugging in the values:
F=(8.99 ×109Nm2/C2)· |(−6×10−6C)·(4 ×10−6C)|
(0.1m)2
Step 3: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(6 ×10−6)·(4 ×10−6)
0.01
F=215.76
0.01
F= 21576 N
Therefore, the magnitude of the electric force between the charges is 21576
N.
Question 22
Question
Two point charges of +4.0 µC and -6.0 µC are placed 0.10 m apart. Calculate
the magnitude of the electric force between them.
19
Solution
Step 1: Convert the charges to SI units. Given: q1= +4.0µC= 4.0×10−6C
q2=−6.0µC=−6.0×10−6C
Step 2: Calculate the distance between the charges. Given: r= 0.10 m
Step 3: Calculate the magnitude of the electric force between the charges
using Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the Coulomb constant with a value of 8.99 ×109N·m2/C2.
Substitute the given values into the formula:
F=(8.99 ×109)·(4.0×10−6)·(6.0×10−6)
(0.10)2
Step 4: Calculate the magnitude of the electric force.
F=35.96 ×103
0.01 = 3.596 ×106N
Therefore, the magnitude of the electric force between the charges is 3.596 ×
106N.
Question 23
Question
Two point charges with magnitudes q1= 3.0µC and q2=−5.0µC are placed
20.0 cm apart in a vacuum. Calculate the magnitude and direction of the electric
force that each charge exerts on the other.
Solution
To calculate the magnitude and direction of the electric force between the two
charges, we will use Coulomb’s law.
Step 1: Calculate the electric force F12 that charge q2exerts on charge q1.
Coulomb’s law is given by:
F12 =k|q1q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the two charges.
Plugging in the values, we get:
F12 = (8.99 ×109)(3.0×10−6)(5.0×10−6)
(0.20)2
F12 = 2.248 N
20
The direction of the force is from charge q2to charge q1.
Step 2: Calculate the electric force F21 that charge q1exerts on charge q2.
Since the forces act along the line joining the charges, the magnitude of F21 is
the same as that of F12.
Therefore, the magnitude of the electric force that each charge exerts on the
other is 2.248 N, and the direction of the force from charge q1to charge q2.
Question 24
Question
Two point charges, q1=−3.0µC and q2= 4.0µC, are placed 6.0 cm apart.
Calculate the magnitude and direction of the electric force exerted on q1.
Solution
Step 1: Convert the given charges to their equivalent in coulombs. Using the
conversion factor 1µC= 10−6C, we have: q1=−3.0µC=−3.0×10−6C, and
q2= 4.0µC= 4.0×10−6C.
Step 2: Given data: q1=−3.0×10−6Cq2= 4.0×10−6Cr= 6.0cm =
0.06 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k·
q1·q2
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging the values into the formula gives:
F= 8.99 ×109·3.0×10−6·4.0×10−6
(0.06)2
F= 8.99 ×109·12 ×10−12
0.0036
F= 8.99 ×109·0.000003333
F= 29.97 N
Step 4: Determine the direction of the electric force. Since q1is negative and
q2is positive, the electric force exerted on q1is attractive, pulling q1towards
q2. Therefore, the direction of the electric force is towards q2.
Question 25
Question
Two point charges, q1= 4.0µC and q2=−6.0µC, are placed 12.0cm apart in
air. Calculate the magnitude of the electric force between these charges.
21
Solution
Step 1: Let’s first convert the charges to Coulombs by using the relationship
1µC= 10−6C. Thus, q1= 4.0×10−6C and q2=−6.0×10−6C.
Step 2: Next, let’s calculate the distance between the charges in meters.
This can be done by converting 12.0cm to meters: 12.0cm = 0.12 m.
Step 3: Now, we can use Coulomb’s law to find the magnitude of the electric
force between the two charges. Coulomb’s law states that the magnitude of the
electric force between two point charges is given by:
F=k
q1q2
r2
,
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 4: Substituting the given values into Coulomb’s law, we have:
F= (8.99 ×109N m2/C2)|4.0×10−6C· −6.0×10−6C|
(0.12 m)2
Step 5: Calculating the magnitude of the electric force:
F= (8.99 ×109)(2.4×10−11)/(0.0144) N
F≈1.495 ×10−3N
Therefore, the magnitude of the electric force between the charges q1=
4.0µC and q2=−6.0µC, separated by a distance of 12.0cm, is approximately
1.495 ×10−3N.
22
q1
q2q3
23
Question 2
Question
Three point charges are arranged on the corners of an equilateral triangle as
shown below. Charge 1 (q1= +3µC) is located at the top vertex, charge 2
(q2=−2µC) is at the bottom left vertex, and charge 3 (q3= +4µC) is at the
bottom right vertex. The side length of the triangle is 2 meters. Calculate the
net electric force acting on charge 2 due to charges 1 and 3.
q1= +3µC
q3= +4µC
q2=−2µC
Solution
Step 1: Calculate the distance between charges 2 and 1 (top vertex). The
distance between neighboring charges can be calculated using the side length of
the equilateral triangle. Since the triangle is equilateral, each side is equal to
2 meters. The distance between charge 2 and charge 1 is the diagonal of the
triangle, which can be calculated using Pythagorean theorem. Let d21 be the
distance between charge 2 and 1.
d21 =p22+ 22=√8 = 2√2meters
Step 2: Calculate the magnitude and direction of the electric force on charge
2 due to charge 1. The electric force between two charges can be calculated using
Coulomb’s law:
F12 =k|q1q2|
r2
12
where - kis Coulomb’s constant (8.99 ×109Nm2/C2), - q1and q2are the
magnitudes of the charges, and - r12 is the distance between the charges.
The direction is attractive if charges have opposite signs and repulsive if the
charges have the same sign.
F12 =(8.99 ×109)× |3×10−6|×|−2×10−6|
(2√2)2=26.97 ×10−15
8= 3.37×10−15 N
The force is attractive because q1and q2have opposite signs.
Step 3: Calculate the distance between charges 2 and 3 (bottom left and
right vertices). Since the triangle is equilateral, the distance between these two
charges is also equal to 2√2meters.
Step 4: Calculate the magnitude and direction of the electric force on charge
2 due to charge 3.
2
Question 3
Question
Three point charges are arranged as shown in the diagram:
+3q+2q+q
d d
If the charges are separated by a distance d(which is the same between all
the pairs of charges), what is the magnitude of the net electrostatic force acting
on the charge +2qdue to the other two charges?
Given: - Electric force between two point charges q1and q2:F=k|q1q2|
r2-
Coulomb’s constant k= 8.99 ×109N m2/C2
Solution
Step 1: Calculate the force F1,2between charge +3qand charge +2q:
F1,2=k|+ 3q·+2q|
d2
F1,2=k6q2
d2
Step 2: Calculate the force F2,3between charge +2qand charge +q:
F2,3=k|+ 2q·+q|
d2
F2,3=k2q2
d2
Step 3: Calculate the net force acting on charge +2q: The forces F1,2and
F2,3are in the same direction, so we add their magnitudes:
Fnet =F1,2+F2,3
Fnet =k6q2
d2+k2q2
d2
Fnet =k8q2
d2
Therefore, the magnitude of the net electrostatic force acting on the charge
+2qdue to the other two charges is 8kq2/d2.
3
Question 4
Question
Two point charges, q1and q2, are separated by a distance of 4 cm. If q1= 5 µC
and q2=−3µC, calculate the magnitude and direction of the electric force
between the charges.
Solution
Let’s start by using Coulomb’s Law to calculate the electric force between the
two charges.
Step 1: Identify the given information
• Charge q1= 5 µC
• Charge q2=−3µC
• Distance between the charges r= 4 cm = 0.04 m
Step 2: Calculate the electric force The magnitude of the electric force
between two charges is given by Coulomb’s Law:
F=k|q1q2|
r2
where - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the charges, and - ris
the distance between the charges.
Plugging in the values:
F= (8.99 ×109)|5×10−6× −3×10−6|
(0.04)2
Step 3: Calculate the magnitude of the electric force
F= 8.99 ×109×15 ×10−12
0.0016
F= 8.99 ×15
1.6×10−3
F≈8.43 ×10−2N
Step 4: Determine the direction of the electric force Since q1and q2
have opposite charges, the electric force is attractive and directed from q1to q2.
Therefore, the magnitude of the electric force between the charges is approx-
imately 8.43 ×10−2N, and it is directed from q1to q2.
4
Question 5
Question
A small charge of +3.0µC is placed 20 cm to the right of a larger charge of
−6.0µC. Calculate the magnitude and direction of the electric force that the
larger charge exerts on the smaller charge.
Solution
Step 1: Determine the distance between the charges in meters. Given that the
smaller charge is placed 20 cm to the right of the larger charge, the distance
between the charges is d= 20 cm = 0.20 m.
Step 2: Calculate the magnitude of the electric force using Coulomb’s Law.
The magnitude of the electric force Fbetween two point charges q1and q2
separated by a distance ris given by:
F=k
q1·q2
r2
where kis Coulomb’s constant, 8.99 ×109N m2/C2.
Plugging in the given values:
F= (8.99 ×109)
(3.0×10−6)·(−6.0×10−6)
(0.20)2
Step 3: Calculate the magnitude of the electric force.
F= (8.99 ×109)
−18 ×10−12
0.04
F= 8.99 ×109×4.5×10−10
F= 4.0455 N
Step 4: Determine the direction of the electric force. The electric force
between the two charges is attractive since they have opposite signs. Therefore,
the larger charge of −6.0µC exerts a force to the right on the smaller charge of
+3.0µC.
Thus, the magnitude of the electric force that the larger charge exerts on
the smaller charge is 4.0455 N to the right.
Question 6
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. Charge q1=−2nC is located at the top vertex, charge q2= 4
nC is located at the bottom left vertex, and charge q3=−3nC is located at the
5
bottom right vertex. Calculate the magnitude and direction of the net electric
force experienced by the charge at the top vertex.
q2= 4 nC
q1=−2nC q3=−3nC
Solution
Step 1: We start by determining the direction and magnitude of the electric
force between q1and q2. The electric force between two point charges is given
by Coulomb’s Law:
F=k· |q1|·|q2|
r2
The direction of the force is along the line connecting the charges, and the
force is repulsive if both charges have the same sign and attractive if they have
opposite signs.
Step 2: Calculate the distance rbetween q1and q2. Since the charges are
at the vertices of an equilateral triangle,
r=2
√3a
where ais the side length of the triangle.
Step 3: Calculate the magnitude of the electric force between q1and q2using
Coulomb’s Law:
F12 =k· |q1|·|q2|
r2
Step 4: Next, we determine the direction and magnitude of the electric force
between q1and q3. Similarly to Step 1, calculate the distance between q1and
q3and then the force between them.
Step 5: Finally, calculate the net electric force experienced by the charge at
the top vertex by considering the vector sum of the forces between q1and q2and
between q1and q3. Remember to consider both the magnitude and direction of
each force.
Question 7
Question
Two point charges, Q1=−3.0µC and Q2= 6.0µC, are placed 10.0 cm apart in
a vacuum. Calculate the magnitude of the electric force that Q1exerts on Q2.
Solution
Let’s first write down the formula for the electric force between two point
charges:
F=k|Q1Q2|
r2
6
where F= magnitude of the electric force, k= Coulomb’s constant (8.99 ×
109N m2/C2), Q1and Q2= magnitudes of the charges, and r= separation
distance between the charges.
Step 1: Convert the charges to Coulombs. Since 1µC= 10−6C, Q1=
−3.0µC=−3.0×10−6C and Q2= 6.0µC= 6.0×10−6C.
Step 2: Calculate the force using the formula.
F=(8.99 ×109)×|−3.0×10−6×6.0×10−6|
(0.10)2
F=8.99 ×109×18 ×10−12
0.01
F=161.82 ×10−3
0.01 = 16.182 N
So, the magnitude of the electric force that Q1exerts on Q2is 16.182 N.
Question 8
Question
Three point charges are arranged in a line as shown below:
+2 µC−5µC+3 µC
The charges are located at positions x=−5cm, x= 0 cm, and x= 5 cm
respectively. Determine the magnitude and direction of the net electrostatic
force acting on the +3 µC charge.
Solution
Step 1: Calculate the distance between each pair of charges. First, we will
determine the distances r1,r2, and r3between the charges +2 µC and −5µC,
−5µC and +3 µC, and +2 µC and +3 µC respectively.
r1= 5 cm −(−5cm) = 10 cm = 0.10 m
r2= 0 cm −(−5cm) = 5 cm = 0.05 m
r3= 5 cm −0cm = 5 cm = 0.05 m
Step 2: Calculate the magnitude of the force between each pair of charges.
The magnitude of the electrostatic force between two point charges q1and
q2separated by a distance ris given by Coulomb’s Law:
F=k|q1q2|
r2
Where k= 8.99 ×109N m2/C2is Coulomb’s constant.
7
For charges +2 µC and −5µC:
F1= 8.99 ×109|2×10−6C×(−5×10−6C)|
(0.10 m)2
F1= 8.99 ×109×1×10−11 N= 8.99 ×10−2N= 0.0899 N
For charges −5µC and +3 µC:
F2= 8.99 ×109| − 5×10−6C×3×10−6C|
(0.05 m)2
F2= 8.99 ×109×1.5×10−11 N= 1.35 ×10−2N= 0.0135 N
For charges +2 µC and +3 µC:
F3= 8.99 ×109|2×10−6C×3×10−6C|
(0.05 m)2
F3= 8.99
Question 9
Question
Three point charges are placed at the corners of an equilateral triangle of side
d. The charges are +q,−2q, and +3q. What is the net force on the charge +q
due to the other two charges? Assume the charges are placed at the vertices of
the triangle and the sides are of equal length.
Solution
Let’s denote the charge +qas Q1, the charge −2qas Q2, and the charge +3qas
Q3. We will first find the force on Q1due to Q2and then the force on Q1due
to Q3. Finally, we will find the net force by summing up the forces.
Step 1: Calculate the force on Q1due to Q2. The electric force between
two charges Q1and Q2is given by Coulomb’s law:
F12 =k|Q1Q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2) and ris the distance
between the charges. The distance between Q1and Q2in this case, considering
the equilateral triangle, is d. So, the force F12 is given by:
F12 =k|q·(−2q)|
d2=k|q2·2|
d2=2kq2
d2
8
Step 2: Calculate the force on Q1due to Q3. Following the same steps as
above, we find the force F13 on Q1due to Q3:
F13 =k|q·3q|
d2=k|q2·3|
d2=3kq2
d2
Step 3: Calculate the net force on Q1. The net force is the vector sum of
F12 and F13, where the direction of F12 is opposite to F13 (since Q2is negative
and Q3is positive).
Fnet =F13 −F12 =3kq2
d2−2kq2
d2=kq2
d2
Therefore, the net force on the charge +qdue to the other two charges is
kq2
d2directed towards the charge +3q.
Question 10
Question
Three point charges are arranged along the x-axis as follows: Q1=−2µC at
the origin, Q2= 4 µC at x= 3 m, and Q3=−3µC at x= 5 m. Calculate the
net force on Q2due to Q1and Q3.
Solution
Step 1: Determine the force on Q2due to Q1. The force F12 on Q2due to Q1
can be calculated using the electric force equation:
F12 =k|Q1Q2|
r2
12
where: - k= 8.99 ×109Nm2/C2is the Coulomb constant, - Q1=−2×10−6C,
-Q2= 4 ×10−6C, - r12 = 3 mis the distance between Q1and Q2.
Plugging in the values:
F12 =8.99 ×109×|−2×10−6×4×10−6|
(3)2
F12 =8.99 ×109×8×10−12
9= 7.99 ×10−3N
Step 2: Determine the force on Q2due to Q3. The force F23 on Q2due to
Q3can be calculated using the electric force equation:
F23 =k|Q2Q3|
r2
23
9
where: - Q2= 4 ×10−6C, - Q3=−3×10−6C, - r23 = 2 mis the distance
between Q2and Q3.
Plugging in the values:
F23 =8.99 ×109× |4×10−6× −3×10−6|
(2)2
F23 =8.99 ×109×12 ×10−12
4= 26.97 ×10−3= 2.697 ×10−2N
Step 3: Calculate the net force on Q2. The net force on Q2is the vector
sum of forces F12 and F23.
Fnet =F12 +F23 = 7.99 ×10−3+ 2.697 ×10−2
Fnet ≈3.49 ×10−2N
Therefore, the net force on Q2due to Q1and Q3is approximately 3.49 ×
10−2Nin the positive x-direction.
Question 12
Question
Three point charges are arranged in the vertices of an equilateral triangle of side
length a= 2 m as shown in the figure below. Charge q1=−3µC is at the top
vertex, charge q2= 5 µC is at the bottom left vertex, and charge q3=−4µC is
at the bottom right vertex. Calculate the net electric force acting on q3.
Solution
Step 1: Calculate the electric force acting on charge q3due to charges q1and
q2separately.
The electric force between two charges q1and q3is given by Coulomb’s Law:
F13 =k|q1||q3|
r2
13
where kis Coulomb’s constant, r13 is the distance between charges q1and q3,
and |q1|= 3 ×10−6C, |q3|= 4 ×10−6C.
Given that the side length of the equilateral triangle is a= 2 m, the distance
r13 between charges q1and q3is:
r13 =a
Plugging the given values into Coulomb’s Law, we get:
F13 =9×109×3×10−6×4×10−6
(2)2
10
F13 =108
4×10−9×103
F13 = 27 ×10−3N
So, the electric force between charges q1and q3is 27 ×10−3N.
Using the same method, we can calculate the electric force between charges
q2and q3as:
F23 =9×109×5×10−6×4×10−6
(2)2
F23 =180
4×10−9×103
F23 = 45 ×10−3N
So, the electric force between charges q2and q3is 45 ×10−3N.
Step 2: Calculate the net electric force acting on charge q3.
To find the net force, we need to consider the direction of the forces. Since
the charges q1and q3have opposite signs, their forces will be in opposite direc-
tions. However, the forces between q2and q3will both be pointing towards q3.
Calculating the net force:
Fnet =F23 −F13
Fnet = 45 ×10−3−27 ×10−3
Fnet = 18 ×10−3
Question 13
Question
Three point charges are placed at the vertices of an equilateral triangle of side
2.0 cm. Two of the charges are +3.0 nC, and the third charge is +5.0 nC.
a) What is the net electric force exerted on the charge of +3.0 nC?
b) What is the net electric force exerted on the charge of +5.0 nC?
Solution
a) To find the net electric force on the +3.0 nC charge, we need to consider the
forces exerted by the other two charges.
Let’s label the charges at the vertices of the triangle: A (+3.0 nC), B (+3.0
nC), and C (+5.0 nC).
Step 1: Calculate the distance between charges A and B. Since they are
at the vertices of an equilateral triangle of side 2.0 cm, the distance between A
and B is 2.0cm.
11
Step 2: Calculate the magnitude of the electric force between charges A
and B using Coulomb’s law:
FAB =k·q1·q2
r2
where k= 8.988×109N m2/C2is the electrostatic constant, q1= +3.0×10−9C
is the charge of A, q2= +3.0×10−9C is the charge of B, and r= 0.02 m is the
distance between A and B.
FAB = (8.988 ×109)·(3.0×10−9)2
(0.02)2
FAB ≈1.343 ×10−3N
This force is repulsive since the charges have the same sign.
Step 3: Find the net electric force on charge A. The force
FAB acts along
the line joining charges A and B, and it has the direction away from charge B.
Since A and C are equidistant from A, the force
FAC also acts along the line
joining A and C, but in the opposite direction to that of
FAB .
The net force on charge A will be the vector sum of
FAB and
FAC .
Fnet =
FAB +
FAC
Since the forces are along the same line, we can add them as scalars.
Fnet = 1.343 ×10−3N+ 1.343 ×10−3N
Fnet ≈2.686 ×10−3N
Therefore, the net electric force on the +3.0 nC charge is approximately
2.686 ×10−3N away from the charge at C.
b) The net electric force on the +5.0 nC charge can be calculated in a similar
way considering the forces from charges A and B.
Question 14
Question
Two point charges, q1=−4µC and q2= 2 µC, are located 5 cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
To find the electric force between the charges, we can use Coulomb’s law, which
states that the magnitude of the electric force between two point charges is given
by:
F=k·|q1·q2|
r2
12
where kis the Coulomb’s constant (8.9875 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 1: Convert the distance between charges from cm to meters:
r= 5 cm = 0.05 m
Step 2: Substitute the given values into Coulomb’s law to find the magni-
tude of the electric force:
F= (8.9875 ×109)·| − 4×10−6·2×10−6|
(0.05)2
Step 3: Calculate the magnitude of the electric force:
F= (8.9875 ×109)·8×10−12
0.0025
F= (8.9875 ×109)·3.2×10−9
F= 2.876 ×10−7N
Therefore, the magnitude of the electric force between the charges is 2.876 ×
10−7N.
Question 15
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length a. Charge q1is at the top vertex, charge q2is at the bottom left
vertex, and charge q3is at the bottom right vertex. Calculate the magnitude of
the net electric force on charge q1.
Solution
Step 1: Calculate the electric force between q1and q2. The distance between q1
and q2is a
√3(half of a side length of an equilateral triangle). The magnitude of
the electric force between two charges q1and q2is given by Coulomb’s law:
F12 =k|q1q2|
r2
12
where - kis Coulomb’s constant, k= 8.99 ×109N m2/C2, - r12 =a
√3is the dis-
tance between q1and q2, - q1=q2=q, as the charges are the same magnitude.
Thus,
F12 =(8.99 ×109N m2/C2)(q2)
a
√32
13
Step 2: Calculate the electric force between q1and q3. The distance between
q1and q3is 2a
√3(the full diagonal length of the equilateral triangle). Similarly,
applying Coulomb’s law:
F13 =(8.99 ×109N m2/C2)(q2)
2a
√32
Step 3: Find the net electric force on q1. The net electric force on charge
q1is in the upward direction due to the forces from q2and q3. The vertical
components of F12 and F13 add up to give the net force on q1. To find the net
force, subtract the force due to q3from the force due to q2.
Fnet = 2F12 sin(30◦)−F13
Solving this expression using the values from Steps 1 and 2 will give the
magnitude of the net electric force on charge q1.
Question 16
Question
Three point charges are placed at the corners of an equilateral triangle with side
length d. The charges are +q,+2q, and −3q. Calculate the magnitude of the
net electrostatic force on the +2qcharge due to the other two charges.
Solution
To find the net electrostatic force on the +2qcharge, we need to calculate
the individual forces on it due to the other two charges and then add them
vectorially.
Step 1: Calculate the force on the +2qcharge due to the +qcharge
The magnitude of the force between two charges q1and q2separated by a
distance ris given by Coulomb’s law:
F=k|q1||q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2). The distance between
the +2qand +qcharges in the triangle can be calculated using the height of
the equilateral triangle. The height of an equilateral triangle with side length d
is √3
2d. So, the distance between the +2qand +qcharges is d. Therefore, the
magnitude of the force on the +2qcharge due to the +qcharge is:
F1=k|2q||q|
d2
Step 2: Calculate the force on the +2qcharge due to the −3qcharge
Using the same reasoning as in Step 1, the distance between the +2qand −3q
14
charges is also d. Therefore, the magnitude of the force on the +2qcharge due
to the −3qcharge is:
F2=k|2q||3q|
d2
Step 3: Determine the net force on the +2qcharge The net force on
the +2qcharge is the vector sum of F1and F2. The forces F1and F2act in
opposite directions along the same line passing through the +2qcharge, so we
just need to subtract their magnitudes:
Fnet =F1−F2=k|2q||q|
d2−k|2q||3q|
d2
Solving this expression will give us the magnitude of the net electrostatic
force on the +2qcharge.
Question 17
Question
Three point charges are arranged in a line. Charge q1=−4.0µC is located at
the origin, charge q2= 6.0µC is located at x= 3.0m, and charge q3=−2.0µC
is located at x= 5.0m. Calculate the net electrostatic force on charge q2due
to charges q1and q3.
Solution
To find the net electrostatic force on charge q2, we first need to calculate the
forces acting on charge q2due to charges q1and q3separately, then sum these
forces vectorially.
Step 1: Calculate force on charge q2due to charge q1
The magnitude of the force on charge q2due to charge q1is given by
Coulomb’s law:
F1 on 2 =k|q1q2|
r2
12
where - kis the electrostatic constant, 8.99 ×109N m2/C2, - q1=−4.0µC=
−4.0×10−6C, - q2= 6.0µC= 6.0×10−6C, - r12 = 3.0m.
Plugging in these values, we get:
F1 on 2 = (8.99 ×109)| − 4.0×10−6×6.0×10−6|
(3.0)2
F1 on 2 = (8.99 ×109)24 ×10−12
9= 23.97 N (repulsive)
So, the force on charge q2due to charge q1is 23.97 N in the positive direction.
Step 2: Calculate force on charge q2due to charge q3
15
Similarly, the magnitude of the force on charge q2due to charge q3is given
by:
F3 on 2 =k|q3q2|
r2
32
where - q3=−2.0µC=−2.0×10−6C, - r32 = 2.0m.
Plugging in these values, we get:
F3 on 2 = (8.99 ×109)| − 2.0×10−6×6.0×10−6|
(2.0)2
F3 on 2 = (8.99 ×109)12 ×10−12
4= 26.97 N (attractive)
Therefore, the force on charge q2due to charge q3is 26.97 N in the negative
direction.
Step 3: Calculate the net force on charge q2
The net force exerted on charge q2is the vector sum of the forces due to
charges q1and q3:
Net force on q2=F1 on 2 +F3 on 2 = 23.97 N+ (−26.97 N)
Question 18
Question
Two point charges, q1= +4.0µC and q2=−2.0µC, are initially separated by
a distance of 10.0cm. If the charges are brought 20.0cm closer together, what
is the change in the magnitude of the electric force between them?
Solution
Step 1: Calculate the initial force between the charges using Coulomb’s law:
The electric force between two point charges can be calculated using Coulomb’s
law:
F=k|q1·q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant.
Substitute the given values:
F1= (8.99 ×109)|4.0×10−6·(−2.0×10−6)|
(0.10)2
Calculating F1gives:
F1≈5.758 N
Step 2: Calculate the final force when the charges are 20.0cm closer together:
After the charges are brought 20.0cm closer together, the new separation dis-
tance is r′= 10.0cm −20.0cm =−10.0cm.
16
Calculate the final force F2using the new separation distance r′:
F2= (8.99 ×109)|4.0×10−6·(−2.0×10−6)|
(0.10 −0.20)2
Calculating F2gives:
F2≈23.032 N
Step 3: Calculate the change in magnitude of the electric force: The change
in the magnitude of the electric force is given by:
∆F=|F2−F1|
Substitute the calculated values to find ∆F:
∆F=|23.032 −5.758|
∆F≈17.274 N
Therefore, the change in the magnitude of the electric force between the
charges is approximately 17.274 N.
Question 19
Question
Three point charges are located on the x-axis: a charge of +3 nC at x = 0, a
charge of -2 nC at x = 60 cm, and a charge of +5 nC at x = 100 cm. Calculate
the net electric force on the +5 nC charge.
Solution
Step 1: Convert all distances to meters. Given: q1= +3 nC = 3 ×10−9C at
x= 0 m, q2=−2nC =−2×10−9C at x= 60 cm = 0.6m, q3= +5 nC =
5×10−9C at x= 100 cm = 1.0m.
Step 2: Calculate the net electric force on the +5 nC charge. The net force
on the +5 nC charge is the sum of the forces due to the other charges. The
force between two charges is given by Coulomb’s Law:
F=k|q1q2|
r2,
where k= 8.99 ×109N m2/C2is the electrostatic constant, q1and q2are the
magnitudes of the two charges, and ris the separation between the charges.
The force between q1and q3is:
F13 =k|q1q3|
r2
13
.
17
The force between q2and q3is:
F23 =k|q2q3|
r2
23
.
The net force on the +5 nC charge due to q1and q2is:
Fnet =F13 +F23.
Substitute the given values and solve for the net force.
Question 20
Question
Two point charges, +3.0µC and −6.0µC, are placed 10.0cm apart in air. Calcu-
late the magnitude and direction of the electric force experienced by the positive
charge due to the negative charge.
Solution
Step 1: Convert the charges to coulombs. Given that 1µC= 10−6C, the first
charge +3.0µC becomes +3.0×10−6C and the second charge −6.0µC becomes
−6.0×10−6C.
Step 2: Calculate the distance between the charges in meters. The distance
between the charges is 10.0cm = 0.10 m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
The magnitude of the electric force Fbetween two point charges is given by
Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Substitute the given values into the formula:
F=(8.99 ×109)× |3.0×10−6×6.0×10−6|
(0.10)2
Step 4: Calculate the magnitude of the electric force.
F=(8.99 ×109)×18 ×10−12
0.01
Step 5: Simplify the expression.
F= 161.82 N
18
Step 6: Determine the direction of the electric force. The electric force is
attractive since the charges are of opposite sign. Therefore, the electric force
experienced by the positive charge is directed towards the negative charge.
Therefore, the magnitude of the electric force experienced by the positive
charge due to the negative charge is 161.82 N and it is directed towards the
negative charge.
Question 21
Question
Two point charges, q1=−6µC and q2= 4 µC, are placed 10 cm apart in air.
Calculate the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−6µC =−6×10−6C
q2= 4 µC = 4 ×10−6C
Step 2: Calculate the electric force. The electric force between two point
charges can be calculated using Coulomb’s Law:
F=k· |q1·q2|
r2
where Fis the magnitude of the electric force, kis the Coulomb constant (8.99×
109Nm2/C2), q1and q2are the magnitudes of the charges, and ris the distance
between the charges.
Plugging in the values:
F=(8.99 ×109Nm2/C2)· |(−6×10−6C)·(4 ×10−6C)|
(0.1m)2
Step 3: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(6 ×10−6)·(4 ×10−6)
0.01
F=215.76
0.01
F= 21576 N
Therefore, the magnitude of the electric force between the charges is 21576
N.
Question 22
Question
Two point charges of +4.0 µC and -6.0 µC are placed 0.10 m apart. Calculate
the magnitude of the electric force between them.
19
Solution
Step 1: Convert the charges to SI units. Given: q1= +4.0µC= 4.0×10−6C
q2=−6.0µC=−6.0×10−6C
Step 2: Calculate the distance between the charges. Given: r= 0.10 m
Step 3: Calculate the magnitude of the electric force between the charges
using Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the Coulomb constant with a value of 8.99 ×109N·m2/C2.
Substitute the given values into the formula:
F=(8.99 ×109)·(4.0×10−6)·(6.0×10−6)
(0.10)2
Step 4: Calculate the magnitude of the electric force.
F=35.96 ×103
0.01 = 3.596 ×106N
Therefore, the magnitude of the electric force between the charges is 3.596 ×
106N.
Question 23
Question
Two point charges with magnitudes q1= 3.0µC and q2=−5.0µC are placed
20.0 cm apart in a vacuum. Calculate the magnitude and direction of the electric
force that each charge exerts on the other.
Solution
To calculate the magnitude and direction of the electric force between the two
charges, we will use Coulomb’s law.
Step 1: Calculate the electric force F12 that charge q2exerts on charge q1.
Coulomb’s law is given by:
F12 =k|q1q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the two charges.
Plugging in the values, we get:
F12 = (8.99 ×109)(3.0×10−6)(5.0×10−6)
(0.20)2
F12 = 2.248 N
20
The direction of the force is from charge q2to charge q1.
Step 2: Calculate the electric force F21 that charge q1exerts on charge q2.
Since the forces act along the line joining the charges, the magnitude of F21 is
the same as that of F12.
Therefore, the magnitude of the electric force that each charge exerts on the
other is 2.248 N, and the direction of the force from charge q1to charge q2.
Question 24
Question
Two point charges, q1=−3.0µC and q2= 4.0µC, are placed 6.0 cm apart.
Calculate the magnitude and direction of the electric force exerted on q1.
Solution
Step 1: Convert the given charges to their equivalent in coulombs. Using the
conversion factor 1µC= 10−6C, we have: q1=−3.0µC=−3.0×10−6C, and
q2= 4.0µC= 4.0×10−6C.
Step 2: Given data: q1=−3.0×10−6Cq2= 4.0×10−6Cr= 6.0cm =
0.06 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k·
q1·q2
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging the values into the formula gives:
F= 8.99 ×109·3.0×10−6·4.0×10−6
(0.06)2
F= 8.99 ×109·12 ×10−12
0.0036
F= 8.99 ×109·0.000003333
F= 29.97 N
Step 4: Determine the direction of the electric force. Since q1is negative and
q2is positive, the electric force exerted on q1is attractive, pulling q1towards
q2. Therefore, the direction of the electric force is towards q2.
Question 25
Question
Two point charges, q1= 4.0µC and q2=−6.0µC, are placed 12.0cm apart in
air. Calculate the magnitude of the electric force between these charges.
21
Solution
Step 1: Let’s first convert the charges to Coulombs by using the relationship
1µC= 10−6C. Thus, q1= 4.0×10−6C and q2=−6.0×10−6C.
Step 2: Next, let’s calculate the distance between the charges in meters.
This can be done by converting 12.0cm to meters: 12.0cm = 0.12 m.
Step 3: Now, we can use Coulomb’s law to find the magnitude of the electric
force between the two charges. Coulomb’s law states that the magnitude of the
electric force between two point charges is given by:
F=k
q1q2
r2
,
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 4: Substituting the given values into Coulomb’s law, we have:
F= (8.99 ×109N m2/C2)|4.0×10−6C· −6.0×10−6C|
(0.12 m)2
Step 5: Calculating the magnitude of the electric force:
F= (8.99 ×109)(2.4×10−11)/(0.0144) N
F≈1.495 ×10−3N
Therefore, the magnitude of the electric force between the charges q1=
4.0µC and q2=−6.0µC, separated by a distance of 12.0cm, is approximately
1.495 ×10−3N.
22
q1
q2q3
23
Question 2
Question
Three point charges are arranged on the corners of an equilateral triangle as
shown below. Charge 1 (q1= +3µC) is located at the top vertex, charge 2
(q2=−2µC) is at the bottom left vertex, and charge 3 (q3= +4µC) is at the
bottom right vertex. The side length of the triangle is 2 meters. Calculate the
net electric force acting on charge 2 due to charges 1 and 3.
q1= +3µC
q3= +4µC
q2=−2µC
Solution
Step 1: Calculate the distance between charges 2 and 1 (top vertex). The
distance between neighboring charges can be calculated using the side length of
the equilateral triangle. Since the triangle is equilateral, each side is equal to
2 meters. The distance between charge 2 and charge 1 is the diagonal of the
triangle, which can be calculated using Pythagorean theorem. Let d21 be the
distance between charge 2 and 1.
d21 =p22+ 22=√8 = 2√2meters
Step 2: Calculate the magnitude and direction of the electric force on charge
2 due to charge 1. The electric force between two charges can be calculated using
Coulomb’s law:
F12 =k|q1q2|
r2
12
where - kis Coulomb’s constant (8.99 ×109Nm2/C2), - q1and q2are the
magnitudes of the charges, and - r12 is the distance between the charges.
The direction is attractive if charges have opposite signs and repulsive if the
charges have the same sign.
F12 =(8.99 ×109)× |3×10−6|×|−2×10−6|
(2√2)2=26.97 ×10−15
8= 3.37×10−15 N
The force is attractive because q1and q2have opposite signs.
Step 3: Calculate the distance between charges 2 and 3 (bottom left and
right vertices). Since the triangle is equilateral, the distance between these two
charges is also equal to 2√2meters.
Step 4: Calculate the magnitude and direction of the electric force on charge
2 due to charge 3.
2
Question 3
Question
Three point charges are arranged as shown in the diagram:
+3q+2q+q
d d
If the charges are separated by a distance d(which is the same between all
the pairs of charges), what is the magnitude of the net electrostatic force acting
on the charge +2qdue to the other two charges?
Given: - Electric force between two point charges q1and q2:F=k|q1q2|
r2-
Coulomb’s constant k= 8.99 ×109N m2/C2
Solution
Step 1: Calculate the force F1,2between charge +3qand charge +2q:
F1,2=k|+ 3q·+2q|
d2
F1,2=k6q2
d2
Step 2: Calculate the force F2,3between charge +2qand charge +q:
F2,3=k|+ 2q·+q|
d2
F2,3=k2q2
d2
Step 3: Calculate the net force acting on charge +2q: The forces F1,2and
F2,3are in the same direction, so we add their magnitudes:
Fnet =F1,2+F2,3
Fnet =k6q2
d2+k2q2
d2
Fnet =k8q2
d2
Therefore, the magnitude of the net electrostatic force acting on the charge
+2qdue to the other two charges is 8kq2/d2.
3
Question 4
Question
Two point charges, q1and q2, are separated by a distance of 4 cm. If q1= 5 µC
and q2=−3µC, calculate the magnitude and direction of the electric force
between the charges.
Solution
Let’s start by using Coulomb’s Law to calculate the electric force between the
two charges.
Step 1: Identify the given information
• Charge q1= 5 µC
• Charge q2=−3µC
• Distance between the charges r= 4 cm = 0.04 m
Step 2: Calculate the electric force The magnitude of the electric force
between two charges is given by Coulomb’s Law:
F=k|q1q2|
r2
where - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the charges, and - ris
the distance between the charges.
Plugging in the values:
F= (8.99 ×109)|5×10−6× −3×10−6|
(0.04)2
Step 3: Calculate the magnitude of the electric force
F= 8.99 ×109×15 ×10−12
0.0016
F= 8.99 ×15
1.6×10−3
F≈8.43 ×10−2N
Step 4: Determine the direction of the electric force Since q1and q2
have opposite charges, the electric force is attractive and directed from q1to q2.
Therefore, the magnitude of the electric force between the charges is approx-
imately 8.43 ×10−2N, and it is directed from q1to q2.
4
Question 5
Question
A small charge of +3.0µC is placed 20 cm to the right of a larger charge of
−6.0µC. Calculate the magnitude and direction of the electric force that the
larger charge exerts on the smaller charge.
Solution
Step 1: Determine the distance between the charges in meters. Given that the
smaller charge is placed 20 cm to the right of the larger charge, the distance
between the charges is d= 20 cm = 0.20 m.
Step 2: Calculate the magnitude of the electric force using Coulomb’s Law.
The magnitude of the electric force Fbetween two point charges q1and q2
separated by a distance ris given by:
F=k
q1·q2
r2
where kis Coulomb’s constant, 8.99 ×109N m2/C2.
Plugging in the given values:
F= (8.99 ×109)
(3.0×10−6)·(−6.0×10−6)
(0.20)2
Step 3: Calculate the magnitude of the electric force.
F= (8.99 ×109)
−18 ×10−12
0.04
F= 8.99 ×109×4.5×10−10
F= 4.0455 N
Step 4: Determine the direction of the electric force. The electric force
between the two charges is attractive since they have opposite signs. Therefore,
the larger charge of −6.0µC exerts a force to the right on the smaller charge of
+3.0µC.
Thus, the magnitude of the electric force that the larger charge exerts on
the smaller charge is 4.0455 N to the right.
Question 6
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. Charge q1=−2nC is located at the top vertex, charge q2= 4
nC is located at the bottom left vertex, and charge q3=−3nC is located at the
5
bottom right vertex. Calculate the magnitude and direction of the net electric
force experienced by the charge at the top vertex.
q2= 4 nC
q1=−2nC q3=−3nC
Solution
Step 1: We start by determining the direction and magnitude of the electric
force between q1and q2. The electric force between two point charges is given
by Coulomb’s Law:
F=k· |q1|·|q2|
r2
The direction of the force is along the line connecting the charges, and the
force is repulsive if both charges have the same sign and attractive if they have
opposite signs.
Step 2: Calculate the distance rbetween q1and q2. Since the charges are
at the vertices of an equilateral triangle,
r=2
√3a
where ais the side length of the triangle.
Step 3: Calculate the magnitude of the electric force between q1and q2using
Coulomb’s Law:
F12 =k· |q1|·|q2|
r2
Step 4: Next, we determine the direction and magnitude of the electric force
between q1and q3. Similarly to Step 1, calculate the distance between q1and
q3and then the force between them.
Step 5: Finally, calculate the net electric force experienced by the charge at
the top vertex by considering the vector sum of the forces between q1and q2and
between q1and q3. Remember to consider both the magnitude and direction of
each force.
Question 7
Question
Two point charges, Q1=−3.0µC and Q2= 6.0µC, are placed 10.0 cm apart in
a vacuum. Calculate the magnitude of the electric force that Q1exerts on Q2.
Solution
Let’s first write down the formula for the electric force between two point
charges:
F=k|Q1Q2|
r2
6
where F= magnitude of the electric force, k= Coulomb’s constant (8.99 ×
109N m2/C2), Q1and Q2= magnitudes of the charges, and r= separation
distance between the charges.
Step 1: Convert the charges to Coulombs. Since 1µC= 10−6C, Q1=
−3.0µC=−3.0×10−6C and Q2= 6.0µC= 6.0×10−6C.
Step 2: Calculate the force using the formula.
F=(8.99 ×109)×|−3.0×10−6×6.0×10−6|
(0.10)2
F=8.99 ×109×18 ×10−12
0.01
F=161.82 ×10−3
0.01 = 16.182 N
So, the magnitude of the electric force that Q1exerts on Q2is 16.182 N.
Question 8
Question
Three point charges are arranged in a line as shown below:
+2 µC−5µC+3 µC
The charges are located at positions x=−5cm, x= 0 cm, and x= 5 cm
respectively. Determine the magnitude and direction of the net electrostatic
force acting on the +3 µC charge.
Solution
Step 1: Calculate the distance between each pair of charges. First, we will
determine the distances r1,r2, and r3between the charges +2 µC and −5µC,
−5µC and +3 µC, and +2 µC and +3 µC respectively.
r1= 5 cm −(−5cm) = 10 cm = 0.10 m
r2= 0 cm −(−5cm) = 5 cm = 0.05 m
r3= 5 cm −0cm = 5 cm = 0.05 m
Step 2: Calculate the magnitude of the force between each pair of charges.
The magnitude of the electrostatic force between two point charges q1and
q2separated by a distance ris given by Coulomb’s Law:
F=k|q1q2|
r2
Where k= 8.99 ×109N m2/C2is Coulomb’s constant.
7
For charges +2 µC and −5µC:
F1= 8.99 ×109|2×10−6C×(−5×10−6C)|
(0.10 m)2
F1= 8.99 ×109×1×10−11 N= 8.99 ×10−2N= 0.0899 N
For charges −5µC and +3 µC:
F2= 8.99 ×109| − 5×10−6C×3×10−6C|
(0.05 m)2
F2= 8.99 ×109×1.5×10−11 N= 1.35 ×10−2N= 0.0135 N
For charges +2 µC and +3 µC:
F3= 8.99 ×109|2×10−6C×3×10−6C|
(0.05 m)2
F3= 8.99
Question 9
Question
Three point charges are placed at the corners of an equilateral triangle of side
d. The charges are +q,−2q, and +3q. What is the net force on the charge +q
due to the other two charges? Assume the charges are placed at the vertices of
the triangle and the sides are of equal length.
Solution
Let’s denote the charge +qas Q1, the charge −2qas Q2, and the charge +3qas
Q3. We will first find the force on Q1due to Q2and then the force on Q1due
to Q3. Finally, we will find the net force by summing up the forces.
Step 1: Calculate the force on Q1due to Q2. The electric force between
two charges Q1and Q2is given by Coulomb’s law:
F12 =k|Q1Q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2) and ris the distance
between the charges. The distance between Q1and Q2in this case, considering
the equilateral triangle, is d. So, the force F12 is given by:
F12 =k|q·(−2q)|
d2=k|q2·2|
d2=2kq2
d2
8
Step 2: Calculate the force on Q1due to Q3. Following the same steps as
above, we find the force F13 on Q1due to Q3:
F13 =k|q·3q|
d2=k|q2·3|
d2=3kq2
d2
Step 3: Calculate the net force on Q1. The net force is the vector sum of
F12 and F13, where the direction of F12 is opposite to F13 (since Q2is negative
and Q3is positive).
Fnet =F13 −F12 =3kq2
d2−2kq2
d2=kq2
d2
Therefore, the net force on the charge +qdue to the other two charges is
kq2
d2directed towards the charge +3q.
Question 10
Question
Three point charges are arranged along the x-axis as follows: Q1=−2µC at
the origin, Q2= 4 µC at x= 3 m, and Q3=−3µC at x= 5 m. Calculate the
net force on Q2due to Q1and Q3.
Solution
Step 1: Determine the force on Q2due to Q1. The force F12 on Q2due to Q1
can be calculated using the electric force equation:
F12 =k|Q1Q2|
r2
12
where: - k= 8.99 ×109Nm2/C2is the Coulomb constant, - Q1=−2×10−6C,
-Q2= 4 ×10−6C, - r12 = 3 mis the distance between Q1and Q2.
Plugging in the values:
F12 =8.99 ×109×|−2×10−6×4×10−6|
(3)2
F12 =8.99 ×109×8×10−12
9= 7.99 ×10−3N
Step 2: Determine the force on Q2due to Q3. The force F23 on Q2due to
Q3can be calculated using the electric force equation:
F23 =k|Q2Q3|
r2
23
9
where: - Q2= 4 ×10−6C, - Q3=−3×10−6C, - r23 = 2 mis the distance
between Q2and Q3.
Plugging in the values:
F23 =8.99 ×109× |4×10−6× −3×10−6|
(2)2
F23 =8.99 ×109×12 ×10−12
4= 26.97 ×10−3= 2.697 ×10−2N
Step 3: Calculate the net force on Q2. The net force on Q2is the vector
sum of forces F12 and F23.
Fnet =F12 +F23 = 7.99 ×10−3+ 2.697 ×10−2
Fnet ≈3.49 ×10−2N
Therefore, the net force on Q2due to Q1and Q3is approximately 3.49 ×
10−2Nin the positive x-direction.
Question 12
Question
Three point charges are arranged in the vertices of an equilateral triangle of side
length a= 2 m as shown in the figure below. Charge q1=−3µC is at the top
vertex, charge q2= 5 µC is at the bottom left vertex, and charge q3=−4µC is
at the bottom right vertex. Calculate the net electric force acting on q3.
Solution
Step 1: Calculate the electric force acting on charge q3due to charges q1and
q2separately.
The electric force between two charges q1and q3is given by Coulomb’s Law:
F13 =k|q1||q3|
r2
13
where kis Coulomb’s constant, r13 is the distance between charges q1and q3,
and |q1|= 3 ×10−6C, |q3|= 4 ×10−6C.
Given that the side length of the equilateral triangle is a= 2 m, the distance
r13 between charges q1and q3is:
r13 =a
Plugging the given values into Coulomb’s Law, we get:
F13 =9×109×3×10−6×4×10−6
(2)2
10
F13 =108
4×10−9×103
F13 = 27 ×10−3N
So, the electric force between charges q1and q3is 27 ×10−3N.
Using the same method, we can calculate the electric force between charges
q2and q3as:
F23 =9×109×5×10−6×4×10−6
(2)2
F23 =180
4×10−9×103
F23 = 45 ×10−3N
So, the electric force between charges q2and q3is 45 ×10−3N.
Step 2: Calculate the net electric force acting on charge q3.
To find the net force, we need to consider the direction of the forces. Since
the charges q1and q3have opposite signs, their forces will be in opposite direc-
tions. However, the forces between q2and q3will both be pointing towards q3.
Calculating the net force:
Fnet =F23 −F13
Fnet = 45 ×10−3−27 ×10−3
Fnet = 18 ×10−3
Question 13
Question
Three point charges are placed at the vertices of an equilateral triangle of side
2.0 cm. Two of the charges are +3.0 nC, and the third charge is +5.0 nC.
a) What is the net electric force exerted on the charge of +3.0 nC?
b) What is the net electric force exerted on the charge of +5.0 nC?
Solution
a) To find the net electric force on the +3.0 nC charge, we need to consider the
forces exerted by the other two charges.
Let’s label the charges at the vertices of the triangle: A (+3.0 nC), B (+3.0
nC), and C (+5.0 nC).
Step 1: Calculate the distance between charges A and B. Since they are
at the vertices of an equilateral triangle of side 2.0 cm, the distance between A
and B is 2.0cm.
11
Step 2: Calculate the magnitude of the electric force between charges A
and B using Coulomb’s law:
FAB =k·q1·q2
r2
where k= 8.988×109N m2/C2is the electrostatic constant, q1= +3.0×10−9C
is the charge of A, q2= +3.0×10−9C is the charge of B, and r= 0.02 m is the
distance between A and B.
FAB = (8.988 ×109)·(3.0×10−9)2
(0.02)2
FAB ≈1.343 ×10−3N
This force is repulsive since the charges have the same sign.
Step 3: Find the net electric force on charge A. The force
FAB acts along
the line joining charges A and B, and it has the direction away from charge B.
Since A and C are equidistant from A, the force
FAC also acts along the line
joining A and C, but in the opposite direction to that of
FAB .
The net force on charge A will be the vector sum of
FAB and
FAC .
Fnet =
FAB +
FAC
Since the forces are along the same line, we can add them as scalars.
Fnet = 1.343 ×10−3N+ 1.343 ×10−3N
Fnet ≈2.686 ×10−3N
Therefore, the net electric force on the +3.0 nC charge is approximately
2.686 ×10−3N away from the charge at C.
b) The net electric force on the +5.0 nC charge can be calculated in a similar
way considering the forces from charges A and B.
Question 14
Question
Two point charges, q1=−4µC and q2= 2 µC, are located 5 cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
To find the electric force between the charges, we can use Coulomb’s law, which
states that the magnitude of the electric force between two point charges is given
by:
F=k·|q1·q2|
r2
12
where kis the Coulomb’s constant (8.9875 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 1: Convert the distance between charges from cm to meters:
r= 5 cm = 0.05 m
Step 2: Substitute the given values into Coulomb’s law to find the magni-
tude of the electric force:
F= (8.9875 ×109)·| − 4×10−6·2×10−6|
(0.05)2
Step 3: Calculate the magnitude of the electric force:
F= (8.9875 ×109)·8×10−12
0.0025
F= (8.9875 ×109)·3.2×10−9
F= 2.876 ×10−7N
Therefore, the magnitude of the electric force between the charges is 2.876 ×
10−7N.
Question 15
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length a. Charge q1is at the top vertex, charge q2is at the bottom left
vertex, and charge q3is at the bottom right vertex. Calculate the magnitude of
the net electric force on charge q1.
Solution
Step 1: Calculate the electric force between q1and q2. The distance between q1
and q2is a
√3(half of a side length of an equilateral triangle). The magnitude of
the electric force between two charges q1and q2is given by Coulomb’s law:
F12 =k|q1q2|
r2
12
where - kis Coulomb’s constant, k= 8.99 ×109N m2/C2, - r12 =a
√3is the dis-
tance between q1and q2, - q1=q2=q, as the charges are the same magnitude.
Thus,
F12 =(8.99 ×109N m2/C2)(q2)
a
√32
13
Step 2: Calculate the electric force between q1and q3. The distance between
q1and q3is 2a
√3(the full diagonal length of the equilateral triangle). Similarly,
applying Coulomb’s law:
F13 =(8.99 ×109N m2/C2)(q2)
2a
√32
Step 3: Find the net electric force on q1. The net electric force on charge
q1is in the upward direction due to the forces from q2and q3. The vertical
components of F12 and F13 add up to give the net force on q1. To find the net
force, subtract the force due to q3from the force due to q2.
Fnet = 2F12 sin(30◦)−F13
Solving this expression using the values from Steps 1 and 2 will give the
magnitude of the net electric force on charge q1.
Question 16
Question
Three point charges are placed at the corners of an equilateral triangle with side
length d. The charges are +q,+2q, and −3q. Calculate the magnitude of the
net electrostatic force on the +2qcharge due to the other two charges.
Solution
To find the net electrostatic force on the +2qcharge, we need to calculate
the individual forces on it due to the other two charges and then add them
vectorially.
Step 1: Calculate the force on the +2qcharge due to the +qcharge
The magnitude of the force between two charges q1and q2separated by a
distance ris given by Coulomb’s law:
F=k|q1||q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2). The distance between
the +2qand +qcharges in the triangle can be calculated using the height of
the equilateral triangle. The height of an equilateral triangle with side length d
is √3
2d. So, the distance between the +2qand +qcharges is d. Therefore, the
magnitude of the force on the +2qcharge due to the +qcharge is:
F1=k|2q||q|
d2
Step 2: Calculate the force on the +2qcharge due to the −3qcharge
Using the same reasoning as in Step 1, the distance between the +2qand −3q
14
charges is also d. Therefore, the magnitude of the force on the +2qcharge due
to the −3qcharge is:
F2=k|2q||3q|
d2
Step 3: Determine the net force on the +2qcharge The net force on
the +2qcharge is the vector sum of F1and F2. The forces F1and F2act in
opposite directions along the same line passing through the +2qcharge, so we
just need to subtract their magnitudes:
Fnet =F1−F2=k|2q||q|
d2−k|2q||3q|
d2
Solving this expression will give us the magnitude of the net electrostatic
force on the +2qcharge.
Question 17
Question
Three point charges are arranged in a line. Charge q1=−4.0µC is located at
the origin, charge q2= 6.0µC is located at x= 3.0m, and charge q3=−2.0µC
is located at x= 5.0m. Calculate the net electrostatic force on charge q2due
to charges q1and q3.
Solution
To find the net electrostatic force on charge q2, we first need to calculate the
forces acting on charge q2due to charges q1and q3separately, then sum these
forces vectorially.
Step 1: Calculate force on charge q2due to charge q1
The magnitude of the force on charge q2due to charge q1is given by
Coulomb’s law:
F1 on 2 =k|q1q2|
r2
12
where - kis the electrostatic constant, 8.99 ×109N m2/C2, - q1=−4.0µC=
−4.0×10−6C, - q2= 6.0µC= 6.0×10−6C, - r12 = 3.0m.
Plugging in these values, we get:
F1 on 2 = (8.99 ×109)| − 4.0×10−6×6.0×10−6|
(3.0)2
F1 on 2 = (8.99 ×109)24 ×10−12
9= 23.97 N (repulsive)
So, the force on charge q2due to charge q1is 23.97 N in the positive direction.
Step 2: Calculate force on charge q2due to charge q3
15
Similarly, the magnitude of the force on charge q2due to charge q3is given
by:
F3 on 2 =k|q3q2|
r2
32
where - q3=−2.0µC=−2.0×10−6C, - r32 = 2.0m.
Plugging in these values, we get:
F3 on 2 = (8.99 ×109)| − 2.0×10−6×6.0×10−6|
(2.0)2
F3 on 2 = (8.99 ×109)12 ×10−12
4= 26.97 N (attractive)
Therefore, the force on charge q2due to charge q3is 26.97 N in the negative
direction.
Step 3: Calculate the net force on charge q2
The net force exerted on charge q2is the vector sum of the forces due to
charges q1and q3:
Net force on q2=F1 on 2 +F3 on 2 = 23.97 N+ (−26.97 N)
Question 18
Question
Two point charges, q1= +4.0µC and q2=−2.0µC, are initially separated by
a distance of 10.0cm. If the charges are brought 20.0cm closer together, what
is the change in the magnitude of the electric force between them?
Solution
Step 1: Calculate the initial force between the charges using Coulomb’s law:
The electric force between two point charges can be calculated using Coulomb’s
law:
F=k|q1·q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant.
Substitute the given values:
F1= (8.99 ×109)|4.0×10−6·(−2.0×10−6)|
(0.10)2
Calculating F1gives:
F1≈5.758 N
Step 2: Calculate the final force when the charges are 20.0cm closer together:
After the charges are brought 20.0cm closer together, the new separation dis-
tance is r′= 10.0cm −20.0cm =−10.0cm.
16
Calculate the final force F2using the new separation distance r′:
F2= (8.99 ×109)|4.0×10−6·(−2.0×10−6)|
(0.10 −0.20)2
Calculating F2gives:
F2≈23.032 N
Step 3: Calculate the change in magnitude of the electric force: The change
in the magnitude of the electric force is given by:
∆F=|F2−F1|
Substitute the calculated values to find ∆F:
∆F=|23.032 −5.758|
∆F≈17.274 N
Therefore, the change in the magnitude of the electric force between the
charges is approximately 17.274 N.
Question 19
Question
Three point charges are located on the x-axis: a charge of +3 nC at x = 0, a
charge of -2 nC at x = 60 cm, and a charge of +5 nC at x = 100 cm. Calculate
the net electric force on the +5 nC charge.
Solution
Step 1: Convert all distances to meters. Given: q1= +3 nC = 3 ×10−9C at
x= 0 m, q2=−2nC =−2×10−9C at x= 60 cm = 0.6m, q3= +5 nC =
5×10−9C at x= 100 cm = 1.0m.
Step 2: Calculate the net electric force on the +5 nC charge. The net force
on the +5 nC charge is the sum of the forces due to the other charges. The
force between two charges is given by Coulomb’s Law:
F=k|q1q2|
r2,
where k= 8.99 ×109N m2/C2is the electrostatic constant, q1and q2are the
magnitudes of the two charges, and ris the separation between the charges.
The force between q1and q3is:
F13 =k|q1q3|
r2
13
.
17
The force between q2and q3is:
F23 =k|q2q3|
r2
23
.
The net force on the +5 nC charge due to q1and q2is:
Fnet =F13 +F23.
Substitute the given values and solve for the net force.
Question 20
Question
Two point charges, +3.0µC and −6.0µC, are placed 10.0cm apart in air. Calcu-
late the magnitude and direction of the electric force experienced by the positive
charge due to the negative charge.
Solution
Step 1: Convert the charges to coulombs. Given that 1µC= 10−6C, the first
charge +3.0µC becomes +3.0×10−6C and the second charge −6.0µC becomes
−6.0×10−6C.
Step 2: Calculate the distance between the charges in meters. The distance
between the charges is 10.0cm = 0.10 m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
The magnitude of the electric force Fbetween two point charges is given by
Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Substitute the given values into the formula:
F=(8.99 ×109)× |3.0×10−6×6.0×10−6|
(0.10)2
Step 4: Calculate the magnitude of the electric force.
F=(8.99 ×109)×18 ×10−12
0.01
Step 5: Simplify the expression.
F= 161.82 N
18
Step 6: Determine the direction of the electric force. The electric force is
attractive since the charges are of opposite sign. Therefore, the electric force
experienced by the positive charge is directed towards the negative charge.
Therefore, the magnitude of the electric force experienced by the positive
charge due to the negative charge is 161.82 N and it is directed towards the
negative charge.
Question 21
Question
Two point charges, q1=−6µC and q2= 4 µC, are placed 10 cm apart in air.
Calculate the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−6µC =−6×10−6C
q2= 4 µC = 4 ×10−6C
Step 2: Calculate the electric force. The electric force between two point
charges can be calculated using Coulomb’s Law:
F=k· |q1·q2|
r2
where Fis the magnitude of the electric force, kis the Coulomb constant (8.99×
109Nm2/C2), q1and q2are the magnitudes of the charges, and ris the distance
between the charges.
Plugging in the values:
F=(8.99 ×109Nm2/C2)· |(−6×10−6C)·(4 ×10−6C)|
(0.1m)2
Step 3: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(6 ×10−6)·(4 ×10−6)
0.01
F=215.76
0.01
F= 21576 N
Therefore, the magnitude of the electric force between the charges is 21576
N.
Question 22
Question
Two point charges of +4.0 µC and -6.0 µC are placed 0.10 m apart. Calculate
the magnitude of the electric force between them.
19
Solution
Step 1: Convert the charges to SI units. Given: q1= +4.0µC= 4.0×10−6C
q2=−6.0µC=−6.0×10−6C
Step 2: Calculate the distance between the charges. Given: r= 0.10 m
Step 3: Calculate the magnitude of the electric force between the charges
using Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the Coulomb constant with a value of 8.99 ×109N·m2/C2.
Substitute the given values into the formula:
F=(8.99 ×109)·(4.0×10−6)·(6.0×10−6)
(0.10)2
Step 4: Calculate the magnitude of the electric force.
F=35.96 ×103
0.01 = 3.596 ×106N
Therefore, the magnitude of the electric force between the charges is 3.596 ×
106N.
Question 23
Question
Two point charges with magnitudes q1= 3.0µC and q2=−5.0µC are placed
20.0 cm apart in a vacuum. Calculate the magnitude and direction of the electric
force that each charge exerts on the other.
Solution
To calculate the magnitude and direction of the electric force between the two
charges, we will use Coulomb’s law.
Step 1: Calculate the electric force F12 that charge q2exerts on charge q1.
Coulomb’s law is given by:
F12 =k|q1q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the two charges.
Plugging in the values, we get:
F12 = (8.99 ×109)(3.0×10−6)(5.0×10−6)
(0.20)2
F12 = 2.248 N
20
The direction of the force is from charge q2to charge q1.
Step 2: Calculate the electric force F21 that charge q1exerts on charge q2.
Since the forces act along the line joining the charges, the magnitude of F21 is
the same as that of F12.
Therefore, the magnitude of the electric force that each charge exerts on the
other is 2.248 N, and the direction of the force from charge q1to charge q2.
Question 24
Question
Two point charges, q1=−3.0µC and q2= 4.0µC, are placed 6.0 cm apart.
Calculate the magnitude and direction of the electric force exerted on q1.
Solution
Step 1: Convert the given charges to their equivalent in coulombs. Using the
conversion factor 1µC= 10−6C, we have: q1=−3.0µC=−3.0×10−6C, and
q2= 4.0µC= 4.0×10−6C.
Step 2: Given data: q1=−3.0×10−6Cq2= 4.0×10−6Cr= 6.0cm =
0.06 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k·
q1·q2
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging the values into the formula gives:
F= 8.99 ×109·3.0×10−6·4.0×10−6
(0.06)2
F= 8.99 ×109·12 ×10−12
0.0036
F= 8.99 ×109·0.000003333
F= 29.97 N
Step 4: Determine the direction of the electric force. Since q1is negative and
q2is positive, the electric force exerted on q1is attractive, pulling q1towards
q2. Therefore, the direction of the electric force is towards q2.
Question 25
Question
Two point charges, q1= 4.0µC and q2=−6.0µC, are placed 12.0cm apart in
air. Calculate the magnitude of the electric force between these charges.
21
Solution
Step 1: Let’s first convert the charges to Coulombs by using the relationship
1µC= 10−6C. Thus, q1= 4.0×10−6C and q2=−6.0×10−6C.
Step 2: Next, let’s calculate the distance between the charges in meters.
This can be done by converting 12.0cm to meters: 12.0cm = 0.12 m.
Step 3: Now, we can use Coulomb’s law to find the magnitude of the electric
force between the two charges. Coulomb’s law states that the magnitude of the
electric force between two point charges is given by:
F=k
q1q2
r2
,
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 4: Substituting the given values into Coulomb’s law, we have:
F= (8.99 ×109N m2/C2)|4.0×10−6C· −6.0×10−6C|
(0.12 m)2
Step 5: Calculating the magnitude of the electric force:
F= (8.99 ×109)(2.4×10−11)/(0.0144) N
F≈1.495 ×10−3N
Therefore, the magnitude of the electric force between the charges q1=
4.0µC and q2=−6.0µC, separated by a distance of 12.0cm, is approximately
1.495 ×10−3N.
22
q1
q2q3
23
