PHYS 232 - UNIVERSITY PHYSICS
II - Calculation of electric forces
between point charges
Question Bank - Set 9
Liberty University
Question 1
Question
Two point charges, q1=−2nC and q2= 4 nC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to SI units.
Given: q1=−2nC =−2×10−9C
q2= 4 nC = 4 ×10−9C
Step 2: Calculate the distance between the charges in meters.
Given: Distance d= 10 cm = 0.1m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99×109N m2/C2), q1and q2are the charges,
and ris the distance between the charges.
Step 4: Substituting the given values into the formula:
F= 8.99 ×109N m2/C2·| − 2×10−9C·4×10−9C|
(0.1m)2
Step 5: Calculate the magnitude of the electric force:
F= 8.99 ×109N m2/C2·8×10−18 C2
0.01 m2
F= 8.99 ×109N m2/C2·8×10−16 C2
F= 7.192 ×10−6N
Therefore, the magnitude of the electric force between the charges is 7.192 ×
10−6N.
Question 2
Question
Two point charges, q1=−4µC and q2= 8 µC, are separated by a distance of
0.5m. Calculate the magnitude of the electric force that q1exerts on q2.
Solution
Step 1: Determine the electric force between the two charges using Coulomb’s
Law:
F=k·|q1·q2|
r2,
where k= 8.99 ×109N m2/C2is the Coulomb’s constant, q1and q2are the
charges, and ris the distance between the charges.
Step 2: Calculate the force by plugging in the given values:
F= (8.99 ×109)·|(−4×10−6)·(8 ×10−6)|
(0.5)2.
Step 3: Simplify the expression:
F= (8.99 ×109)·32 ×10−12
0.25 .
F= (8.99 ×109)·128 ×10−12.
Step 4: Perform the multiplication:
F= 1150.72 ×10−3.
Step 5: Express the result in Newtons by converting the magnitude to sci-
entific notation:
F= 1.15072 N.
Therefore, the magnitude of the electric force that q1exerts on q2is 1.15072 N.
Question 3
Question
Two point charges are fixed in place, with charge q1=−2.0µC at coordinates
(0,0) and charge q2= 4.0µC at coordinates (2.0m,0). Calculate the magnitude
and direction of the electric force that q1exerts on q2.
2
Solution
Step 1: Calculate the distance between the two charges.
Let rbe the distance between charge q1and charge q2. Using the distance
formula in 2D space:
r=√(2.0m)2+ (0 −0)2=√4.0m2= 2.0m
Step 2: Calculate the magnitude of the electric force.
The electric force between two point charges is given by Coulomb’s law:
F=k|q1q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the equation:
F= (8.99 ×109)|(−2.0×10−6)(4.0×10−6)|
(2.0)2
F= (8.99 ×109)8.0×10−12
4.0= (8.99 ×109)×2.0×10−12 = 1.80 ×10−2N
The magnitude of the electric force is 1.80 ×10−2N.
Step 3: Determine the direction of the electric force.
The direction of the electric force is along the line joining the two charges. Since
charge q1has a negative charge, the force on q2will be attractive towards q1in
the negative xdirection.
Therefore, the electric force that q1exerts on q2is 1.80 ×10−2N in the
negative xdirection.
Question 4
Question
Two point charges, q1=−5µC and q2= 8 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges in
Newtons.
Solution
Step 1: Convert the charges to coulombs (µC to C). Given charges: q1=−5µC
and q2= 8 µC
Converting to coulombs: q1=−5×10−6C and q2= 8 ×10−6C
Step 2: Calculate the electric force using Coulomb’s Law: The magnitude of
the electric force between two point charges is given by Coulomb’s Law:
F=k· |q1·q2|
r2
3
where: F= electric force between the charges, k= Coulomb’s constant (8.99 ×
109N m2/C2), q1and q2= magnitudes of the point charges, r= distance be-
tween the charges.
In this case: k= 8.99 ×109N m2/C2,q1=−5×10−6C, q2= 8 ×10−6C,
r= 0.1m (since 10 cm is 0.1 m).
Step 3: Plug the values into Coulomb’s Law and calculate the electric force:
F=8.99 ×109×|−5×10−6×8×10−6|
(0.1)2
F=8.99 ×109×5×8×10−12
0.01
F=8.99 ×5×8
10 ×10−3
F= 8.99 ×5×0.8×10−3
F= 35.96 ×10−3
F= 0.03596 N
Therefore, the magnitude of the electric force between the charges is 0.03596
N.
Question 5
Question
Three point charges are placed on the vertices of an equilateral triangle. The
charges are +2µC,−4µC, and +3µC. Calculate the net electric force on the
charge at the vertex with the +2µC charge.
Solution
Step 1: We first need to calculate the distance between the charges. Since
the charges are placed on the vertices of an equilateral triangle, the distances
between them are equal. Let’s denote this distance as d.
Step 2: The electric force between two charges q1and q2separated by dis-
tance ris given by Coulomb’s law:
F=k·|q1·q2|
r2,
where k≈8.99 ×109N m2/C2is the Coulomb’s constant.
4
Step 3: Let’s first calculate the net electric force on the +2µC charge due
to the −4µC and +3µC charges. The force on +2µC due to −4µC can be
calculated using Coulomb’s law, and the direction of this force will be attractive.
Step 4: The force on +2µC due to +3µC can be calculated using Coulomb’s
law as well, and the direction of this force will be repulsive.
Step 5: Once we have the magnitudes and directions of the forces due to
each charge, we can find the net force on the +2µC charge by vector addition
of individual forces.
Step 6: Calculate the net electric force on the charge at the vertex with the
+2µC charge.
Question 7
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below:
+Q
+Q
−2Q
The charges are +Qat each of the two upper vertices and −2Qat the lower
vertex. Find the net electrostatic force on the charge at the lower vertex due to
the other two charges.
Solution
Step 1: Calculate the magnitude of the force between the charge at the lower
vertex (−2Q) and the charge at the upper left vertex (+Q).
F1=k· | + 2Q||Q|
r2
where kis the Coulomb’s constant (8.99×109N·m2/C2), |+2Q|= 2Q,|Q|=Q,
and ris the distance between the charges. Since the charges are at the vertices
of an equilateral triangle, the distance between the charges is equal to the side
length of the triangle.
5
Step 2: Calculate the magnitude of the force between the charge at the lower
vertex and the charge at the upper right vertex (+Q) using the same formula
as in Step 1.
F2=k· | + 2Q||Q|
r2
Step 3: Determine the direction of the forces. The force between the charge
at the lower vertex and the charge at the upper left vertex is repulsive, and the
force between the charge at the lower vertex and the charge at the upper right
vertex is also repulsive. Thus, both forces act away from the charge at the lower
vertex.
Step 4: Calculate the net force on the charge at the lower vertex by summing
the individual forces vectorially. The net force will be directed away from the
charge at the lower vertex and can be found using the Pythagorean theorem:
Fnet =√F2
1+F2
2
Question 8
Question
Three point charges are arranged along a straight line as shown below:
+3µC−2µC+5µC
The charges are spaced 10 cm apart with the +3 µC charge at the origin,
the -2 µC charge at 10 cm, and the +5 µC charge at 20 cm. Calculate the net
electric force on the -2 µC charge due to the other two charges.
Solution
Let’s denote the charges as follows: - Charge at 0cm: q1= +3µC - Charge at
10 cm: q2=−2µC - Charge at 20 cm: q3= +5µC
The net electric force on the -2 µC charge (q2) is the sum of the forces due to
the other two charges. We will first calculate the force between q1and q2, and
then the force between q3and q2, and finally sum them to find the net force.
Step 1: Calculate the force between q1and q2. The formula for the magni-
tude of the force between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
Where: - kis the Coulomb’s constant (8.99 ×109N m2/C2), - q1and q2are
the magnitudes of the charges, and - ris the distance between the charges.
Given q1= +3µC, q2=−2µC, and r= 10 cm = 0.1m, we have:
6
F12 =8.99 ×109· |3×10−6·(−2×10−6)|
(0.1)2
F12 =8.99 ×109·6×10−12
0.01
F12 =53.94 ×10−3
0.01
F12 = 5.394 N
So, the force between q1and q2is 5.394 N (pointing to the left).
Step 2: Calculate the force between q2and q3. This calculation follows the
same process as in Step 1.
Step 3: Sum the forces to find the net force on q2.
Thus, the net force on the -2 µC charge due to the other two charges is 5.394
N to the left.
Question 9
Question
Two point charges, +2µC and −3µC, are placed 10 cm apart in air. Calculate
the magnitude of the electric force between them.
Solution
Step 1: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k|q1·q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
charges (+2µC and −3µC), and ris the distance between the charges (0.1 m).
Step 2: Substitute the given values into the equation:
F= (8.99 ×109)|2×10−6· −3×10−6|
(0.1)2
Step 3: Calculate the magnitude of the electric force:
F= (8.99 ×109)6×10−12
0.01
F= 8.99 ×109×6×10−10
F= 53.94 ×10−1
F= 5.394N
Therefore, the magnitude of the electric force between the charges +2µC
and −3µC is 5.394N.
7
Question 10
Question
Three point charges are arranged in the following configuration: a charge of
+4.0µC at the origin, a charge of −2.0µC at coordinates (0,0,3m), and a
charge of +3.0µC at coordinates (4 m,0,0). Calculate the net electric force on
the charge at the origin due to the other two charges.
Solution
The electric force
Felec between two charges q1and q2separated by a distance
ris given by Coulomb’s law:
Felec =kq1q2
r2ˆr
where k≈8.99×109N·m2/C2is the Coulomb constant and ˆris the unit vector
pointing from q2to q1.
Step 1: Find the force on the charge at the origin due to the charge at
coordinates (0,0,3m). The distance between these two charges is r1= 3 m.
The electric force can be calculated as:
|
F1|=k|(+4.0µC)(−2.0µC)|
(3 m)2=8.99 ×109×4.0×2.0×10−12
9= 8.0×10−3N
Step 2: Find the force on the charge at the origin due to the charge at
coordinates (4 m,0,0). The distance between these two charges is r2= 4 m.
The electric force can be calculated as:
|
F2|=k|(+4.0µC)(+3.0µC)|
(4 m)2=8.99 ×109×4.0×3.0×10−12
16 = 6.7×10−3N
Step 3: Find the net force on the charge at the origin. The net force is the
vector sum of the individual forces. Since one force acts in the z-direction and
the other in the x-direction, the net force will have both xand zcomponents.
Fnet =
F1+
F2Calculating the magnitudes of the components:
Fnet,x =F2= 6.7×10−3N
Fnet,z =F1= 8.0×10−3N
Therefore, the net electric force on the charge at the origin is 6.7×10−3N
in the x-direction and 8.0×10−3N in the z-direction.
Question 11
Question
Three point charges are arranged in an equilateral triangle with sides of length d.
The charges are +qat point A,−2qat point B, and +3qat point C. Calculate
the net electric force on the charge at point Adue to the other two charges.
8
Solution
To find the net electric force on the charge at point A, we need to calculate the
electric forces between Aand B, and between Aand C, and then add these
forces as vectors to get the net force.
Step 1: Find the force between Aand B.
The electric force between two point charges q1and q2separated by a dis-
tance ris given by Coulomb’s Law:
F=kq1q2
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
The distance between Aand Bis also dbecause they are at opposite ends of
the side of length dof the equilateral triangle. Therefore, the net electric force
between Aand Bis:
FAB =k|+q|·|−2q|
d2=k2q2
d2
Step 2: Find the force between Aand C.
The distance between Aand Cis das well due to the equilateral triangle
arrangement. Therefore, the net electric force between Aand Cis:
FAC =k|+q|·|3q|
d2=k3q2
d2
Step 3: Find the net force on A.
The net force on the charge at point Ais the vector sum of FAB and FAC .
Since these forces are in opposite directions and have the same magnitude, the
net force is simply given by their difference:
Net Force on A =|FAC |−|FAB |=k3q2
d2−k2q2
d2=kq2
d2
Therefore, the net electric force on the charge at point Adue to the other
two charges is kq2
d2.
Question 12
Question
Two point charges, q1=−3.0µC and q2= 5.0µC, are placed 20 cm apart in a
vacuum. Calculate the magnitude of the electric force between them.
9
Solution
Step 1: Convert the given charges to coulombs.
q1=−3.0µC =−3.0×10−6C
q2= 5.0µC = 5.0×10−6C
Step 2: Write down the formula for the electric force between two point
charges. The magnitude of the electric force between two point charges is given
by Coulomb’s law:
F=k· |q1·q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant, q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 3: Substitute the given values into the formula.
F=(8.99 ×109N·m2/C2)· |(−3.0×10−6C)·(5.0×10−6C)|
(0.20 m)2
Step 4: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(3.0×5.0) ×10−12
0.04
=1.35 ×10−11 ×8.99 ×109
0.04
=1.21365 ×10−1
0.04
= 3.034 ×10−10 N
Step 5: Therefore, the magnitude of the electric force between the two
charges is 3.034 ×10−10 N.
Question 13
Question
Three point charges are arranged along the x-axis as follows: +3.0 nC at x = 0
m, -6.0 nC at x = 2.0 m, and +4.0 nC at x = 3.0 m. Calculate the net electric
force on the +4.0 nC charge due to the other two charges.
10
Solution
Step 1: Calculate the electric force on the +4.0 nC charge due to the +3.0 nC
charge. The electric force between two charges is given by Coulomb’s Law:
F=kq1q2
r2
where: F= electric force k= Coulomb’s constant = 8.9875×109Nm²/C² q1, q2
= magnitudes of the charges r= distance between the charges
Plugging in the values: k= 8.9875 ×109Nm²/C² q1= +3.0×10−9C
q2= +4.0×10−9Cr= 3.0m
F1= 8.9875 ×109×(3.0×10−9)(4.0×10−9)
3.02
F1= 8.9875 ×109×12 ×10−18
9
F1= 8.9875 ×109×1.3333 ×10−18
F1= 11.98 ×10−9
F1= 1.198 ×10−8N
Step 2: Calculate the electric force on the +4.0 nC charge due to the -6.0
nC charge. Following the same steps as above with q1=−6.0×10−9C:
F2= 8.9875 ×109×(6.0×10−9)(4.0×10−9)
1
F2= 8.9875 ×109×24.0×10−18
F2= 215.7×10−9
F2= 2.157 ×10−7N
Step 3: Calculate the net electric force on the +4.0 nC charge. Since the
forces are in opposite directions (one repelling and the other attracting), we
need to find the difference between them:
Net force =F1−F2= 1.198 ×10−8−2.157 ×10−7
Net force =−1.037 ×10−7N
Therefore, the net electric force on the +4.0 nC charge due to the other two
charges is −1.037 ×10−7N, directed towards the -6.0 nC charge.
Question 14
Question
Two point charges are placed at the two opposite corners of a square with sides
of length a. The charges have magnitudes qand 2q. Calculate the magnitude of
the electric force between the charges along one of the diagonals of the square.
11
Solution
Let’s denote the magnitude of the charge qas q1=qand the magnitude of the
charge 2qas q2= 2q.
Step 1: Calculate the electric force between q1and q2along one of the
diagonals of the square. Let’s choose the diagonal from q1to q2.
Using the formula for the magnitude of the electric force between two point
charges, we have:
F=k|q1q2|
r2
where - kis the Coulomb constant, - q1and q2are the magnitudes of the charges,
and - ris the distance between the charges.
First, let’s find the distance ralong the diagonal of the square: Since the
charges are at opposite corners of a square with sides of length a, the diagonal
length can be calculated using the Pythagorean theorem:
r=√a2+a2=√2a
Step 2: Substitute the values of q1,q2, and rinto the formula for the electric
force:
F=k|q1q2|
r2=k|q(2q)|
(√2a)2=k2q2
2a2=kq2
a2
Therefore, the magnitude of the electric force between the charges qand 2q
along one of the diagonals of the square is kq2
a2.
Question 15
Question
Two point charges, q1=−3µC and q2= 5 µC, are placed 10 cm apart along
the x-axis. Calculate the magnitude and direction of the electric force that q2
exerts on q1.
Solution
Step 1: First, calculate the electric force between the two charges using Coulomb’s
Law:
F=k·|q1·q2|
r2
where Fis the magnitude of the electric force, kis the electrostatic constant
(8.99 ×109N m2/C2), q1and q2are the charges, and ris the distance between
the charges.
Step 2: Substitute the given values into the formula:
F= (8.99 ×109)·|(−3×10−6)(5 ×10−6)|
(0.1)2
12
Step 3: Calculate the electric force:
F= (8.99 ×109)·15 ×10−12
0.01
F= (8.99 ×109)·1.5×10−9
F= 13.485 ×100N
F= 13.485 N
Step 4: Now, determine the direction of the force. The force will be attractive
since the charges are of opposite signs. Therefore, the force will act along the
line joining the charges, i.e., along the negative x-axis.
Therefore, the magnitude of the electric force that q2exerts on q1is 13.485
N along the negative x-axis.
Question 16
Question
Two point charges, q1=−2.5µC and q2= 3.0µC, are placed 10 cm apart in
air. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to standard units. Step 2: Calculate the distance
between the charges. Step 3: Use Coulomb’s Law to calculate the electric force.
Step 4: Provide the final answer with appropriate units.
Step 1: Convert the charges to standard units. q1=−2.5µC =−2.5×
10−6C
q2= 3.0µC = 3.0×10−6C
Step 2: Calculate the distance between the charges. Given that the charges
are placed 10 cm apart, the distance rbetween them is r= 10 cm = 0.10 m.
Step 3: Use Coulomb’s Law to calculate the electric force. The magnitude
of the electric force between two charges is given by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109Nm2/C2).
Plugging in the values:
F= 8.99 ×109Nm2/C2·| − 2.5×10−6C·3.0×10−6C|
(0.10 m)2
Calculating the force will give us the magnitude of the electric force between
the charges.
Step 4: Provide the final answer with appropriate units.
F= 53.7×10−6N= 53.7µN
Therefore, the magnitude of the electric force between the charges is 53.7µN .
13
Question 17
Question
Two point charges, q1= 4.0×10−6C and q2=−6.0×10−6C, are placed 0.30
m apart. Calculate the magnitude and direction of the electric force that q2
exerts on q1.
Solution
Step 1: Calculate the electric force magnitude using Coulomb’s Law.
F=k|q1||q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Step 2: Substitute the given values into the formula.
F=(8.99 ×109N m2/C2)(4.0×10−6C)(6.0×10−6C)
(0.30 m)2
Step 3: Calculate the electric force magnitude.
F=(8.99 ×109)(4.0×10−6)(6.0×10−6)
(0.30)2
F=215.76
0.09
F= 2397.33 N
Step 4: Determine the direction of the force. The force is attractive in this
case because the charges q1and q2have opposite signs. Thus, the force points
from q2to q1.
Therefore, the magnitude of the electric force that q2exerts on q1is 2397.33 N
and the direction is from q2to q1.
Question 18
Question
Two point charges, q1= 3.0µC and q2=−2.0µC, are located 10.0 cm apart.
Calculate the magnitude of the electric force between these charges.
14
Solution
Step 1: Convert the charges to Coulombs. Given: q1= 3.0µC,q2=−2.0µC
1 microCoulomb (µC) = 1×10−6Coulombs So, q1= 3.0×10−6C and q2=
−2.0×10−6C.
Step 2: Calculate the magnitude of the electric force using Coulomb’s Law.
The electric force Fbetween two point charges q1and q2separated by a distance
ris given by Coulomb’s Law:
F=k·|q1·q2|
r2
where k= 8.99 ×109Nm2/C2is Coulomb’s constant.
Plugging in the given values: F= (8.99 ×109)·|3.0×10−6·−2.0×10−6|
(0.10)2
Step 3: Calculate the magnitude of the electric force.
F= 8.99 ×109·6.0×10−12
0.01
F= 8.99 ×109·6.0×10−10
F= 53.94 ×10−1
F= 5.39 ×100N
Therefore, the magnitude of the electric force between these charges is 5.39
N.
Question 19
Question
Three point charges are arranged along the x-axis as follows: +3.00 µC at the
origin, −4.00 µC at x= 50.0cm, and +5.00 µC at x= 70.0cm. What is the net
electric force on the −4.00 µC charge?
Solution
1. Calculate the electric force on the −4.00 µC charge due to the +3.00 µC
charge at the origin. The electric force F1between two point charges is given
by Coulomb’s Law:
F1=kq1q2
r2
where - kis the Coulomb constant (8.99 ×109N·m2/C2), - q1and q2are
the magnitudes of the point charges (3.00 µC and −4.00 µC, respectively), - r
is the distance between the charges (origin to x= 50.0cm = 0.50 m).
Plugging in the values, we get:
F1= (8.99 ×109)(3.00 ×10−6)·(4.00 ×10−6)
(0.50)2
15
Calculating the result gives F1=−4.79 N, directed towards the origin.
2. Calculate the electric force on the −4.00 µC charge due to the +5.00 µC
charge at x= 70.0cm. Using Coulomb’s Law again, this time with q2=
+5.00 µC and r= 0.20 m, we have:
F2= (8.99 ×109)(5.00 ×10−6)·(4.00 ×10−6)
(0.20)2
Calculating the result gives F2= 44.94 N, directed towards the positive
charge.
3. The net force is the vector sum of F1and F2.
Fnet =F2−F1= 44.94 N−4.79 N= 40.15 N
Therefore, the net electric force on the −4.00 µC charge is 40.15 N directed
towards the +5.00 µC charge at x= 70.0cm.
Question 20
Question
Two point charges, q1=−2.5µC and q2= 4.0µC, are located at points
A(−2.0m,0) and B(3.0m,0) respectively on the x-axis. Calculate the electric
force exerted on q2due to q1.
Solution
Step 1: First, we need to calculate the distance between the two charges. Given:
xA=−2.0m
xB= 3.0m
Let dbe the distance between the charges. d=xB−xA= 3.0m−(−2.0m) =
5.0m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law:
F=k|q1·q2|
d2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Step 3: Substituting the given values into the formula:
F= 8.99 ×109N m2/C2·| − 2.5×10−6C·4.0×10−6C|
(5.0m)2
F= 8.99 ×109·10 ×10−12
25
F= 8.99 ×109·4×10−12
F= 35.96 ×10−3
16
F= 3.596 mN
Therefore, the electric force exerted on q2due to q1is 3.596 mN directed
towards q1.
Question 21
Question
Three point charges are fixed on the x-axis. The charges are located at x=−2
m, x= 0 m, and x= 3 m, with magnitudes q1= 4 µC, q2=−6µC, and q3= 2
µC, respectively. Calculate the net electrostatic force acting on q1due to the
other two charges.
Solution
Step 1: Calculate the force on q1due to q2. The magnitude of the force F12 can
be calculated using Coulomb’s Law:
F12 =k|q1||q2|
r2
12
where kis the Coulomb’s constant, r12 is the distance between q1and q2, and
|q1|and |q2|are the magnitudes of the charges.
Substitute the given values into the equation:
F12 =(9 ×109N·m2/C2)(4 ×10−6C)(6 ×10−6C)
(−2m)2
F12 =216 ×10−12
4= 54 ×10−12 = 5.4×10−11 N
Step 2: Calculate the force on q1due to q3. Similarly, calculate the force
F13 on q1due to q3using Coulomb’s Law:
F13 =k|q1||q3|
r2
13
where r13 is the distance between q1and q3.
Substitute the given values into the equation:
F13 =(9 ×109N·m2/C2)(4 ×10−6C)(2 ×10−6C)
(3 m−(−2m))2
F13 =72 ×10−12
25 = 2.88 ×10−12 N
Step 3: Calculate the net force on q1. The net force on q1is the vector
sum of F12 and F13. Since F12 and F13 are in opposite directions, we need to
subtract them:
Fnet =F12 −F13 = 5.4×10−11 N−2.88 ×10−12 N
17
Fnet = 5.112 ×10−11 N
Therefore, the net electrostatic force acting on q1due to the other two
charges is 5.112 ×10−11 N.
Question 22
Question
Two point charges, q1=−4µC and q2= 2 µC, are placed 10 cm apart along the
x-axis. Calculate the magnitude and direction of the electric force experienced
by each charge.
Solution
Step 1: Determine the distance between the charges Given that the charges are
10 cm apart along the x-axis, the distance between them is r= 10 cm = 0.10
m.
Step 2: Calculate the magnitude of the electric force experienced by q1The
magnitude of the electric force between two point charges q1and q2is given by
Coulomb’s Law:
F=k|q1q2|
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant.
Substitute q1=−4µC, q2= 2 µC, and r= 0.10 m into the formula:
F=(8.99 ×109N·m2/C2)|(−4×10−6C)(2 ×10−6C)|
(0.10 m)2
Calculating this expression gives the magnitude of the electric force experi-
enced by q1.
Step 3: Determine the direction of the electric force experienced by q1The
direction of the electric force experienced by q1will be attractive towards q2.
Step 4: Calculate the magnitude of the electric force experienced by q2Using
the same formula as in Step 2, but switching the signs of q1and q2, calculate
the electric force experienced by q2.
Step 5: Determine the direction of the electric force experienced by q2The
direction of the electric force experienced by q2will be repulsive away from q1.
Question 23
Question
A charge of −2.0µC is located at the origin, and a charge of 4.0µC is located
at the point (0,3.0) m. Calculate the magnitude and direction of the electric
force on the positive charge due to the negative charge.
18
Solution
To find the electric force, we will use Coulomb’s Law, given by:
Electric force =k·|q1·q2|
r2
where kis Coulomb’s constant (8.99 ×109Nm2/C2), q1and q2are the two
charges, and ris the distance between the charges.
Step 1: Calculate the distance rbetween the charges
The distance rbetween the charges can be found using the distance formula
in two dimensions:
r=√(0 −0)2+ (3.0−0)2
r=√0+9.0
r=√9.0
r= 3.0m
Step 2: Calculate the magnitude of the electric force
Now, we can substitute the given values into Coulomb’s Law:
Electric force = (8.99 ×109)·| − 2.0×10−6×4.0×10−6|
(3.0)2
Electric force = 8.99 ×109·8.0×10−12
9.0
Electric force = 8.99 ×109·8.89 ×10−13
Electric force = 7.99 ×10−3N
Step 3: Determine the direction of the electric force
The electric force will be attractive since the charges are of opposite signs.
Therefore, the direction of the electric force is towards the negative charge (from
the positive charge to the negative charge) along the line connecting the charges.
Question 24
Question
Two point charges, q1= +5 µC and q2=−3µC, are placed 10 cm apart. What
is the magnitude of the electric force between these two charges?
19
Solution
Step 1: Convert the charges to Coulombs. The charges given are q1= +5 µC
and q2=−3µC. To convert microcoulombs to coulombs, we use the following
conversion factor: 1µC= 10−6C. Therefore, q1= +5 ×10−6C= 5 ×10−6C
and q2=−3×10−6C=−3×10−6C.
Step 2: Calculate the distance between the charges in meters. The charges
are placed 10 cm apart, which is 0.10 meters.
Step 3: Calculate the electric force between the charges using Coulomb’s
Law. Coulomb’s Law states that the magnitude of the electric force between
two point charges is given by:
F=k· |q1·q2|
r2
where F= electric force, k= Coulomb’s constant (8.99 ×109N m2/C2), q1, q2
= magnitudes of the two charges, and r= distance between the charges.
Substitute the known values into the formula:
F=(8.99 ×109N m2/C2)· |(5 ×10−6C)·(−3×10−6C)|
(0.10 m)2
Step 4: Calculate the electric force.
F=(8.99 ×109)·(5 ×10−6)·(3 ×10−6)
0.01
F=134850
100
F= 1348.50 N
Therefore, the magnitude of the electric force between the two charges is
1348.50 N.
Question 25
Question
Two point charges, q1=−3.00 µC and q2= 5.00 µC, are placed 10.0 cm apart
in vacuum.
Calculate the magnitude of the electric force between the charges, and de-
termine the direction of the force exerted on each charge.
Solution
Step 1: Convert the charges to coulombs. The charges are given in micro-
coulombs (µC), so we need to convert them to coulombs. q1=−3.00 µC=
−3.00 ×10−6Cq2= 5.00 µC= 5.00 ×10−6C
20
Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force between two point charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant and r= 0.10 m is
the separation distance between the charges. Substituting in the given values:
F=8.99 ×109·3.00 ×10−6·5.00 ×10−6
(0.10)2
Step 3: Calculate the numerical value of the electric force.
F=8.99 ×3×5
103=134.85
103= 0.13485 N
Step 4: Determine the direction of the force. The force between the charges
is attractive since the charges have opposite signs.
Therefore, the magnitude of the electric force between the charges is 0.13485 N,
and the force exerted on q1is attractive towards q2, while the force exerted on
q2is attractive towards q1.
21
F= 8.99 ×109N m2/C2·8×10−16 C2
F= 7.192 ×10−6N
Therefore, the magnitude of the electric force between the charges is 7.192 ×
10−6N.
Question 2
Question
Two point charges, q1=−4µC and q2= 8 µC, are separated by a distance of
0.5m. Calculate the magnitude of the electric force that q1exerts on q2.
Solution
Step 1: Determine the electric force between the two charges using Coulomb’s
Law:
F=k·|q1·q2|
r2,
where k= 8.99 ×109N m2/C2is the Coulomb’s constant, q1and q2are the
charges, and ris the distance between the charges.
Step 2: Calculate the force by plugging in the given values:
F= (8.99 ×109)·|(−4×10−6)·(8 ×10−6)|
(0.5)2.
Step 3: Simplify the expression:
F= (8.99 ×109)·32 ×10−12
0.25 .
F= (8.99 ×109)·128 ×10−12.
Step 4: Perform the multiplication:
F= 1150.72 ×10−3.
Step 5: Express the result in Newtons by converting the magnitude to sci-
entific notation:
F= 1.15072 N.
Therefore, the magnitude of the electric force that q1exerts on q2is 1.15072 N.
Question 3
Question
Two point charges are fixed in place, with charge q1=−2.0µC at coordinates
(0,0) and charge q2= 4.0µC at coordinates (2.0m,0). Calculate the magnitude
and direction of the electric force that q1exerts on q2.
2
Solution
Step 1: Calculate the distance between the two charges.
Let rbe the distance between charge q1and charge q2. Using the distance
formula in 2D space:
r=√(2.0m)2+ (0 −0)2=√4.0m2= 2.0m
Step 2: Calculate the magnitude of the electric force.
The electric force between two point charges is given by Coulomb’s law:
F=k|q1q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the equation:
F= (8.99 ×109)|(−2.0×10−6)(4.0×10−6)|
(2.0)2
F= (8.99 ×109)8.0×10−12
4.0= (8.99 ×109)×2.0×10−12 = 1.80 ×10−2N
The magnitude of the electric force is 1.80 ×10−2N.
Step 3: Determine the direction of the electric force.
The direction of the electric force is along the line joining the two charges. Since
charge q1has a negative charge, the force on q2will be attractive towards q1in
the negative xdirection.
Therefore, the electric force that q1exerts on q2is 1.80 ×10−2N in the
negative xdirection.
Question 4
Question
Two point charges, q1=−5µC and q2= 8 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges in
Newtons.
Solution
Step 1: Convert the charges to coulombs (µC to C). Given charges: q1=−5µC
and q2= 8 µC
Converting to coulombs: q1=−5×10−6C and q2= 8 ×10−6C
Step 2: Calculate the electric force using Coulomb’s Law: The magnitude of
the electric force between two point charges is given by Coulomb’s Law:
F=k· |q1·q2|
r2
3
where: F= electric force between the charges, k= Coulomb’s constant (8.99 ×
109N m2/C2), q1and q2= magnitudes of the point charges, r= distance be-
tween the charges.
In this case: k= 8.99 ×109N m2/C2,q1=−5×10−6C, q2= 8 ×10−6C,
r= 0.1m (since 10 cm is 0.1 m).
Step 3: Plug the values into Coulomb’s Law and calculate the electric force:
F=8.99 ×109×|−5×10−6×8×10−6|
(0.1)2
F=8.99 ×109×5×8×10−12
0.01
F=8.99 ×5×8
10 ×10−3
F= 8.99 ×5×0.8×10−3
F= 35.96 ×10−3
F= 0.03596 N
Therefore, the magnitude of the electric force between the charges is 0.03596
N.
Question 5
Question
Three point charges are placed on the vertices of an equilateral triangle. The
charges are +2µC,−4µC, and +3µC. Calculate the net electric force on the
charge at the vertex with the +2µC charge.
Solution
Step 1: We first need to calculate the distance between the charges. Since
the charges are placed on the vertices of an equilateral triangle, the distances
between them are equal. Let’s denote this distance as d.
Step 2: The electric force between two charges q1and q2separated by dis-
tance ris given by Coulomb’s law:
F=k·|q1·q2|
r2,
where k≈8.99 ×109N m2/C2is the Coulomb’s constant.
4
Step 3: Let’s first calculate the net electric force on the +2µC charge due
to the −4µC and +3µC charges. The force on +2µC due to −4µC can be
calculated using Coulomb’s law, and the direction of this force will be attractive.
Step 4: The force on +2µC due to +3µC can be calculated using Coulomb’s
law as well, and the direction of this force will be repulsive.
Step 5: Once we have the magnitudes and directions of the forces due to
each charge, we can find the net force on the +2µC charge by vector addition
of individual forces.
Step 6: Calculate the net electric force on the charge at the vertex with the
+2µC charge.
Question 7
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below:
+Q
+Q
−2Q
The charges are +Qat each of the two upper vertices and −2Qat the lower
vertex. Find the net electrostatic force on the charge at the lower vertex due to
the other two charges.
Solution
Step 1: Calculate the magnitude of the force between the charge at the lower
vertex (−2Q) and the charge at the upper left vertex (+Q).
F1=k· | + 2Q||Q|
r2
where kis the Coulomb’s constant (8.99×109N·m2/C2), |+2Q|= 2Q,|Q|=Q,
and ris the distance between the charges. Since the charges are at the vertices
of an equilateral triangle, the distance between the charges is equal to the side
length of the triangle.
5
Step 2: Calculate the magnitude of the force between the charge at the lower
vertex and the charge at the upper right vertex (+Q) using the same formula
as in Step 1.
F2=k· | + 2Q||Q|
r2
Step 3: Determine the direction of the forces. The force between the charge
at the lower vertex and the charge at the upper left vertex is repulsive, and the
force between the charge at the lower vertex and the charge at the upper right
vertex is also repulsive. Thus, both forces act away from the charge at the lower
vertex.
Step 4: Calculate the net force on the charge at the lower vertex by summing
the individual forces vectorially. The net force will be directed away from the
charge at the lower vertex and can be found using the Pythagorean theorem:
Fnet =√F2
1+F2
2
Question 8
Question
Three point charges are arranged along a straight line as shown below:
+3µC−2µC+5µC
The charges are spaced 10 cm apart with the +3 µC charge at the origin,
the -2 µC charge at 10 cm, and the +5 µC charge at 20 cm. Calculate the net
electric force on the -2 µC charge due to the other two charges.
Solution
Let’s denote the charges as follows: - Charge at 0cm: q1= +3µC - Charge at
10 cm: q2=−2µC - Charge at 20 cm: q3= +5µC
The net electric force on the -2 µC charge (q2) is the sum of the forces due to
the other two charges. We will first calculate the force between q1and q2, and
then the force between q3and q2, and finally sum them to find the net force.
Step 1: Calculate the force between q1and q2. The formula for the magni-
tude of the force between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
Where: - kis the Coulomb’s constant (8.99 ×109N m2/C2), - q1and q2are
the magnitudes of the charges, and - ris the distance between the charges.
Given q1= +3µC, q2=−2µC, and r= 10 cm = 0.1m, we have:
6
F12 =8.99 ×109· |3×10−6·(−2×10−6)|
(0.1)2
F12 =8.99 ×109·6×10−12
0.01
F12 =53.94 ×10−3
0.01
F12 = 5.394 N
So, the force between q1and q2is 5.394 N (pointing to the left).
Step 2: Calculate the force between q2and q3. This calculation follows the
same process as in Step 1.
Step 3: Sum the forces to find the net force on q2.
Thus, the net force on the -2 µC charge due to the other two charges is 5.394
N to the left.
Question 9
Question
Two point charges, +2µC and −3µC, are placed 10 cm apart in air. Calculate
the magnitude of the electric force between them.
Solution
Step 1: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k|q1·q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
charges (+2µC and −3µC), and ris the distance between the charges (0.1 m).
Step 2: Substitute the given values into the equation:
F= (8.99 ×109)|2×10−6· −3×10−6|
(0.1)2
Step 3: Calculate the magnitude of the electric force:
F= (8.99 ×109)6×10−12
0.01
F= 8.99 ×109×6×10−10
F= 53.94 ×10−1
F= 5.394N
Therefore, the magnitude of the electric force between the charges +2µC
and −3µC is 5.394N.
7
Question 10
Question
Three point charges are arranged in the following configuration: a charge of
+4.0µC at the origin, a charge of −2.0µC at coordinates (0,0,3m), and a
charge of +3.0µC at coordinates (4 m,0,0). Calculate the net electric force on
the charge at the origin due to the other two charges.
Solution
The electric force
Felec between two charges q1and q2separated by a distance
ris given by Coulomb’s law:
Felec =kq1q2
r2ˆr
where k≈8.99×109N·m2/C2is the Coulomb constant and ˆris the unit vector
pointing from q2to q1.
Step 1: Find the force on the charge at the origin due to the charge at
coordinates (0,0,3m). The distance between these two charges is r1= 3 m.
The electric force can be calculated as:
|
F1|=k|(+4.0µC)(−2.0µC)|
(3 m)2=8.99 ×109×4.0×2.0×10−12
9= 8.0×10−3N
Step 2: Find the force on the charge at the origin due to the charge at
coordinates (4 m,0,0). The distance between these two charges is r2= 4 m.
The electric force can be calculated as:
|
F2|=k|(+4.0µC)(+3.0µC)|
(4 m)2=8.99 ×109×4.0×3.0×10−12
16 = 6.7×10−3N
Step 3: Find the net force on the charge at the origin. The net force is the
vector sum of the individual forces. Since one force acts in the z-direction and
the other in the x-direction, the net force will have both xand zcomponents.
Fnet =
F1+
F2Calculating the magnitudes of the components:
Fnet,x =F2= 6.7×10−3N
Fnet,z =F1= 8.0×10−3N
Therefore, the net electric force on the charge at the origin is 6.7×10−3N
in the x-direction and 8.0×10−3N in the z-direction.
Question 11
Question
Three point charges are arranged in an equilateral triangle with sides of length d.
The charges are +qat point A,−2qat point B, and +3qat point C. Calculate
the net electric force on the charge at point Adue to the other two charges.
8
Solution
To find the net electric force on the charge at point A, we need to calculate the
electric forces between Aand B, and between Aand C, and then add these
forces as vectors to get the net force.
Step 1: Find the force between Aand B.
The electric force between two point charges q1and q2separated by a dis-
tance ris given by Coulomb’s Law:
F=kq1q2
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
The distance between Aand Bis also dbecause they are at opposite ends of
the side of length dof the equilateral triangle. Therefore, the net electric force
between Aand Bis:
FAB =k|+q|·|−2q|
d2=k2q2
d2
Step 2: Find the force between Aand C.
The distance between Aand Cis das well due to the equilateral triangle
arrangement. Therefore, the net electric force between Aand Cis:
FAC =k|+q|·|3q|
d2=k3q2
d2
Step 3: Find the net force on A.
The net force on the charge at point Ais the vector sum of FAB and FAC .
Since these forces are in opposite directions and have the same magnitude, the
net force is simply given by their difference:
Net Force on A =|FAC |−|FAB |=k3q2
d2−k2q2
d2=kq2
d2
Therefore, the net electric force on the charge at point Adue to the other
two charges is kq2
d2.
Question 12
Question
Two point charges, q1=−3.0µC and q2= 5.0µC, are placed 20 cm apart in a
vacuum. Calculate the magnitude of the electric force between them.
9
Solution
Step 1: Convert the given charges to coulombs.
q1=−3.0µC =−3.0×10−6C
q2= 5.0µC = 5.0×10−6C
Step 2: Write down the formula for the electric force between two point
charges. The magnitude of the electric force between two point charges is given
by Coulomb’s law:
F=k· |q1·q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant, q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 3: Substitute the given values into the formula.
F=(8.99 ×109N·m2/C2)· |(−3.0×10−6C)·(5.0×10−6C)|
(0.20 m)2
Step 4: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(3.0×5.0) ×10−12
0.04
=1.35 ×10−11 ×8.99 ×109
0.04
=1.21365 ×10−1
0.04
= 3.034 ×10−10 N
Step 5: Therefore, the magnitude of the electric force between the two
charges is 3.034 ×10−10 N.
Question 13
Question
Three point charges are arranged along the x-axis as follows: +3.0 nC at x = 0
m, -6.0 nC at x = 2.0 m, and +4.0 nC at x = 3.0 m. Calculate the net electric
force on the +4.0 nC charge due to the other two charges.
10
Solution
Step 1: Calculate the electric force on the +4.0 nC charge due to the +3.0 nC
charge. The electric force between two charges is given by Coulomb’s Law:
F=kq1q2
r2
where: F= electric force k= Coulomb’s constant = 8.9875×109Nm²/C² q1, q2
= magnitudes of the charges r= distance between the charges
Plugging in the values: k= 8.9875 ×109Nm²/C² q1= +3.0×10−9C
q2= +4.0×10−9Cr= 3.0m
F1= 8.9875 ×109×(3.0×10−9)(4.0×10−9)
3.02
F1= 8.9875 ×109×12 ×10−18
9
F1= 8.9875 ×109×1.3333 ×10−18
F1= 11.98 ×10−9
F1= 1.198 ×10−8N
Step 2: Calculate the electric force on the +4.0 nC charge due to the -6.0
nC charge. Following the same steps as above with q1=−6.0×10−9C:
F2= 8.9875 ×109×(6.0×10−9)(4.0×10−9)
1
F2= 8.9875 ×109×24.0×10−18
F2= 215.7×10−9
F2= 2.157 ×10−7N
Step 3: Calculate the net electric force on the +4.0 nC charge. Since the
forces are in opposite directions (one repelling and the other attracting), we
need to find the difference between them:
Net force =F1−F2= 1.198 ×10−8−2.157 ×10−7
Net force =−1.037 ×10−7N
Therefore, the net electric force on the +4.0 nC charge due to the other two
charges is −1.037 ×10−7N, directed towards the -6.0 nC charge.
Question 14
Question
Two point charges are placed at the two opposite corners of a square with sides
of length a. The charges have magnitudes qand 2q. Calculate the magnitude of
the electric force between the charges along one of the diagonals of the square.
11
Solution
Let’s denote the magnitude of the charge qas q1=qand the magnitude of the
charge 2qas q2= 2q.
Step 1: Calculate the electric force between q1and q2along one of the
diagonals of the square. Let’s choose the diagonal from q1to q2.
Using the formula for the magnitude of the electric force between two point
charges, we have:
F=k|q1q2|
r2
where - kis the Coulomb constant, - q1and q2are the magnitudes of the charges,
and - ris the distance between the charges.
First, let’s find the distance ralong the diagonal of the square: Since the
charges are at opposite corners of a square with sides of length a, the diagonal
length can be calculated using the Pythagorean theorem:
r=√a2+a2=√2a
Step 2: Substitute the values of q1,q2, and rinto the formula for the electric
force:
F=k|q1q2|
r2=k|q(2q)|
(√2a)2=k2q2
2a2=kq2
a2
Therefore, the magnitude of the electric force between the charges qand 2q
along one of the diagonals of the square is kq2
a2.
Question 15
Question
Two point charges, q1=−3µC and q2= 5 µC, are placed 10 cm apart along
the x-axis. Calculate the magnitude and direction of the electric force that q2
exerts on q1.
Solution
Step 1: First, calculate the electric force between the two charges using Coulomb’s
Law:
F=k·|q1·q2|
r2
where Fis the magnitude of the electric force, kis the electrostatic constant
(8.99 ×109N m2/C2), q1and q2are the charges, and ris the distance between
the charges.
Step 2: Substitute the given values into the formula:
F= (8.99 ×109)·|(−3×10−6)(5 ×10−6)|
(0.1)2
12
Step 3: Calculate the electric force:
F= (8.99 ×109)·15 ×10−12
0.01
F= (8.99 ×109)·1.5×10−9
F= 13.485 ×100N
F= 13.485 N
Step 4: Now, determine the direction of the force. The force will be attractive
since the charges are of opposite signs. Therefore, the force will act along the
line joining the charges, i.e., along the negative x-axis.
Therefore, the magnitude of the electric force that q2exerts on q1is 13.485
N along the negative x-axis.
Question 16
Question
Two point charges, q1=−2.5µC and q2= 3.0µC, are placed 10 cm apart in
air. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to standard units. Step 2: Calculate the distance
between the charges. Step 3: Use Coulomb’s Law to calculate the electric force.
Step 4: Provide the final answer with appropriate units.
Step 1: Convert the charges to standard units. q1=−2.5µC =−2.5×
10−6C
q2= 3.0µC = 3.0×10−6C
Step 2: Calculate the distance between the charges. Given that the charges
are placed 10 cm apart, the distance rbetween them is r= 10 cm = 0.10 m.
Step 3: Use Coulomb’s Law to calculate the electric force. The magnitude
of the electric force between two charges is given by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109Nm2/C2).
Plugging in the values:
F= 8.99 ×109Nm2/C2·| − 2.5×10−6C·3.0×10−6C|
(0.10 m)2
Calculating the force will give us the magnitude of the electric force between
the charges.
Step 4: Provide the final answer with appropriate units.
F= 53.7×10−6N= 53.7µN
Therefore, the magnitude of the electric force between the charges is 53.7µN .
13
Question 17
Question
Two point charges, q1= 4.0×10−6C and q2=−6.0×10−6C, are placed 0.30
m apart. Calculate the magnitude and direction of the electric force that q2
exerts on q1.
Solution
Step 1: Calculate the electric force magnitude using Coulomb’s Law.
F=k|q1||q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Step 2: Substitute the given values into the formula.
F=(8.99 ×109N m2/C2)(4.0×10−6C)(6.0×10−6C)
(0.30 m)2
Step 3: Calculate the electric force magnitude.
F=(8.99 ×109)(4.0×10−6)(6.0×10−6)
(0.30)2
F=215.76
0.09
F= 2397.33 N
Step 4: Determine the direction of the force. The force is attractive in this
case because the charges q1and q2have opposite signs. Thus, the force points
from q2to q1.
Therefore, the magnitude of the electric force that q2exerts on q1is 2397.33 N
and the direction is from q2to q1.
Question 18
Question
Two point charges, q1= 3.0µC and q2=−2.0µC, are located 10.0 cm apart.
Calculate the magnitude of the electric force between these charges.
14
Solution
Step 1: Convert the charges to Coulombs. Given: q1= 3.0µC,q2=−2.0µC
1 microCoulomb (µC) = 1×10−6Coulombs So, q1= 3.0×10−6C and q2=
−2.0×10−6C.
Step 2: Calculate the magnitude of the electric force using Coulomb’s Law.
The electric force Fbetween two point charges q1and q2separated by a distance
ris given by Coulomb’s Law:
F=k·|q1·q2|
r2
where k= 8.99 ×109Nm2/C2is Coulomb’s constant.
Plugging in the given values: F= (8.99 ×109)·|3.0×10−6·−2.0×10−6|
(0.10)2
Step 3: Calculate the magnitude of the electric force.
F= 8.99 ×109·6.0×10−12
0.01
F= 8.99 ×109·6.0×10−10
F= 53.94 ×10−1
F= 5.39 ×100N
Therefore, the magnitude of the electric force between these charges is 5.39
N.
Question 19
Question
Three point charges are arranged along the x-axis as follows: +3.00 µC at the
origin, −4.00 µC at x= 50.0cm, and +5.00 µC at x= 70.0cm. What is the net
electric force on the −4.00 µC charge?
Solution
1. Calculate the electric force on the −4.00 µC charge due to the +3.00 µC
charge at the origin. The electric force F1between two point charges is given
by Coulomb’s Law:
F1=kq1q2
r2
where - kis the Coulomb constant (8.99 ×109N·m2/C2), - q1and q2are
the magnitudes of the point charges (3.00 µC and −4.00 µC, respectively), - r
is the distance between the charges (origin to x= 50.0cm = 0.50 m).
Plugging in the values, we get:
F1= (8.99 ×109)(3.00 ×10−6)·(4.00 ×10−6)
(0.50)2
15
Calculating the result gives F1=−4.79 N, directed towards the origin.
2. Calculate the electric force on the −4.00 µC charge due to the +5.00 µC
charge at x= 70.0cm. Using Coulomb’s Law again, this time with q2=
+5.00 µC and r= 0.20 m, we have:
F2= (8.99 ×109)(5.00 ×10−6)·(4.00 ×10−6)
(0.20)2
Calculating the result gives F2= 44.94 N, directed towards the positive
charge.
3. The net force is the vector sum of F1and F2.
Fnet =F2−F1= 44.94 N−4.79 N= 40.15 N
Therefore, the net electric force on the −4.00 µC charge is 40.15 N directed
towards the +5.00 µC charge at x= 70.0cm.
Question 20
Question
Two point charges, q1=−2.5µC and q2= 4.0µC, are located at points
A(−2.0m,0) and B(3.0m,0) respectively on the x-axis. Calculate the electric
force exerted on q2due to q1.
Solution
Step 1: First, we need to calculate the distance between the two charges. Given:
xA=−2.0m
xB= 3.0m
Let dbe the distance between the charges. d=xB−xA= 3.0m−(−2.0m) =
5.0m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law:
F=k|q1·q2|
d2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Step 3: Substituting the given values into the formula:
F= 8.99 ×109N m2/C2·| − 2.5×10−6C·4.0×10−6C|
(5.0m)2
F= 8.99 ×109·10 ×10−12
25
F= 8.99 ×109·4×10−12
F= 35.96 ×10−3
16
F= 3.596 mN
Therefore, the electric force exerted on q2due to q1is 3.596 mN directed
towards q1.
Question 21
Question
Three point charges are fixed on the x-axis. The charges are located at x=−2
m, x= 0 m, and x= 3 m, with magnitudes q1= 4 µC, q2=−6µC, and q3= 2
µC, respectively. Calculate the net electrostatic force acting on q1due to the
other two charges.
Solution
Step 1: Calculate the force on q1due to q2. The magnitude of the force F12 can
be calculated using Coulomb’s Law:
F12 =k|q1||q2|
r2
12
where kis the Coulomb’s constant, r12 is the distance between q1and q2, and
|q1|and |q2|are the magnitudes of the charges.
Substitute the given values into the equation:
F12 =(9 ×109N·m2/C2)(4 ×10−6C)(6 ×10−6C)
(−2m)2
F12 =216 ×10−12
4= 54 ×10−12 = 5.4×10−11 N
Step 2: Calculate the force on q1due to q3. Similarly, calculate the force
F13 on q1due to q3using Coulomb’s Law:
F13 =k|q1||q3|
r2
13
where r13 is the distance between q1and q3.
Substitute the given values into the equation:
F13 =(9 ×109N·m2/C2)(4 ×10−6C)(2 ×10−6C)
(3 m−(−2m))2
F13 =72 ×10−12
25 = 2.88 ×10−12 N
Step 3: Calculate the net force on q1. The net force on q1is the vector
sum of F12 and F13. Since F12 and F13 are in opposite directions, we need to
subtract them:
Fnet =F12 −F13 = 5.4×10−11 N−2.88 ×10−12 N
17
Fnet = 5.112 ×10−11 N
Therefore, the net electrostatic force acting on q1due to the other two
charges is 5.112 ×10−11 N.
Question 22
Question
Two point charges, q1=−4µC and q2= 2 µC, are placed 10 cm apart along the
x-axis. Calculate the magnitude and direction of the electric force experienced
by each charge.
Solution
Step 1: Determine the distance between the charges Given that the charges are
10 cm apart along the x-axis, the distance between them is r= 10 cm = 0.10
m.
Step 2: Calculate the magnitude of the electric force experienced by q1The
magnitude of the electric force between two point charges q1and q2is given by
Coulomb’s Law:
F=k|q1q2|
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant.
Substitute q1=−4µC, q2= 2 µC, and r= 0.10 m into the formula:
F=(8.99 ×109N·m2/C2)|(−4×10−6C)(2 ×10−6C)|
(0.10 m)2
Calculating this expression gives the magnitude of the electric force experi-
enced by q1.
Step 3: Determine the direction of the electric force experienced by q1The
direction of the electric force experienced by q1will be attractive towards q2.
Step 4: Calculate the magnitude of the electric force experienced by q2Using
the same formula as in Step 2, but switching the signs of q1and q2, calculate
the electric force experienced by q2.
Step 5: Determine the direction of the electric force experienced by q2The
direction of the electric force experienced by q2will be repulsive away from q1.
Question 23
Question
A charge of −2.0µC is located at the origin, and a charge of 4.0µC is located
at the point (0,3.0) m. Calculate the magnitude and direction of the electric
force on the positive charge due to the negative charge.
18
Solution
To find the electric force, we will use Coulomb’s Law, given by:
Electric force =k·|q1·q2|
r2
where kis Coulomb’s constant (8.99 ×109Nm2/C2), q1and q2are the two
charges, and ris the distance between the charges.
Step 1: Calculate the distance rbetween the charges
The distance rbetween the charges can be found using the distance formula
in two dimensions:
r=√(0 −0)2+ (3.0−0)2
r=√0+9.0
r=√9.0
r= 3.0m
Step 2: Calculate the magnitude of the electric force
Now, we can substitute the given values into Coulomb’s Law:
Electric force = (8.99 ×109)·| − 2.0×10−6×4.0×10−6|
(3.0)2
Electric force = 8.99 ×109·8.0×10−12
9.0
Electric force = 8.99 ×109·8.89 ×10−13
Electric force = 7.99 ×10−3N
Step 3: Determine the direction of the electric force
The electric force will be attractive since the charges are of opposite signs.
Therefore, the direction of the electric force is towards the negative charge (from
the positive charge to the negative charge) along the line connecting the charges.
Question 24
Question
Two point charges, q1= +5 µC and q2=−3µC, are placed 10 cm apart. What
is the magnitude of the electric force between these two charges?
19
Solution
Step 1: Convert the charges to Coulombs. The charges given are q1= +5 µC
and q2=−3µC. To convert microcoulombs to coulombs, we use the following
conversion factor: 1µC= 10−6C. Therefore, q1= +5 ×10−6C= 5 ×10−6C
and q2=−3×10−6C=−3×10−6C.
Step 2: Calculate the distance between the charges in meters. The charges
are placed 10 cm apart, which is 0.10 meters.
Step 3: Calculate the electric force between the charges using Coulomb’s
Law. Coulomb’s Law states that the magnitude of the electric force between
two point charges is given by:
F=k· |q1·q2|
r2
where F= electric force, k= Coulomb’s constant (8.99 ×109N m2/C2), q1, q2
= magnitudes of the two charges, and r= distance between the charges.
Substitute the known values into the formula:
F=(8.99 ×109N m2/C2)· |(5 ×10−6C)·(−3×10−6C)|
(0.10 m)2
Step 4: Calculate the electric force.
F=(8.99 ×109)·(5 ×10−6)·(3 ×10−6)
0.01
F=134850
100
F= 1348.50 N
Therefore, the magnitude of the electric force between the two charges is
1348.50 N.
Question 25
Question
Two point charges, q1=−3.00 µC and q2= 5.00 µC, are placed 10.0 cm apart
in vacuum.
Calculate the magnitude of the electric force between the charges, and de-
termine the direction of the force exerted on each charge.
Solution
Step 1: Convert the charges to coulombs. The charges are given in micro-
coulombs (µC), so we need to convert them to coulombs. q1=−3.00 µC=
−3.00 ×10−6Cq2= 5.00 µC= 5.00 ×10−6C
20
Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force between two point charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant and r= 0.10 m is
the separation distance between the charges. Substituting in the given values:
F=8.99 ×109·3.00 ×10−6·5.00 ×10−6
(0.10)2
Step 3: Calculate the numerical value of the electric force.
F=8.99 ×3×5
103=134.85
103= 0.13485 N
Step 4: Determine the direction of the force. The force between the charges
is attractive since the charges have opposite signs.
Therefore, the magnitude of the electric force between the charges is 0.13485 N,
and the force exerted on q1is attractive towards q2, while the force exerted on
q2is attractive towards q1.
21
F= 8.99 ×109N m2/C2·8×10−16 C2
F= 7.192 ×10−6N
Therefore, the magnitude of the electric force between the charges is 7.192 ×
10−6N.
Question 2
Question
Two point charges, q1=−4µC and q2= 8 µC, are separated by a distance of
0.5m. Calculate the magnitude of the electric force that q1exerts on q2.
Solution
Step 1: Determine the electric force between the two charges using Coulomb’s
Law:
F=k·|q1·q2|
r2,
where k= 8.99 ×109N m2/C2is the Coulomb’s constant, q1and q2are the
charges, and ris the distance between the charges.
Step 2: Calculate the force by plugging in the given values:
F= (8.99 ×109)·|(−4×10−6)·(8 ×10−6)|
(0.5)2.
Step 3: Simplify the expression:
F= (8.99 ×109)·32 ×10−12
0.25 .
F= (8.99 ×109)·128 ×10−12.
Step 4: Perform the multiplication:
F= 1150.72 ×10−3.
Step 5: Express the result in Newtons by converting the magnitude to sci-
entific notation:
F= 1.15072 N.
Therefore, the magnitude of the electric force that q1exerts on q2is 1.15072 N.
Question 3
Question
Two point charges are fixed in place, with charge q1=−2.0µC at coordinates
(0,0) and charge q2= 4.0µC at coordinates (2.0m,0). Calculate the magnitude
and direction of the electric force that q1exerts on q2.
2
Solution
Step 1: Calculate the distance between the two charges.
Let rbe the distance between charge q1and charge q2. Using the distance
formula in 2D space:
r=√(2.0m)2+ (0 −0)2=√4.0m2= 2.0m
Step 2: Calculate the magnitude of the electric force.
The electric force between two point charges is given by Coulomb’s law:
F=k|q1q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the equation:
F= (8.99 ×109)|(−2.0×10−6)(4.0×10−6)|
(2.0)2
F= (8.99 ×109)8.0×10−12
4.0= (8.99 ×109)×2.0×10−12 = 1.80 ×10−2N
The magnitude of the electric force is 1.80 ×10−2N.
Step 3: Determine the direction of the electric force.
The direction of the electric force is along the line joining the two charges. Since
charge q1has a negative charge, the force on q2will be attractive towards q1in
the negative xdirection.
Therefore, the electric force that q1exerts on q2is 1.80 ×10−2N in the
negative xdirection.
Question 4
Question
Two point charges, q1=−5µC and q2= 8 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges in
Newtons.
Solution
Step 1: Convert the charges to coulombs (µC to C). Given charges: q1=−5µC
and q2= 8 µC
Converting to coulombs: q1=−5×10−6C and q2= 8 ×10−6C
Step 2: Calculate the electric force using Coulomb’s Law: The magnitude of
the electric force between two point charges is given by Coulomb’s Law:
F=k· |q1·q2|
r2
3
where: F= electric force between the charges, k= Coulomb’s constant (8.99 ×
109N m2/C2), q1and q2= magnitudes of the point charges, r= distance be-
tween the charges.
In this case: k= 8.99 ×109N m2/C2,q1=−5×10−6C, q2= 8 ×10−6C,
r= 0.1m (since 10 cm is 0.1 m).
Step 3: Plug the values into Coulomb’s Law and calculate the electric force:
F=8.99 ×109×|−5×10−6×8×10−6|
(0.1)2
F=8.99 ×109×5×8×10−12
0.01
F=8.99 ×5×8
10 ×10−3
F= 8.99 ×5×0.8×10−3
F= 35.96 ×10−3
F= 0.03596 N
Therefore, the magnitude of the electric force between the charges is 0.03596
N.
Question 5
Question
Three point charges are placed on the vertices of an equilateral triangle. The
charges are +2µC,−4µC, and +3µC. Calculate the net electric force on the
charge at the vertex with the +2µC charge.
Solution
Step 1: We first need to calculate the distance between the charges. Since
the charges are placed on the vertices of an equilateral triangle, the distances
between them are equal. Let’s denote this distance as d.
Step 2: The electric force between two charges q1and q2separated by dis-
tance ris given by Coulomb’s law:
F=k·|q1·q2|
r2,
where k≈8.99 ×109N m2/C2is the Coulomb’s constant.
4
Step 3: Let’s first calculate the net electric force on the +2µC charge due
to the −4µC and +3µC charges. The force on +2µC due to −4µC can be
calculated using Coulomb’s law, and the direction of this force will be attractive.
Step 4: The force on +2µC due to +3µC can be calculated using Coulomb’s
law as well, and the direction of this force will be repulsive.
Step 5: Once we have the magnitudes and directions of the forces due to
each charge, we can find the net force on the +2µC charge by vector addition
of individual forces.
Step 6: Calculate the net electric force on the charge at the vertex with the
+2µC charge.
Question 7
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below:
+Q
+Q
−2Q
The charges are +Qat each of the two upper vertices and −2Qat the lower
vertex. Find the net electrostatic force on the charge at the lower vertex due to
the other two charges.
Solution
Step 1: Calculate the magnitude of the force between the charge at the lower
vertex (−2Q) and the charge at the upper left vertex (+Q).
F1=k· | + 2Q||Q|
r2
where kis the Coulomb’s constant (8.99×109N·m2/C2), |+2Q|= 2Q,|Q|=Q,
and ris the distance between the charges. Since the charges are at the vertices
of an equilateral triangle, the distance between the charges is equal to the side
length of the triangle.
5
Step 2: Calculate the magnitude of the force between the charge at the lower
vertex and the charge at the upper right vertex (+Q) using the same formula
as in Step 1.
F2=k· | + 2Q||Q|
r2
Step 3: Determine the direction of the forces. The force between the charge
at the lower vertex and the charge at the upper left vertex is repulsive, and the
force between the charge at the lower vertex and the charge at the upper right
vertex is also repulsive. Thus, both forces act away from the charge at the lower
vertex.
Step 4: Calculate the net force on the charge at the lower vertex by summing
the individual forces vectorially. The net force will be directed away from the
charge at the lower vertex and can be found using the Pythagorean theorem:
Fnet =√F2
1+F2
2
Question 8
Question
Three point charges are arranged along a straight line as shown below:
+3µC−2µC+5µC
The charges are spaced 10 cm apart with the +3 µC charge at the origin,
the -2 µC charge at 10 cm, and the +5 µC charge at 20 cm. Calculate the net
electric force on the -2 µC charge due to the other two charges.
Solution
Let’s denote the charges as follows: - Charge at 0cm: q1= +3µC - Charge at
10 cm: q2=−2µC - Charge at 20 cm: q3= +5µC
The net electric force on the -2 µC charge (q2) is the sum of the forces due to
the other two charges. We will first calculate the force between q1and q2, and
then the force between q3and q2, and finally sum them to find the net force.
Step 1: Calculate the force between q1and q2. The formula for the magni-
tude of the force between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
Where: - kis the Coulomb’s constant (8.99 ×109N m2/C2), - q1and q2are
the magnitudes of the charges, and - ris the distance between the charges.
Given q1= +3µC, q2=−2µC, and r= 10 cm = 0.1m, we have:
6
F12 =8.99 ×109· |3×10−6·(−2×10−6)|
(0.1)2
F12 =8.99 ×109·6×10−12
0.01
F12 =53.94 ×10−3
0.01
F12 = 5.394 N
So, the force between q1and q2is 5.394 N (pointing to the left).
Step 2: Calculate the force between q2and q3. This calculation follows the
same process as in Step 1.
Step 3: Sum the forces to find the net force on q2.
Thus, the net force on the -2 µC charge due to the other two charges is 5.394
N to the left.
Question 9
Question
Two point charges, +2µC and −3µC, are placed 10 cm apart in air. Calculate
the magnitude of the electric force between them.
Solution
Step 1: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k|q1·q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
charges (+2µC and −3µC), and ris the distance between the charges (0.1 m).
Step 2: Substitute the given values into the equation:
F= (8.99 ×109)|2×10−6· −3×10−6|
(0.1)2
Step 3: Calculate the magnitude of the electric force:
F= (8.99 ×109)6×10−12
0.01
F= 8.99 ×109×6×10−10
F= 53.94 ×10−1
F= 5.394N
Therefore, the magnitude of the electric force between the charges +2µC
and −3µC is 5.394N.
7
Question 10
Question
Three point charges are arranged in the following configuration: a charge of
+4.0µC at the origin, a charge of −2.0µC at coordinates (0,0,3m), and a
charge of +3.0µC at coordinates (4 m,0,0). Calculate the net electric force on
the charge at the origin due to the other two charges.
Solution
The electric force
Felec between two charges q1and q2separated by a distance
ris given by Coulomb’s law:
Felec =kq1q2
r2ˆr
where k≈8.99×109N·m2/C2is the Coulomb constant and ˆris the unit vector
pointing from q2to q1.
Step 1: Find the force on the charge at the origin due to the charge at
coordinates (0,0,3m). The distance between these two charges is r1= 3 m.
The electric force can be calculated as:
|
F1|=k|(+4.0µC)(−2.0µC)|
(3 m)2=8.99 ×109×4.0×2.0×10−12
9= 8.0×10−3N
Step 2: Find the force on the charge at the origin due to the charge at
coordinates (4 m,0,0). The distance between these two charges is r2= 4 m.
The electric force can be calculated as:
|
F2|=k|(+4.0µC)(+3.0µC)|
(4 m)2=8.99 ×109×4.0×3.0×10−12
16 = 6.7×10−3N
Step 3: Find the net force on the charge at the origin. The net force is the
vector sum of the individual forces. Since one force acts in the z-direction and
the other in the x-direction, the net force will have both xand zcomponents.
Fnet =
F1+
F2Calculating the magnitudes of the components:
Fnet,x =F2= 6.7×10−3N
Fnet,z =F1= 8.0×10−3N
Therefore, the net electric force on the charge at the origin is 6.7×10−3N
in the x-direction and 8.0×10−3N in the z-direction.
Question 11
Question
Three point charges are arranged in an equilateral triangle with sides of length d.
The charges are +qat point A,−2qat point B, and +3qat point C. Calculate
the net electric force on the charge at point Adue to the other two charges.
8
Solution
To find the net electric force on the charge at point A, we need to calculate the
electric forces between Aand B, and between Aand C, and then add these
forces as vectors to get the net force.
Step 1: Find the force between Aand B.
The electric force between two point charges q1and q2separated by a dis-
tance ris given by Coulomb’s Law:
F=kq1q2
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
The distance between Aand Bis also dbecause they are at opposite ends of
the side of length dof the equilateral triangle. Therefore, the net electric force
between Aand Bis:
FAB =k|+q|·|−2q|
d2=k2q2
d2
Step 2: Find the force between Aand C.
The distance between Aand Cis das well due to the equilateral triangle
arrangement. Therefore, the net electric force between Aand Cis:
FAC =k|+q|·|3q|
d2=k3q2
d2
Step 3: Find the net force on A.
The net force on the charge at point Ais the vector sum of FAB and FAC .
Since these forces are in opposite directions and have the same magnitude, the
net force is simply given by their difference:
Net Force on A =|FAC |−|FAB |=k3q2
d2−k2q2
d2=kq2
d2
Therefore, the net electric force on the charge at point Adue to the other
two charges is kq2
d2.
Question 12
Question
Two point charges, q1=−3.0µC and q2= 5.0µC, are placed 20 cm apart in a
vacuum. Calculate the magnitude of the electric force between them.
9
Solution
Step 1: Convert the given charges to coulombs.
q1=−3.0µC =−3.0×10−6C
q2= 5.0µC = 5.0×10−6C
Step 2: Write down the formula for the electric force between two point
charges. The magnitude of the electric force between two point charges is given
by Coulomb’s law:
F=k· |q1·q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant, q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 3: Substitute the given values into the formula.
F=(8.99 ×109N·m2/C2)· |(−3.0×10−6C)·(5.0×10−6C)|
(0.20 m)2
Step 4: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(3.0×5.0) ×10−12
0.04
=1.35 ×10−11 ×8.99 ×109
0.04
=1.21365 ×10−1
0.04
= 3.034 ×10−10 N
Step 5: Therefore, the magnitude of the electric force between the two
charges is 3.034 ×10−10 N.
Question 13
Question
Three point charges are arranged along the x-axis as follows: +3.0 nC at x = 0
m, -6.0 nC at x = 2.0 m, and +4.0 nC at x = 3.0 m. Calculate the net electric
force on the +4.0 nC charge due to the other two charges.
10
Solution
Step 1: Calculate the electric force on the +4.0 nC charge due to the +3.0 nC
charge. The electric force between two charges is given by Coulomb’s Law:
F=kq1q2
r2
where: F= electric force k= Coulomb’s constant = 8.9875×109Nm²/C² q1, q2
= magnitudes of the charges r= distance between the charges
Plugging in the values: k= 8.9875 ×109Nm²/C² q1= +3.0×10−9C
q2= +4.0×10−9Cr= 3.0m
F1= 8.9875 ×109×(3.0×10−9)(4.0×10−9)
3.02
F1= 8.9875 ×109×12 ×10−18
9
F1= 8.9875 ×109×1.3333 ×10−18
F1= 11.98 ×10−9
F1= 1.198 ×10−8N
Step 2: Calculate the electric force on the +4.0 nC charge due to the -6.0
nC charge. Following the same steps as above with q1=−6.0×10−9C:
F2= 8.9875 ×109×(6.0×10−9)(4.0×10−9)
1
F2= 8.9875 ×109×24.0×10−18
F2= 215.7×10−9
F2= 2.157 ×10−7N
Step 3: Calculate the net electric force on the +4.0 nC charge. Since the
forces are in opposite directions (one repelling and the other attracting), we
need to find the difference between them:
Net force =F1−F2= 1.198 ×10−8−2.157 ×10−7
Net force =−1.037 ×10−7N
Therefore, the net electric force on the +4.0 nC charge due to the other two
charges is −1.037 ×10−7N, directed towards the -6.0 nC charge.
Question 14
Question
Two point charges are placed at the two opposite corners of a square with sides
of length a. The charges have magnitudes qand 2q. Calculate the magnitude of
the electric force between the charges along one of the diagonals of the square.
11
Solution
Let’s denote the magnitude of the charge qas q1=qand the magnitude of the
charge 2qas q2= 2q.
Step 1: Calculate the electric force between q1and q2along one of the
diagonals of the square. Let’s choose the diagonal from q1to q2.
Using the formula for the magnitude of the electric force between two point
charges, we have:
F=k|q1q2|
r2
where - kis the Coulomb constant, - q1and q2are the magnitudes of the charges,
and - ris the distance between the charges.
First, let’s find the distance ralong the diagonal of the square: Since the
charges are at opposite corners of a square with sides of length a, the diagonal
length can be calculated using the Pythagorean theorem:
r=√a2+a2=√2a
Step 2: Substitute the values of q1,q2, and rinto the formula for the electric
force:
F=k|q1q2|
r2=k|q(2q)|
(√2a)2=k2q2
2a2=kq2
a2
Therefore, the magnitude of the electric force between the charges qand 2q
along one of the diagonals of the square is kq2
a2.
Question 15
Question
Two point charges, q1=−3µC and q2= 5 µC, are placed 10 cm apart along
the x-axis. Calculate the magnitude and direction of the electric force that q2
exerts on q1.
Solution
Step 1: First, calculate the electric force between the two charges using Coulomb’s
Law:
F=k·|q1·q2|
r2
where Fis the magnitude of the electric force, kis the electrostatic constant
(8.99 ×109N m2/C2), q1and q2are the charges, and ris the distance between
the charges.
Step 2: Substitute the given values into the formula:
F= (8.99 ×109)·|(−3×10−6)(5 ×10−6)|
(0.1)2
12
Step 3: Calculate the electric force:
F= (8.99 ×109)·15 ×10−12
0.01
F= (8.99 ×109)·1.5×10−9
F= 13.485 ×100N
F= 13.485 N
Step 4: Now, determine the direction of the force. The force will be attractive
since the charges are of opposite signs. Therefore, the force will act along the
line joining the charges, i.e., along the negative x-axis.
Therefore, the magnitude of the electric force that q2exerts on q1is 13.485
N along the negative x-axis.
Question 16
Question
Two point charges, q1=−2.5µC and q2= 3.0µC, are placed 10 cm apart in
air. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to standard units. Step 2: Calculate the distance
between the charges. Step 3: Use Coulomb’s Law to calculate the electric force.
Step 4: Provide the final answer with appropriate units.
Step 1: Convert the charges to standard units. q1=−2.5µC =−2.5×
10−6C
q2= 3.0µC = 3.0×10−6C
Step 2: Calculate the distance between the charges. Given that the charges
are placed 10 cm apart, the distance rbetween them is r= 10 cm = 0.10 m.
Step 3: Use Coulomb’s Law to calculate the electric force. The magnitude
of the electric force between two charges is given by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109Nm2/C2).
Plugging in the values:
F= 8.99 ×109Nm2/C2·| − 2.5×10−6C·3.0×10−6C|
(0.10 m)2
Calculating the force will give us the magnitude of the electric force between
the charges.
Step 4: Provide the final answer with appropriate units.
F= 53.7×10−6N= 53.7µN
Therefore, the magnitude of the electric force between the charges is 53.7µN .
13
Question 17
Question
Two point charges, q1= 4.0×10−6C and q2=−6.0×10−6C, are placed 0.30
m apart. Calculate the magnitude and direction of the electric force that q2
exerts on q1.
Solution
Step 1: Calculate the electric force magnitude using Coulomb’s Law.
F=k|q1||q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Step 2: Substitute the given values into the formula.
F=(8.99 ×109N m2/C2)(4.0×10−6C)(6.0×10−6C)
(0.30 m)2
Step 3: Calculate the electric force magnitude.
F=(8.99 ×109)(4.0×10−6)(6.0×10−6)
(0.30)2
F=215.76
0.09
F= 2397.33 N
Step 4: Determine the direction of the force. The force is attractive in this
case because the charges q1and q2have opposite signs. Thus, the force points
from q2to q1.
Therefore, the magnitude of the electric force that q2exerts on q1is 2397.33 N
and the direction is from q2to q1.
Question 18
Question
Two point charges, q1= 3.0µC and q2=−2.0µC, are located 10.0 cm apart.
Calculate the magnitude of the electric force between these charges.
14
Solution
Step 1: Convert the charges to Coulombs. Given: q1= 3.0µC,q2=−2.0µC
1 microCoulomb (µC) = 1×10−6Coulombs So, q1= 3.0×10−6C and q2=
−2.0×10−6C.
Step 2: Calculate the magnitude of the electric force using Coulomb’s Law.
The electric force Fbetween two point charges q1and q2separated by a distance
ris given by Coulomb’s Law:
F=k·|q1·q2|
r2
where k= 8.99 ×109Nm2/C2is Coulomb’s constant.
Plugging in the given values: F= (8.99 ×109)·|3.0×10−6·−2.0×10−6|
(0.10)2
Step 3: Calculate the magnitude of the electric force.
F= 8.99 ×109·6.0×10−12
0.01
F= 8.99 ×109·6.0×10−10
F= 53.94 ×10−1
F= 5.39 ×100N
Therefore, the magnitude of the electric force between these charges is 5.39
N.
Question 19
Question
Three point charges are arranged along the x-axis as follows: +3.00 µC at the
origin, −4.00 µC at x= 50.0cm, and +5.00 µC at x= 70.0cm. What is the net
electric force on the −4.00 µC charge?
Solution
1. Calculate the electric force on the −4.00 µC charge due to the +3.00 µC
charge at the origin. The electric force F1between two point charges is given
by Coulomb’s Law:
F1=kq1q2
r2
where - kis the Coulomb constant (8.99 ×109N·m2/C2), - q1and q2are
the magnitudes of the point charges (3.00 µC and −4.00 µC, respectively), - r
is the distance between the charges (origin to x= 50.0cm = 0.50 m).
Plugging in the values, we get:
F1= (8.99 ×109)(3.00 ×10−6)·(4.00 ×10−6)
(0.50)2
15
Calculating the result gives F1=−4.79 N, directed towards the origin.
2. Calculate the electric force on the −4.00 µC charge due to the +5.00 µC
charge at x= 70.0cm. Using Coulomb’s Law again, this time with q2=
+5.00 µC and r= 0.20 m, we have:
F2= (8.99 ×109)(5.00 ×10−6)·(4.00 ×10−6)
(0.20)2
Calculating the result gives F2= 44.94 N, directed towards the positive
charge.
3. The net force is the vector sum of F1and F2.
Fnet =F2−F1= 44.94 N−4.79 N= 40.15 N
Therefore, the net electric force on the −4.00 µC charge is 40.15 N directed
towards the +5.00 µC charge at x= 70.0cm.
Question 20
Question
Two point charges, q1=−2.5µC and q2= 4.0µC, are located at points
A(−2.0m,0) and B(3.0m,0) respectively on the x-axis. Calculate the electric
force exerted on q2due to q1.
Solution
Step 1: First, we need to calculate the distance between the two charges. Given:
xA=−2.0m
xB= 3.0m
Let dbe the distance between the charges. d=xB−xA= 3.0m−(−2.0m) =
5.0m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law:
F=k|q1·q2|
d2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Step 3: Substituting the given values into the formula:
F= 8.99 ×109N m2/C2·| − 2.5×10−6C·4.0×10−6C|
(5.0m)2
F= 8.99 ×109·10 ×10−12
25
F= 8.99 ×109·4×10−12
F= 35.96 ×10−3
16
F= 3.596 mN
Therefore, the electric force exerted on q2due to q1is 3.596 mN directed
towards q1.
Question 21
Question
Three point charges are fixed on the x-axis. The charges are located at x=−2
m, x= 0 m, and x= 3 m, with magnitudes q1= 4 µC, q2=−6µC, and q3= 2
µC, respectively. Calculate the net electrostatic force acting on q1due to the
other two charges.
Solution
Step 1: Calculate the force on q1due to q2. The magnitude of the force F12 can
be calculated using Coulomb’s Law:
F12 =k|q1||q2|
r2
12
where kis the Coulomb’s constant, r12 is the distance between q1and q2, and
|q1|and |q2|are the magnitudes of the charges.
Substitute the given values into the equation:
F12 =(9 ×109N·m2/C2)(4 ×10−6C)(6 ×10−6C)
(−2m)2
F12 =216 ×10−12
4= 54 ×10−12 = 5.4×10−11 N
Step 2: Calculate the force on q1due to q3. Similarly, calculate the force
F13 on q1due to q3using Coulomb’s Law:
F13 =k|q1||q3|
r2
13
where r13 is the distance between q1and q3.
Substitute the given values into the equation:
F13 =(9 ×109N·m2/C2)(4 ×10−6C)(2 ×10−6C)
(3 m−(−2m))2
F13 =72 ×10−12
25 = 2.88 ×10−12 N
Step 3: Calculate the net force on q1. The net force on q1is the vector
sum of F12 and F13. Since F12 and F13 are in opposite directions, we need to
subtract them:
Fnet =F12 −F13 = 5.4×10−11 N−2.88 ×10−12 N
17
Fnet = 5.112 ×10−11 N
Therefore, the net electrostatic force acting on q1due to the other two
charges is 5.112 ×10−11 N.
Question 22
Question
Two point charges, q1=−4µC and q2= 2 µC, are placed 10 cm apart along the
x-axis. Calculate the magnitude and direction of the electric force experienced
by each charge.
Solution
Step 1: Determine the distance between the charges Given that the charges are
10 cm apart along the x-axis, the distance between them is r= 10 cm = 0.10
m.
Step 2: Calculate the magnitude of the electric force experienced by q1The
magnitude of the electric force between two point charges q1and q2is given by
Coulomb’s Law:
F=k|q1q2|
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant.
Substitute q1=−4µC, q2= 2 µC, and r= 0.10 m into the formula:
F=(8.99 ×109N·m2/C2)|(−4×10−6C)(2 ×10−6C)|
(0.10 m)2
Calculating this expression gives the magnitude of the electric force experi-
enced by q1.
Step 3: Determine the direction of the electric force experienced by q1The
direction of the electric force experienced by q1will be attractive towards q2.
Step 4: Calculate the magnitude of the electric force experienced by q2Using
the same formula as in Step 2, but switching the signs of q1and q2, calculate
the electric force experienced by q2.
Step 5: Determine the direction of the electric force experienced by q2The
direction of the electric force experienced by q2will be repulsive away from q1.
Question 23
Question
A charge of −2.0µC is located at the origin, and a charge of 4.0µC is located
at the point (0,3.0) m. Calculate the magnitude and direction of the electric
force on the positive charge due to the negative charge.
18
Solution
To find the electric force, we will use Coulomb’s Law, given by:
Electric force =k·|q1·q2|
r2
where kis Coulomb’s constant (8.99 ×109Nm2/C2), q1and q2are the two
charges, and ris the distance between the charges.
Step 1: Calculate the distance rbetween the charges
The distance rbetween the charges can be found using the distance formula
in two dimensions:
r=√(0 −0)2+ (3.0−0)2
r=√0+9.0
r=√9.0
r= 3.0m
Step 2: Calculate the magnitude of the electric force
Now, we can substitute the given values into Coulomb’s Law:
Electric force = (8.99 ×109)·| − 2.0×10−6×4.0×10−6|
(3.0)2
Electric force = 8.99 ×109·8.0×10−12
9.0
Electric force = 8.99 ×109·8.89 ×10−13
Electric force = 7.99 ×10−3N
Step 3: Determine the direction of the electric force
The electric force will be attractive since the charges are of opposite signs.
Therefore, the direction of the electric force is towards the negative charge (from
the positive charge to the negative charge) along the line connecting the charges.
Question 24
Question
Two point charges, q1= +5 µC and q2=−3µC, are placed 10 cm apart. What
is the magnitude of the electric force between these two charges?
19
Solution
Step 1: Convert the charges to Coulombs. The charges given are q1= +5 µC
and q2=−3µC. To convert microcoulombs to coulombs, we use the following
conversion factor: 1µC= 10−6C. Therefore, q1= +5 ×10−6C= 5 ×10−6C
and q2=−3×10−6C=−3×10−6C.
Step 2: Calculate the distance between the charges in meters. The charges
are placed 10 cm apart, which is 0.10 meters.
Step 3: Calculate the electric force between the charges using Coulomb’s
Law. Coulomb’s Law states that the magnitude of the electric force between
two point charges is given by:
F=k· |q1·q2|
r2
where F= electric force, k= Coulomb’s constant (8.99 ×109N m2/C2), q1, q2
= magnitudes of the two charges, and r= distance between the charges.
Substitute the known values into the formula:
F=(8.99 ×109N m2/C2)· |(5 ×10−6C)·(−3×10−6C)|
(0.10 m)2
Step 4: Calculate the electric force.
F=(8.99 ×109)·(5 ×10−6)·(3 ×10−6)
0.01
F=134850
100
F= 1348.50 N
Therefore, the magnitude of the electric force between the two charges is
1348.50 N.
Question 25
Question
Two point charges, q1=−3.00 µC and q2= 5.00 µC, are placed 10.0 cm apart
in vacuum.
Calculate the magnitude of the electric force between the charges, and de-
termine the direction of the force exerted on each charge.
Solution
Step 1: Convert the charges to coulombs. The charges are given in micro-
coulombs (µC), so we need to convert them to coulombs. q1=−3.00 µC=
−3.00 ×10−6Cq2= 5.00 µC= 5.00 ×10−6C
20
Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force between two point charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant and r= 0.10 m is
the separation distance between the charges. Substituting in the given values:
F=8.99 ×109·3.00 ×10−6·5.00 ×10−6
(0.10)2
Step 3: Calculate the numerical value of the electric force.
F=8.99 ×3×5
103=134.85
103= 0.13485 N
Step 4: Determine the direction of the force. The force between the charges
is attractive since the charges have opposite signs.
Therefore, the magnitude of the electric force between the charges is 0.13485 N,
and the force exerted on q1is attractive towards q2, while the force exerted on
q2is attractive towards q1.
21
F= 8.99 ×109N m2/C2·8×10−16 C2
F= 7.192 ×10−6N
Therefore, the magnitude of the electric force between the charges is 7.192 ×
10−6N.
Question 2
Question
Two point charges, q1=−4µC and q2= 8 µC, are separated by a distance of
0.5m. Calculate the magnitude of the electric force that q1exerts on q2.
Solution
Step 1: Determine the electric force between the two charges using Coulomb’s
Law:
F=k·|q1·q2|
r2,
where k= 8.99 ×109N m2/C2is the Coulomb’s constant, q1and q2are the
charges, and ris the distance between the charges.
Step 2: Calculate the force by plugging in the given values:
F= (8.99 ×109)·|(−4×10−6)·(8 ×10−6)|
(0.5)2.
Step 3: Simplify the expression:
F= (8.99 ×109)·32 ×10−12
0.25 .
F= (8.99 ×109)·128 ×10−12.
Step 4: Perform the multiplication:
F= 1150.72 ×10−3.
Step 5: Express the result in Newtons by converting the magnitude to sci-
entific notation:
F= 1.15072 N.
Therefore, the magnitude of the electric force that q1exerts on q2is 1.15072 N.
Question 3
Question
Two point charges are fixed in place, with charge q1=−2.0µC at coordinates
(0,0) and charge q2= 4.0µC at coordinates (2.0m,0). Calculate the magnitude
and direction of the electric force that q1exerts on q2.
2
Solution
Step 1: Calculate the distance between the two charges.
Let rbe the distance between charge q1and charge q2. Using the distance
formula in 2D space:
r=√(2.0m)2+ (0 −0)2=√4.0m2= 2.0m
Step 2: Calculate the magnitude of the electric force.
The electric force between two point charges is given by Coulomb’s law:
F=k|q1q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the equation:
F= (8.99 ×109)|(−2.0×10−6)(4.0×10−6)|
(2.0)2
F= (8.99 ×109)8.0×10−12
4.0= (8.99 ×109)×2.0×10−12 = 1.80 ×10−2N
The magnitude of the electric force is 1.80 ×10−2N.
Step 3: Determine the direction of the electric force.
The direction of the electric force is along the line joining the two charges. Since
charge q1has a negative charge, the force on q2will be attractive towards q1in
the negative xdirection.
Therefore, the electric force that q1exerts on q2is 1.80 ×10−2N in the
negative xdirection.
Question 4
Question
Two point charges, q1=−5µC and q2= 8 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges in
Newtons.
Solution
Step 1: Convert the charges to coulombs (µC to C). Given charges: q1=−5µC
and q2= 8 µC
Converting to coulombs: q1=−5×10−6C and q2= 8 ×10−6C
Step 2: Calculate the electric force using Coulomb’s Law: The magnitude of
the electric force between two point charges is given by Coulomb’s Law:
F=k· |q1·q2|
r2
3
where: F= electric force between the charges, k= Coulomb’s constant (8.99 ×
109N m2/C2), q1and q2= magnitudes of the point charges, r= distance be-
tween the charges.
In this case: k= 8.99 ×109N m2/C2,q1=−5×10−6C, q2= 8 ×10−6C,
r= 0.1m (since 10 cm is 0.1 m).
Step 3: Plug the values into Coulomb’s Law and calculate the electric force:
F=8.99 ×109×|−5×10−6×8×10−6|
(0.1)2
F=8.99 ×109×5×8×10−12
0.01
F=8.99 ×5×8
10 ×10−3
F= 8.99 ×5×0.8×10−3
F= 35.96 ×10−3
F= 0.03596 N
Therefore, the magnitude of the electric force between the charges is 0.03596
N.
Question 5
Question
Three point charges are placed on the vertices of an equilateral triangle. The
charges are +2µC,−4µC, and +3µC. Calculate the net electric force on the
charge at the vertex with the +2µC charge.
Solution
Step 1: We first need to calculate the distance between the charges. Since
the charges are placed on the vertices of an equilateral triangle, the distances
between them are equal. Let’s denote this distance as d.
Step 2: The electric force between two charges q1and q2separated by dis-
tance ris given by Coulomb’s law:
F=k·|q1·q2|
r2,
where k≈8.99 ×109N m2/C2is the Coulomb’s constant.
4
Step 3: Let’s first calculate the net electric force on the +2µC charge due
to the −4µC and +3µC charges. The force on +2µC due to −4µC can be
calculated using Coulomb’s law, and the direction of this force will be attractive.
Step 4: The force on +2µC due to +3µC can be calculated using Coulomb’s
law as well, and the direction of this force will be repulsive.
Step 5: Once we have the magnitudes and directions of the forces due to
each charge, we can find the net force on the +2µC charge by vector addition
of individual forces.
Step 6: Calculate the net electric force on the charge at the vertex with the
+2µC charge.
Question 7
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below:
+Q
+Q
−2Q
The charges are +Qat each of the two upper vertices and −2Qat the lower
vertex. Find the net electrostatic force on the charge at the lower vertex due to
the other two charges.
Solution
Step 1: Calculate the magnitude of the force between the charge at the lower
vertex (−2Q) and the charge at the upper left vertex (+Q).
F1=k· | + 2Q||Q|
r2
where kis the Coulomb’s constant (8.99×109N·m2/C2), |+2Q|= 2Q,|Q|=Q,
and ris the distance between the charges. Since the charges are at the vertices
of an equilateral triangle, the distance between the charges is equal to the side
length of the triangle.
5
Step 2: Calculate the magnitude of the force between the charge at the lower
vertex and the charge at the upper right vertex (+Q) using the same formula
as in Step 1.
F2=k· | + 2Q||Q|
r2
Step 3: Determine the direction of the forces. The force between the charge
at the lower vertex and the charge at the upper left vertex is repulsive, and the
force between the charge at the lower vertex and the charge at the upper right
vertex is also repulsive. Thus, both forces act away from the charge at the lower
vertex.
Step 4: Calculate the net force on the charge at the lower vertex by summing
the individual forces vectorially. The net force will be directed away from the
charge at the lower vertex and can be found using the Pythagorean theorem:
Fnet =√F2
1+F2
2
Question 8
Question
Three point charges are arranged along a straight line as shown below:
+3µC−2µC+5µC
The charges are spaced 10 cm apart with the +3 µC charge at the origin,
the -2 µC charge at 10 cm, and the +5 µC charge at 20 cm. Calculate the net
electric force on the -2 µC charge due to the other two charges.
Solution
Let’s denote the charges as follows: - Charge at 0cm: q1= +3µC - Charge at
10 cm: q2=−2µC - Charge at 20 cm: q3= +5µC
The net electric force on the -2 µC charge (q2) is the sum of the forces due to
the other two charges. We will first calculate the force between q1and q2, and
then the force between q3and q2, and finally sum them to find the net force.
Step 1: Calculate the force between q1and q2. The formula for the magni-
tude of the force between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
Where: - kis the Coulomb’s constant (8.99 ×109N m2/C2), - q1and q2are
the magnitudes of the charges, and - ris the distance between the charges.
Given q1= +3µC, q2=−2µC, and r= 10 cm = 0.1m, we have:
6
F12 =8.99 ×109· |3×10−6·(−2×10−6)|
(0.1)2
F12 =8.99 ×109·6×10−12
0.01
F12 =53.94 ×10−3
0.01
F12 = 5.394 N
So, the force between q1and q2is 5.394 N (pointing to the left).
Step 2: Calculate the force between q2and q3. This calculation follows the
same process as in Step 1.
Step 3: Sum the forces to find the net force on q2.
Thus, the net force on the -2 µC charge due to the other two charges is 5.394
N to the left.
Question 9
Question
Two point charges, +2µC and −3µC, are placed 10 cm apart in air. Calculate
the magnitude of the electric force between them.
Solution
Step 1: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k|q1·q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
charges (+2µC and −3µC), and ris the distance between the charges (0.1 m).
Step 2: Substitute the given values into the equation:
F= (8.99 ×109)|2×10−6· −3×10−6|
(0.1)2
Step 3: Calculate the magnitude of the electric force:
F= (8.99 ×109)6×10−12
0.01
F= 8.99 ×109×6×10−10
F= 53.94 ×10−1
F= 5.394N
Therefore, the magnitude of the electric force between the charges +2µC
and −3µC is 5.394N.
7
Question 10
Question
Three point charges are arranged in the following configuration: a charge of
+4.0µC at the origin, a charge of −2.0µC at coordinates (0,0,3m), and a
charge of +3.0µC at coordinates (4 m,0,0). Calculate the net electric force on
the charge at the origin due to the other two charges.
Solution
The electric force
Felec between two charges q1and q2separated by a distance
ris given by Coulomb’s law:
Felec =kq1q2
r2ˆr
where k≈8.99×109N·m2/C2is the Coulomb constant and ˆris the unit vector
pointing from q2to q1.
Step 1: Find the force on the charge at the origin due to the charge at
coordinates (0,0,3m). The distance between these two charges is r1= 3 m.
The electric force can be calculated as:
|
F1|=k|(+4.0µC)(−2.0µC)|
(3 m)2=8.99 ×109×4.0×2.0×10−12
9= 8.0×10−3N
Step 2: Find the force on the charge at the origin due to the charge at
coordinates (4 m,0,0). The distance between these two charges is r2= 4 m.
The electric force can be calculated as:
|
F2|=k|(+4.0µC)(+3.0µC)|
(4 m)2=8.99 ×109×4.0×3.0×10−12
16 = 6.7×10−3N
Step 3: Find the net force on the charge at the origin. The net force is the
vector sum of the individual forces. Since one force acts in the z-direction and
the other in the x-direction, the net force will have both xand zcomponents.
Fnet =
F1+
F2Calculating the magnitudes of the components:
Fnet,x =F2= 6.7×10−3N
Fnet,z =F1= 8.0×10−3N
Therefore, the net electric force on the charge at the origin is 6.7×10−3N
in the x-direction and 8.0×10−3N in the z-direction.
Question 11
Question
Three point charges are arranged in an equilateral triangle with sides of length d.
The charges are +qat point A,−2qat point B, and +3qat point C. Calculate
the net electric force on the charge at point Adue to the other two charges.
8
Solution
To find the net electric force on the charge at point A, we need to calculate the
electric forces between Aand B, and between Aand C, and then add these
forces as vectors to get the net force.
Step 1: Find the force between Aand B.
The electric force between two point charges q1and q2separated by a dis-
tance ris given by Coulomb’s Law:
F=kq1q2
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
The distance between Aand Bis also dbecause they are at opposite ends of
the side of length dof the equilateral triangle. Therefore, the net electric force
between Aand Bis:
FAB =k|+q|·|−2q|
d2=k2q2
d2
Step 2: Find the force between Aand C.
The distance between Aand Cis das well due to the equilateral triangle
arrangement. Therefore, the net electric force between Aand Cis:
FAC =k|+q|·|3q|
d2=k3q2
d2
Step 3: Find the net force on A.
The net force on the charge at point Ais the vector sum of FAB and FAC .
Since these forces are in opposite directions and have the same magnitude, the
net force is simply given by their difference:
Net Force on A =|FAC |−|FAB |=k3q2
d2−k2q2
d2=kq2
d2
Therefore, the net electric force on the charge at point Adue to the other
two charges is kq2
d2.
Question 12
Question
Two point charges, q1=−3.0µC and q2= 5.0µC, are placed 20 cm apart in a
vacuum. Calculate the magnitude of the electric force between them.
9
Solution
Step 1: Convert the given charges to coulombs.
q1=−3.0µC =−3.0×10−6C
q2= 5.0µC = 5.0×10−6C
Step 2: Write down the formula for the electric force between two point
charges. The magnitude of the electric force between two point charges is given
by Coulomb’s law:
F=k· |q1·q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant, q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 3: Substitute the given values into the formula.
F=(8.99 ×109N·m2/C2)· |(−3.0×10−6C)·(5.0×10−6C)|
(0.20 m)2
Step 4: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(3.0×5.0) ×10−12
0.04
=1.35 ×10−11 ×8.99 ×109
0.04
=1.21365 ×10−1
0.04
= 3.034 ×10−10 N
Step 5: Therefore, the magnitude of the electric force between the two
charges is 3.034 ×10−10 N.
Question 13
Question
Three point charges are arranged along the x-axis as follows: +3.0 nC at x = 0
m, -6.0 nC at x = 2.0 m, and +4.0 nC at x = 3.0 m. Calculate the net electric
force on the +4.0 nC charge due to the other two charges.
10
Solution
Step 1: Calculate the electric force on the +4.0 nC charge due to the +3.0 nC
charge. The electric force between two charges is given by Coulomb’s Law:
F=kq1q2
r2
where: F= electric force k= Coulomb’s constant = 8.9875×109Nm²/C² q1, q2
= magnitudes of the charges r= distance between the charges
Plugging in the values: k= 8.9875 ×109Nm²/C² q1= +3.0×10−9C
q2= +4.0×10−9Cr= 3.0m
F1= 8.9875 ×109×(3.0×10−9)(4.0×10−9)
3.02
F1= 8.9875 ×109×12 ×10−18
9
F1= 8.9875 ×109×1.3333 ×10−18
F1= 11.98 ×10−9
F1= 1.198 ×10−8N
Step 2: Calculate the electric force on the +4.0 nC charge due to the -6.0
nC charge. Following the same steps as above with q1=−6.0×10−9C:
F2= 8.9875 ×109×(6.0×10−9)(4.0×10−9)
1
F2= 8.9875 ×109×24.0×10−18
F2= 215.7×10−9
F2= 2.157 ×10−7N
Step 3: Calculate the net electric force on the +4.0 nC charge. Since the
forces are in opposite directions (one repelling and the other attracting), we
need to find the difference between them:
Net force =F1−F2= 1.198 ×10−8−2.157 ×10−7
Net force =−1.037 ×10−7N
Therefore, the net electric force on the +4.0 nC charge due to the other two
charges is −1.037 ×10−7N, directed towards the -6.0 nC charge.
Question 14
Question
Two point charges are placed at the two opposite corners of a square with sides
of length a. The charges have magnitudes qand 2q. Calculate the magnitude of
the electric force between the charges along one of the diagonals of the square.
11
Solution
Let’s denote the magnitude of the charge qas q1=qand the magnitude of the
charge 2qas q2= 2q.
Step 1: Calculate the electric force between q1and q2along one of the
diagonals of the square. Let’s choose the diagonal from q1to q2.
Using the formula for the magnitude of the electric force between two point
charges, we have:
F=k|q1q2|
r2
where - kis the Coulomb constant, - q1and q2are the magnitudes of the charges,
and - ris the distance between the charges.
First, let’s find the distance ralong the diagonal of the square: Since the
charges are at opposite corners of a square with sides of length a, the diagonal
length can be calculated using the Pythagorean theorem:
r=√a2+a2=√2a
Step 2: Substitute the values of q1,q2, and rinto the formula for the electric
force:
F=k|q1q2|
r2=k|q(2q)|
(√2a)2=k2q2
2a2=kq2
a2
Therefore, the magnitude of the electric force between the charges qand 2q
along one of the diagonals of the square is kq2
a2.
Question 15
Question
Two point charges, q1=−3µC and q2= 5 µC, are placed 10 cm apart along
the x-axis. Calculate the magnitude and direction of the electric force that q2
exerts on q1.
Solution
Step 1: First, calculate the electric force between the two charges using Coulomb’s
Law:
F=k·|q1·q2|
r2
where Fis the magnitude of the electric force, kis the electrostatic constant
(8.99 ×109N m2/C2), q1and q2are the charges, and ris the distance between
the charges.
Step 2: Substitute the given values into the formula:
F= (8.99 ×109)·|(−3×10−6)(5 ×10−6)|
(0.1)2
12
Step 3: Calculate the electric force:
F= (8.99 ×109)·15 ×10−12
0.01
F= (8.99 ×109)·1.5×10−9
F= 13.485 ×100N
F= 13.485 N
Step 4: Now, determine the direction of the force. The force will be attractive
since the charges are of opposite signs. Therefore, the force will act along the
line joining the charges, i.e., along the negative x-axis.
Therefore, the magnitude of the electric force that q2exerts on q1is 13.485
N along the negative x-axis.
Question 16
Question
Two point charges, q1=−2.5µC and q2= 3.0µC, are placed 10 cm apart in
air. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to standard units. Step 2: Calculate the distance
between the charges. Step 3: Use Coulomb’s Law to calculate the electric force.
Step 4: Provide the final answer with appropriate units.
Step 1: Convert the charges to standard units. q1=−2.5µC =−2.5×
10−6C
q2= 3.0µC = 3.0×10−6C
Step 2: Calculate the distance between the charges. Given that the charges
are placed 10 cm apart, the distance rbetween them is r= 10 cm = 0.10 m.
Step 3: Use Coulomb’s Law to calculate the electric force. The magnitude
of the electric force between two charges is given by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109Nm2/C2).
Plugging in the values:
F= 8.99 ×109Nm2/C2·| − 2.5×10−6C·3.0×10−6C|
(0.10 m)2
Calculating the force will give us the magnitude of the electric force between
the charges.
Step 4: Provide the final answer with appropriate units.
F= 53.7×10−6N= 53.7µN
Therefore, the magnitude of the electric force between the charges is 53.7µN .
13
Question 17
Question
Two point charges, q1= 4.0×10−6C and q2=−6.0×10−6C, are placed 0.30
m apart. Calculate the magnitude and direction of the electric force that q2
exerts on q1.
Solution
Step 1: Calculate the electric force magnitude using Coulomb’s Law.
F=k|q1||q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Step 2: Substitute the given values into the formula.
F=(8.99 ×109N m2/C2)(4.0×10−6C)(6.0×10−6C)
(0.30 m)2
Step 3: Calculate the electric force magnitude.
F=(8.99 ×109)(4.0×10−6)(6.0×10−6)
(0.30)2
F=215.76
0.09
F= 2397.33 N
Step 4: Determine the direction of the force. The force is attractive in this
case because the charges q1and q2have opposite signs. Thus, the force points
from q2to q1.
Therefore, the magnitude of the electric force that q2exerts on q1is 2397.33 N
and the direction is from q2to q1.
Question 18
Question
Two point charges, q1= 3.0µC and q2=−2.0µC, are located 10.0 cm apart.
Calculate the magnitude of the electric force between these charges.
14
Solution
Step 1: Convert the charges to Coulombs. Given: q1= 3.0µC,q2=−2.0µC
1 microCoulomb (µC) = 1×10−6Coulombs So, q1= 3.0×10−6C and q2=
−2.0×10−6C.
Step 2: Calculate the magnitude of the electric force using Coulomb’s Law.
The electric force Fbetween two point charges q1and q2separated by a distance
ris given by Coulomb’s Law:
F=k·|q1·q2|
r2
where k= 8.99 ×109Nm2/C2is Coulomb’s constant.
Plugging in the given values: F= (8.99 ×109)·|3.0×10−6·−2.0×10−6|
(0.10)2
Step 3: Calculate the magnitude of the electric force.
F= 8.99 ×109·6.0×10−12
0.01
F= 8.99 ×109·6.0×10−10
F= 53.94 ×10−1
F= 5.39 ×100N
Therefore, the magnitude of the electric force between these charges is 5.39
N.
Question 19
Question
Three point charges are arranged along the x-axis as follows: +3.00 µC at the
origin, −4.00 µC at x= 50.0cm, and +5.00 µC at x= 70.0cm. What is the net
electric force on the −4.00 µC charge?
Solution
1. Calculate the electric force on the −4.00 µC charge due to the +3.00 µC
charge at the origin. The electric force F1between two point charges is given
by Coulomb’s Law:
F1=kq1q2
r2
where - kis the Coulomb constant (8.99 ×109N·m2/C2), - q1and q2are
the magnitudes of the point charges (3.00 µC and −4.00 µC, respectively), - r
is the distance between the charges (origin to x= 50.0cm = 0.50 m).
Plugging in the values, we get:
F1= (8.99 ×109)(3.00 ×10−6)·(4.00 ×10−6)
(0.50)2
15
Calculating the result gives F1=−4.79 N, directed towards the origin.
2. Calculate the electric force on the −4.00 µC charge due to the +5.00 µC
charge at x= 70.0cm. Using Coulomb’s Law again, this time with q2=
+5.00 µC and r= 0.20 m, we have:
F2= (8.99 ×109)(5.00 ×10−6)·(4.00 ×10−6)
(0.20)2
Calculating the result gives F2= 44.94 N, directed towards the positive
charge.
3. The net force is the vector sum of F1and F2.
Fnet =F2−F1= 44.94 N−4.79 N= 40.15 N
Therefore, the net electric force on the −4.00 µC charge is 40.15 N directed
towards the +5.00 µC charge at x= 70.0cm.
Question 20
Question
Two point charges, q1=−2.5µC and q2= 4.0µC, are located at points
A(−2.0m,0) and B(3.0m,0) respectively on the x-axis. Calculate the electric
force exerted on q2due to q1.
Solution
Step 1: First, we need to calculate the distance between the two charges. Given:
xA=−2.0m
xB= 3.0m
Let dbe the distance between the charges. d=xB−xA= 3.0m−(−2.0m) =
5.0m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law:
F=k|q1·q2|
d2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Step 3: Substituting the given values into the formula:
F= 8.99 ×109N m2/C2·| − 2.5×10−6C·4.0×10−6C|
(5.0m)2
F= 8.99 ×109·10 ×10−12
25
F= 8.99 ×109·4×10−12
F= 35.96 ×10−3
16
F= 3.596 mN
Therefore, the electric force exerted on q2due to q1is 3.596 mN directed
towards q1.
Question 21
Question
Three point charges are fixed on the x-axis. The charges are located at x=−2
m, x= 0 m, and x= 3 m, with magnitudes q1= 4 µC, q2=−6µC, and q3= 2
µC, respectively. Calculate the net electrostatic force acting on q1due to the
other two charges.
Solution
Step 1: Calculate the force on q1due to q2. The magnitude of the force F12 can
be calculated using Coulomb’s Law:
F12 =k|q1||q2|
r2
12
where kis the Coulomb’s constant, r12 is the distance between q1and q2, and
|q1|and |q2|are the magnitudes of the charges.
Substitute the given values into the equation:
F12 =(9 ×109N·m2/C2)(4 ×10−6C)(6 ×10−6C)
(−2m)2
F12 =216 ×10−12
4= 54 ×10−12 = 5.4×10−11 N
Step 2: Calculate the force on q1due to q3. Similarly, calculate the force
F13 on q1due to q3using Coulomb’s Law:
F13 =k|q1||q3|
r2
13
where r13 is the distance between q1and q3.
Substitute the given values into the equation:
F13 =(9 ×109N·m2/C2)(4 ×10−6C)(2 ×10−6C)
(3 m−(−2m))2
F13 =72 ×10−12
25 = 2.88 ×10−12 N
Step 3: Calculate the net force on q1. The net force on q1is the vector
sum of F12 and F13. Since F12 and F13 are in opposite directions, we need to
subtract them:
Fnet =F12 −F13 = 5.4×10−11 N−2.88 ×10−12 N
17
Fnet = 5.112 ×10−11 N
Therefore, the net electrostatic force acting on q1due to the other two
charges is 5.112 ×10−11 N.
Question 22
Question
Two point charges, q1=−4µC and q2= 2 µC, are placed 10 cm apart along the
x-axis. Calculate the magnitude and direction of the electric force experienced
by each charge.
Solution
Step 1: Determine the distance between the charges Given that the charges are
10 cm apart along the x-axis, the distance between them is r= 10 cm = 0.10
m.
Step 2: Calculate the magnitude of the electric force experienced by q1The
magnitude of the electric force between two point charges q1and q2is given by
Coulomb’s Law:
F=k|q1q2|
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant.
Substitute q1=−4µC, q2= 2 µC, and r= 0.10 m into the formula:
F=(8.99 ×109N·m2/C2)|(−4×10−6C)(2 ×10−6C)|
(0.10 m)2
Calculating this expression gives the magnitude of the electric force experi-
enced by q1.
Step 3: Determine the direction of the electric force experienced by q1The
direction of the electric force experienced by q1will be attractive towards q2.
Step 4: Calculate the magnitude of the electric force experienced by q2Using
the same formula as in Step 2, but switching the signs of q1and q2, calculate
the electric force experienced by q2.
Step 5: Determine the direction of the electric force experienced by q2The
direction of the electric force experienced by q2will be repulsive away from q1.
Question 23
Question
A charge of −2.0µC is located at the origin, and a charge of 4.0µC is located
at the point (0,3.0) m. Calculate the magnitude and direction of the electric
force on the positive charge due to the negative charge.
18
Solution
To find the electric force, we will use Coulomb’s Law, given by:
Electric force =k·|q1·q2|
r2
where kis Coulomb’s constant (8.99 ×109Nm2/C2), q1and q2are the two
charges, and ris the distance between the charges.
Step 1: Calculate the distance rbetween the charges
The distance rbetween the charges can be found using the distance formula
in two dimensions:
r=√(0 −0)2+ (3.0−0)2
r=√0+9.0
r=√9.0
r= 3.0m
Step 2: Calculate the magnitude of the electric force
Now, we can substitute the given values into Coulomb’s Law:
Electric force = (8.99 ×109)·| − 2.0×10−6×4.0×10−6|
(3.0)2
Electric force = 8.99 ×109·8.0×10−12
9.0
Electric force = 8.99 ×109·8.89 ×10−13
Electric force = 7.99 ×10−3N
Step 3: Determine the direction of the electric force
The electric force will be attractive since the charges are of opposite signs.
Therefore, the direction of the electric force is towards the negative charge (from
the positive charge to the negative charge) along the line connecting the charges.
Question 24
Question
Two point charges, q1= +5 µC and q2=−3µC, are placed 10 cm apart. What
is the magnitude of the electric force between these two charges?
19
Solution
Step 1: Convert the charges to Coulombs. The charges given are q1= +5 µC
and q2=−3µC. To convert microcoulombs to coulombs, we use the following
conversion factor: 1µC= 10−6C. Therefore, q1= +5 ×10−6C= 5 ×10−6C
and q2=−3×10−6C=−3×10−6C.
Step 2: Calculate the distance between the charges in meters. The charges
are placed 10 cm apart, which is 0.10 meters.
Step 3: Calculate the electric force between the charges using Coulomb’s
Law. Coulomb’s Law states that the magnitude of the electric force between
two point charges is given by:
F=k· |q1·q2|
r2
where F= electric force, k= Coulomb’s constant (8.99 ×109N m2/C2), q1, q2
= magnitudes of the two charges, and r= distance between the charges.
Substitute the known values into the formula:
F=(8.99 ×109N m2/C2)· |(5 ×10−6C)·(−3×10−6C)|
(0.10 m)2
Step 4: Calculate the electric force.
F=(8.99 ×109)·(5 ×10−6)·(3 ×10−6)
0.01
F=134850
100
F= 1348.50 N
Therefore, the magnitude of the electric force between the two charges is
1348.50 N.
Question 25
Question
Two point charges, q1=−3.00 µC and q2= 5.00 µC, are placed 10.0 cm apart
in vacuum.
Calculate the magnitude of the electric force between the charges, and de-
termine the direction of the force exerted on each charge.
Solution
Step 1: Convert the charges to coulombs. The charges are given in micro-
coulombs (µC), so we need to convert them to coulombs. q1=−3.00 µC=
−3.00 ×10−6Cq2= 5.00 µC= 5.00 ×10−6C
20
Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force between two point charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant and r= 0.10 m is
the separation distance between the charges. Substituting in the given values:
F=8.99 ×109·3.00 ×10−6·5.00 ×10−6
(0.10)2
Step 3: Calculate the numerical value of the electric force.
F=8.99 ×3×5
103=134.85
103= 0.13485 N
Step 4: Determine the direction of the force. The force between the charges
is attractive since the charges have opposite signs.
Therefore, the magnitude of the electric force between the charges is 0.13485 N,
and the force exerted on q1is attractive towards q2, while the force exerted on
q2is attractive towards q1.
21
F= 8.99 ×109N m2/C2·8×10−16 C2
F= 7.192 ×10−6N
Therefore, the magnitude of the electric force between the charges is 7.192 ×
10−6N.
Question 2
Question
Two point charges, q1=−4µC and q2= 8 µC, are separated by a distance of
0.5m. Calculate the magnitude of the electric force that q1exerts on q2.
Solution
Step 1: Determine the electric force between the two charges using Coulomb’s
Law:
F=k·|q1·q2|
r2,
where k= 8.99 ×109N m2/C2is the Coulomb’s constant, q1and q2are the
charges, and ris the distance between the charges.
Step 2: Calculate the force by plugging in the given values:
F= (8.99 ×109)·|(−4×10−6)·(8 ×10−6)|
(0.5)2.
Step 3: Simplify the expression:
F= (8.99 ×109)·32 ×10−12
0.25 .
F= (8.99 ×109)·128 ×10−12.
Step 4: Perform the multiplication:
F= 1150.72 ×10−3.
Step 5: Express the result in Newtons by converting the magnitude to sci-
entific notation:
F= 1.15072 N.
Therefore, the magnitude of the electric force that q1exerts on q2is 1.15072 N.
Question 3
Question
Two point charges are fixed in place, with charge q1=−2.0µC at coordinates
(0,0) and charge q2= 4.0µC at coordinates (2.0m,0). Calculate the magnitude
and direction of the electric force that q1exerts on q2.
2
Solution
Step 1: Calculate the distance between the two charges.
Let rbe the distance between charge q1and charge q2. Using the distance
formula in 2D space:
r=√(2.0m)2+ (0 −0)2=√4.0m2= 2.0m
Step 2: Calculate the magnitude of the electric force.
The electric force between two point charges is given by Coulomb’s law:
F=k|q1q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the equation:
F= (8.99 ×109)|(−2.0×10−6)(4.0×10−6)|
(2.0)2
F= (8.99 ×109)8.0×10−12
4.0= (8.99 ×109)×2.0×10−12 = 1.80 ×10−2N
The magnitude of the electric force is 1.80 ×10−2N.
Step 3: Determine the direction of the electric force.
The direction of the electric force is along the line joining the two charges. Since
charge q1has a negative charge, the force on q2will be attractive towards q1in
the negative xdirection.
Therefore, the electric force that q1exerts on q2is 1.80 ×10−2N in the
negative xdirection.
Question 4
Question
Two point charges, q1=−5µC and q2= 8 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges in
Newtons.
Solution
Step 1: Convert the charges to coulombs (µC to C). Given charges: q1=−5µC
and q2= 8 µC
Converting to coulombs: q1=−5×10−6C and q2= 8 ×10−6C
Step 2: Calculate the electric force using Coulomb’s Law: The magnitude of
the electric force between two point charges is given by Coulomb’s Law:
F=k· |q1·q2|
r2
3
where: F= electric force between the charges, k= Coulomb’s constant (8.99 ×
109N m2/C2), q1and q2= magnitudes of the point charges, r= distance be-
tween the charges.
In this case: k= 8.99 ×109N m2/C2,q1=−5×10−6C, q2= 8 ×10−6C,
r= 0.1m (since 10 cm is 0.1 m).
Step 3: Plug the values into Coulomb’s Law and calculate the electric force:
F=8.99 ×109×|−5×10−6×8×10−6|
(0.1)2
F=8.99 ×109×5×8×10−12
0.01
F=8.99 ×5×8
10 ×10−3
F= 8.99 ×5×0.8×10−3
F= 35.96 ×10−3
F= 0.03596 N
Therefore, the magnitude of the electric force between the charges is 0.03596
N.
Question 5
Question
Three point charges are placed on the vertices of an equilateral triangle. The
charges are +2µC,−4µC, and +3µC. Calculate the net electric force on the
charge at the vertex with the +2µC charge.
Solution
Step 1: We first need to calculate the distance between the charges. Since
the charges are placed on the vertices of an equilateral triangle, the distances
between them are equal. Let’s denote this distance as d.
Step 2: The electric force between two charges q1and q2separated by dis-
tance ris given by Coulomb’s law:
F=k·|q1·q2|
r2,
where k≈8.99 ×109N m2/C2is the Coulomb’s constant.
4
Step 3: Let’s first calculate the net electric force on the +2µC charge due
to the −4µC and +3µC charges. The force on +2µC due to −4µC can be
calculated using Coulomb’s law, and the direction of this force will be attractive.
Step 4: The force on +2µC due to +3µC can be calculated using Coulomb’s
law as well, and the direction of this force will be repulsive.
Step 5: Once we have the magnitudes and directions of the forces due to
each charge, we can find the net force on the +2µC charge by vector addition
of individual forces.
Step 6: Calculate the net electric force on the charge at the vertex with the
+2µC charge.
Question 7
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below:
+Q
+Q
−2Q
The charges are +Qat each of the two upper vertices and −2Qat the lower
vertex. Find the net electrostatic force on the charge at the lower vertex due to
the other two charges.
Solution
Step 1: Calculate the magnitude of the force between the charge at the lower
vertex (−2Q) and the charge at the upper left vertex (+Q).
F1=k· | + 2Q||Q|
r2
where kis the Coulomb’s constant (8.99×109N·m2/C2), |+2Q|= 2Q,|Q|=Q,
and ris the distance between the charges. Since the charges are at the vertices
of an equilateral triangle, the distance between the charges is equal to the side
length of the triangle.
5
Step 2: Calculate the magnitude of the force between the charge at the lower
vertex and the charge at the upper right vertex (+Q) using the same formula
as in Step 1.
F2=k· | + 2Q||Q|
r2
Step 3: Determine the direction of the forces. The force between the charge
at the lower vertex and the charge at the upper left vertex is repulsive, and the
force between the charge at the lower vertex and the charge at the upper right
vertex is also repulsive. Thus, both forces act away from the charge at the lower
vertex.
Step 4: Calculate the net force on the charge at the lower vertex by summing
the individual forces vectorially. The net force will be directed away from the
charge at the lower vertex and can be found using the Pythagorean theorem:
Fnet =√F2
1+F2
2
Question 8
Question
Three point charges are arranged along a straight line as shown below:
+3µC−2µC+5µC
The charges are spaced 10 cm apart with the +3 µC charge at the origin,
the -2 µC charge at 10 cm, and the +5 µC charge at 20 cm. Calculate the net
electric force on the -2 µC charge due to the other two charges.
Solution
Let’s denote the charges as follows: - Charge at 0cm: q1= +3µC - Charge at
10 cm: q2=−2µC - Charge at 20 cm: q3= +5µC
The net electric force on the -2 µC charge (q2) is the sum of the forces due to
the other two charges. We will first calculate the force between q1and q2, and
then the force between q3and q2, and finally sum them to find the net force.
Step 1: Calculate the force between q1and q2. The formula for the magni-
tude of the force between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
Where: - kis the Coulomb’s constant (8.99 ×109N m2/C2), - q1and q2are
the magnitudes of the charges, and - ris the distance between the charges.
Given q1= +3µC, q2=−2µC, and r= 10 cm = 0.1m, we have:
6
F12 =8.99 ×109· |3×10−6·(−2×10−6)|
(0.1)2
F12 =8.99 ×109·6×10−12
0.01
F12 =53.94 ×10−3
0.01
F12 = 5.394 N
So, the force between q1and q2is 5.394 N (pointing to the left).
Step 2: Calculate the force between q2and q3. This calculation follows the
same process as in Step 1.
Step 3: Sum the forces to find the net force on q2.
Thus, the net force on the -2 µC charge due to the other two charges is 5.394
N to the left.
Question 9
Question
Two point charges, +2µC and −3µC, are placed 10 cm apart in air. Calculate
the magnitude of the electric force between them.
Solution
Step 1: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k|q1·q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
charges (+2µC and −3µC), and ris the distance between the charges (0.1 m).
Step 2: Substitute the given values into the equation:
F= (8.99 ×109)|2×10−6· −3×10−6|
(0.1)2
Step 3: Calculate the magnitude of the electric force:
F= (8.99 ×109)6×10−12
0.01
F= 8.99 ×109×6×10−10
F= 53.94 ×10−1
F= 5.394N
Therefore, the magnitude of the electric force between the charges +2µC
and −3µC is 5.394N.
7
Question 10
Question
Three point charges are arranged in the following configuration: a charge of
+4.0µC at the origin, a charge of −2.0µC at coordinates (0,0,3m), and a
charge of +3.0µC at coordinates (4 m,0,0). Calculate the net electric force on
the charge at the origin due to the other two charges.
Solution
The electric force
Felec between two charges q1and q2separated by a distance
ris given by Coulomb’s law:
Felec =kq1q2
r2ˆr
where k≈8.99×109N·m2/C2is the Coulomb constant and ˆris the unit vector
pointing from q2to q1.
Step 1: Find the force on the charge at the origin due to the charge at
coordinates (0,0,3m). The distance between these two charges is r1= 3 m.
The electric force can be calculated as:
|
F1|=k|(+4.0µC)(−2.0µC)|
(3 m)2=8.99 ×109×4.0×2.0×10−12
9= 8.0×10−3N
Step 2: Find the force on the charge at the origin due to the charge at
coordinates (4 m,0,0). The distance between these two charges is r2= 4 m.
The electric force can be calculated as:
|
F2|=k|(+4.0µC)(+3.0µC)|
(4 m)2=8.99 ×109×4.0×3.0×10−12
16 = 6.7×10−3N
Step 3: Find the net force on the charge at the origin. The net force is the
vector sum of the individual forces. Since one force acts in the z-direction and
the other in the x-direction, the net force will have both xand zcomponents.
Fnet =
F1+
F2Calculating the magnitudes of the components:
Fnet,x =F2= 6.7×10−3N
Fnet,z =F1= 8.0×10−3N
Therefore, the net electric force on the charge at the origin is 6.7×10−3N
in the x-direction and 8.0×10−3N in the z-direction.
Question 11
Question
Three point charges are arranged in an equilateral triangle with sides of length d.
The charges are +qat point A,−2qat point B, and +3qat point C. Calculate
the net electric force on the charge at point Adue to the other two charges.
8
Solution
To find the net electric force on the charge at point A, we need to calculate the
electric forces between Aand B, and between Aand C, and then add these
forces as vectors to get the net force.
Step 1: Find the force between Aand B.
The electric force between two point charges q1and q2separated by a dis-
tance ris given by Coulomb’s Law:
F=kq1q2
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
The distance between Aand Bis also dbecause they are at opposite ends of
the side of length dof the equilateral triangle. Therefore, the net electric force
between Aand Bis:
FAB =k|+q|·|−2q|
d2=k2q2
d2
Step 2: Find the force between Aand C.
The distance between Aand Cis das well due to the equilateral triangle
arrangement. Therefore, the net electric force between Aand Cis:
FAC =k|+q|·|3q|
d2=k3q2
d2
Step 3: Find the net force on A.
The net force on the charge at point Ais the vector sum of FAB and FAC .
Since these forces are in opposite directions and have the same magnitude, the
net force is simply given by their difference:
Net Force on A =|FAC |−|FAB |=k3q2
d2−k2q2
d2=kq2
d2
Therefore, the net electric force on the charge at point Adue to the other
two charges is kq2
d2.
Question 12
Question
Two point charges, q1=−3.0µC and q2= 5.0µC, are placed 20 cm apart in a
vacuum. Calculate the magnitude of the electric force between them.
9
Solution
Step 1: Convert the given charges to coulombs.
q1=−3.0µC =−3.0×10−6C
q2= 5.0µC = 5.0×10−6C
Step 2: Write down the formula for the electric force between two point
charges. The magnitude of the electric force between two point charges is given
by Coulomb’s law:
F=k· |q1·q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant, q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 3: Substitute the given values into the formula.
F=(8.99 ×109N·m2/C2)· |(−3.0×10−6C)·(5.0×10−6C)|
(0.20 m)2
Step 4: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(3.0×5.0) ×10−12
0.04
=1.35 ×10−11 ×8.99 ×109
0.04
=1.21365 ×10−1
0.04
= 3.034 ×10−10 N
Step 5: Therefore, the magnitude of the electric force between the two
charges is 3.034 ×10−10 N.
Question 13
Question
Three point charges are arranged along the x-axis as follows: +3.0 nC at x = 0
m, -6.0 nC at x = 2.0 m, and +4.0 nC at x = 3.0 m. Calculate the net electric
force on the +4.0 nC charge due to the other two charges.
10
Solution
Step 1: Calculate the electric force on the +4.0 nC charge due to the +3.0 nC
charge. The electric force between two charges is given by Coulomb’s Law:
F=kq1q2
r2
where: F= electric force k= Coulomb’s constant = 8.9875×109Nm²/C² q1, q2
= magnitudes of the charges r= distance between the charges
Plugging in the values: k= 8.9875 ×109Nm²/C² q1= +3.0×10−9C
q2= +4.0×10−9Cr= 3.0m
F1= 8.9875 ×109×(3.0×10−9)(4.0×10−9)
3.02
F1= 8.9875 ×109×12 ×10−18
9
F1= 8.9875 ×109×1.3333 ×10−18
F1= 11.98 ×10−9
F1= 1.198 ×10−8N
Step 2: Calculate the electric force on the +4.0 nC charge due to the -6.0
nC charge. Following the same steps as above with q1=−6.0×10−9C:
F2= 8.9875 ×109×(6.0×10−9)(4.0×10−9)
1
F2= 8.9875 ×109×24.0×10−18
F2= 215.7×10−9
F2= 2.157 ×10−7N
Step 3: Calculate the net electric force on the +4.0 nC charge. Since the
forces are in opposite directions (one repelling and the other attracting), we
need to find the difference between them:
Net force =F1−F2= 1.198 ×10−8−2.157 ×10−7
Net force =−1.037 ×10−7N
Therefore, the net electric force on the +4.0 nC charge due to the other two
charges is −1.037 ×10−7N, directed towards the -6.0 nC charge.
Question 14
Question
Two point charges are placed at the two opposite corners of a square with sides
of length a. The charges have magnitudes qand 2q. Calculate the magnitude of
the electric force between the charges along one of the diagonals of the square.
11
Solution
Let’s denote the magnitude of the charge qas q1=qand the magnitude of the
charge 2qas q2= 2q.
Step 1: Calculate the electric force between q1and q2along one of the
diagonals of the square. Let’s choose the diagonal from q1to q2.
Using the formula for the magnitude of the electric force between two point
charges, we have:
F=k|q1q2|
r2
where - kis the Coulomb constant, - q1and q2are the magnitudes of the charges,
and - ris the distance between the charges.
First, let’s find the distance ralong the diagonal of the square: Since the
charges are at opposite corners of a square with sides of length a, the diagonal
length can be calculated using the Pythagorean theorem:
r=√a2+a2=√2a
Step 2: Substitute the values of q1,q2, and rinto the formula for the electric
force:
F=k|q1q2|
r2=k|q(2q)|
(√2a)2=k2q2
2a2=kq2
a2
Therefore, the magnitude of the electric force between the charges qand 2q
along one of the diagonals of the square is kq2
a2.
Question 15
Question
Two point charges, q1=−3µC and q2= 5 µC, are placed 10 cm apart along
the x-axis. Calculate the magnitude and direction of the electric force that q2
exerts on q1.
Solution
Step 1: First, calculate the electric force between the two charges using Coulomb’s
Law:
F=k·|q1·q2|
r2
where Fis the magnitude of the electric force, kis the electrostatic constant
(8.99 ×109N m2/C2), q1and q2are the charges, and ris the distance between
the charges.
Step 2: Substitute the given values into the formula:
F= (8.99 ×109)·|(−3×10−6)(5 ×10−6)|
(0.1)2
12
Step 3: Calculate the electric force:
F= (8.99 ×109)·15 ×10−12
0.01
F= (8.99 ×109)·1.5×10−9
F= 13.485 ×100N
F= 13.485 N
Step 4: Now, determine the direction of the force. The force will be attractive
since the charges are of opposite signs. Therefore, the force will act along the
line joining the charges, i.e., along the negative x-axis.
Therefore, the magnitude of the electric force that q2exerts on q1is 13.485
N along the negative x-axis.
Question 16
Question
Two point charges, q1=−2.5µC and q2= 3.0µC, are placed 10 cm apart in
air. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to standard units. Step 2: Calculate the distance
between the charges. Step 3: Use Coulomb’s Law to calculate the electric force.
Step 4: Provide the final answer with appropriate units.
Step 1: Convert the charges to standard units. q1=−2.5µC =−2.5×
10−6C
q2= 3.0µC = 3.0×10−6C
Step 2: Calculate the distance between the charges. Given that the charges
are placed 10 cm apart, the distance rbetween them is r= 10 cm = 0.10 m.
Step 3: Use Coulomb’s Law to calculate the electric force. The magnitude
of the electric force between two charges is given by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109Nm2/C2).
Plugging in the values:
F= 8.99 ×109Nm2/C2·| − 2.5×10−6C·3.0×10−6C|
(0.10 m)2
Calculating the force will give us the magnitude of the electric force between
the charges.
Step 4: Provide the final answer with appropriate units.
F= 53.7×10−6N= 53.7µN
Therefore, the magnitude of the electric force between the charges is 53.7µN .
13
Question 17
Question
Two point charges, q1= 4.0×10−6C and q2=−6.0×10−6C, are placed 0.30
m apart. Calculate the magnitude and direction of the electric force that q2
exerts on q1.
Solution
Step 1: Calculate the electric force magnitude using Coulomb’s Law.
F=k|q1||q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Step 2: Substitute the given values into the formula.
F=(8.99 ×109N m2/C2)(4.0×10−6C)(6.0×10−6C)
(0.30 m)2
Step 3: Calculate the electric force magnitude.
F=(8.99 ×109)(4.0×10−6)(6.0×10−6)
(0.30)2
F=215.76
0.09
F= 2397.33 N
Step 4: Determine the direction of the force. The force is attractive in this
case because the charges q1and q2have opposite signs. Thus, the force points
from q2to q1.
Therefore, the magnitude of the electric force that q2exerts on q1is 2397.33 N
and the direction is from q2to q1.
Question 18
Question
Two point charges, q1= 3.0µC and q2=−2.0µC, are located 10.0 cm apart.
Calculate the magnitude of the electric force between these charges.
14
Solution
Step 1: Convert the charges to Coulombs. Given: q1= 3.0µC,q2=−2.0µC
1 microCoulomb (µC) = 1×10−6Coulombs So, q1= 3.0×10−6C and q2=
−2.0×10−6C.
Step 2: Calculate the magnitude of the electric force using Coulomb’s Law.
The electric force Fbetween two point charges q1and q2separated by a distance
ris given by Coulomb’s Law:
F=k·|q1·q2|
r2
where k= 8.99 ×109Nm2/C2is Coulomb’s constant.
Plugging in the given values: F= (8.99 ×109)·|3.0×10−6·−2.0×10−6|
(0.10)2
Step 3: Calculate the magnitude of the electric force.
F= 8.99 ×109·6.0×10−12
0.01
F= 8.99 ×109·6.0×10−10
F= 53.94 ×10−1
F= 5.39 ×100N
Therefore, the magnitude of the electric force between these charges is 5.39
N.
Question 19
Question
Three point charges are arranged along the x-axis as follows: +3.00 µC at the
origin, −4.00 µC at x= 50.0cm, and +5.00 µC at x= 70.0cm. What is the net
electric force on the −4.00 µC charge?
Solution
1. Calculate the electric force on the −4.00 µC charge due to the +3.00 µC
charge at the origin. The electric force F1between two point charges is given
by Coulomb’s Law:
F1=kq1q2
r2
where - kis the Coulomb constant (8.99 ×109N·m2/C2), - q1and q2are
the magnitudes of the point charges (3.00 µC and −4.00 µC, respectively), - r
is the distance between the charges (origin to x= 50.0cm = 0.50 m).
Plugging in the values, we get:
F1= (8.99 ×109)(3.00 ×10−6)·(4.00 ×10−6)
(0.50)2
15
Calculating the result gives F1=−4.79 N, directed towards the origin.
2. Calculate the electric force on the −4.00 µC charge due to the +5.00 µC
charge at x= 70.0cm. Using Coulomb’s Law again, this time with q2=
+5.00 µC and r= 0.20 m, we have:
F2= (8.99 ×109)(5.00 ×10−6)·(4.00 ×10−6)
(0.20)2
Calculating the result gives F2= 44.94 N, directed towards the positive
charge.
3. The net force is the vector sum of F1and F2.
Fnet =F2−F1= 44.94 N−4.79 N= 40.15 N
Therefore, the net electric force on the −4.00 µC charge is 40.15 N directed
towards the +5.00 µC charge at x= 70.0cm.
Question 20
Question
Two point charges, q1=−2.5µC and q2= 4.0µC, are located at points
A(−2.0m,0) and B(3.0m,0) respectively on the x-axis. Calculate the electric
force exerted on q2due to q1.
Solution
Step 1: First, we need to calculate the distance between the two charges. Given:
xA=−2.0m
xB= 3.0m
Let dbe the distance between the charges. d=xB−xA= 3.0m−(−2.0m) =
5.0m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law:
F=k|q1·q2|
d2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Step 3: Substituting the given values into the formula:
F= 8.99 ×109N m2/C2·| − 2.5×10−6C·4.0×10−6C|
(5.0m)2
F= 8.99 ×109·10 ×10−12
25
F= 8.99 ×109·4×10−12
F= 35.96 ×10−3
16
F= 3.596 mN
Therefore, the electric force exerted on q2due to q1is 3.596 mN directed
towards q1.
Question 21
Question
Three point charges are fixed on the x-axis. The charges are located at x=−2
m, x= 0 m, and x= 3 m, with magnitudes q1= 4 µC, q2=−6µC, and q3= 2
µC, respectively. Calculate the net electrostatic force acting on q1due to the
other two charges.
Solution
Step 1: Calculate the force on q1due to q2. The magnitude of the force F12 can
be calculated using Coulomb’s Law:
F12 =k|q1||q2|
r2
12
where kis the Coulomb’s constant, r12 is the distance between q1and q2, and
|q1|and |q2|are the magnitudes of the charges.
Substitute the given values into the equation:
F12 =(9 ×109N·m2/C2)(4 ×10−6C)(6 ×10−6C)
(−2m)2
F12 =216 ×10−12
4= 54 ×10−12 = 5.4×10−11 N
Step 2: Calculate the force on q1due to q3. Similarly, calculate the force
F13 on q1due to q3using Coulomb’s Law:
F13 =k|q1||q3|
r2
13
where r13 is the distance between q1and q3.
Substitute the given values into the equation:
F13 =(9 ×109N·m2/C2)(4 ×10−6C)(2 ×10−6C)
(3 m−(−2m))2
F13 =72 ×10−12
25 = 2.88 ×10−12 N
Step 3: Calculate the net force on q1. The net force on q1is the vector
sum of F12 and F13. Since F12 and F13 are in opposite directions, we need to
subtract them:
Fnet =F12 −F13 = 5.4×10−11 N−2.88 ×10−12 N
17
Fnet = 5.112 ×10−11 N
Therefore, the net electrostatic force acting on q1due to the other two
charges is 5.112 ×10−11 N.
Question 22
Question
Two point charges, q1=−4µC and q2= 2 µC, are placed 10 cm apart along the
x-axis. Calculate the magnitude and direction of the electric force experienced
by each charge.
Solution
Step 1: Determine the distance between the charges Given that the charges are
10 cm apart along the x-axis, the distance between them is r= 10 cm = 0.10
m.
Step 2: Calculate the magnitude of the electric force experienced by q1The
magnitude of the electric force between two point charges q1and q2is given by
Coulomb’s Law:
F=k|q1q2|
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant.
Substitute q1=−4µC, q2= 2 µC, and r= 0.10 m into the formula:
F=(8.99 ×109N·m2/C2)|(−4×10−6C)(2 ×10−6C)|
(0.10 m)2
Calculating this expression gives the magnitude of the electric force experi-
enced by q1.
Step 3: Determine the direction of the electric force experienced by q1The
direction of the electric force experienced by q1will be attractive towards q2.
Step 4: Calculate the magnitude of the electric force experienced by q2Using
the same formula as in Step 2, but switching the signs of q1and q2, calculate
the electric force experienced by q2.
Step 5: Determine the direction of the electric force experienced by q2The
direction of the electric force experienced by q2will be repulsive away from q1.
Question 23
Question
A charge of −2.0µC is located at the origin, and a charge of 4.0µC is located
at the point (0,3.0) m. Calculate the magnitude and direction of the electric
force on the positive charge due to the negative charge.
18
Solution
To find the electric force, we will use Coulomb’s Law, given by:
Electric force =k·|q1·q2|
r2
where kis Coulomb’s constant (8.99 ×109Nm2/C2), q1and q2are the two
charges, and ris the distance between the charges.
Step 1: Calculate the distance rbetween the charges
The distance rbetween the charges can be found using the distance formula
in two dimensions:
r=√(0 −0)2+ (3.0−0)2
r=√0+9.0
r=√9.0
r= 3.0m
Step 2: Calculate the magnitude of the electric force
Now, we can substitute the given values into Coulomb’s Law:
Electric force = (8.99 ×109)·| − 2.0×10−6×4.0×10−6|
(3.0)2
Electric force = 8.99 ×109·8.0×10−12
9.0
Electric force = 8.99 ×109·8.89 ×10−13
Electric force = 7.99 ×10−3N
Step 3: Determine the direction of the electric force
The electric force will be attractive since the charges are of opposite signs.
Therefore, the direction of the electric force is towards the negative charge (from
the positive charge to the negative charge) along the line connecting the charges.
Question 24
Question
Two point charges, q1= +5 µC and q2=−3µC, are placed 10 cm apart. What
is the magnitude of the electric force between these two charges?
19
Solution
Step 1: Convert the charges to Coulombs. The charges given are q1= +5 µC
and q2=−3µC. To convert microcoulombs to coulombs, we use the following
conversion factor: 1µC= 10−6C. Therefore, q1= +5 ×10−6C= 5 ×10−6C
and q2=−3×10−6C=−3×10−6C.
Step 2: Calculate the distance between the charges in meters. The charges
are placed 10 cm apart, which is 0.10 meters.
Step 3: Calculate the electric force between the charges using Coulomb’s
Law. Coulomb’s Law states that the magnitude of the electric force between
two point charges is given by:
F=k· |q1·q2|
r2
where F= electric force, k= Coulomb’s constant (8.99 ×109N m2/C2), q1, q2
= magnitudes of the two charges, and r= distance between the charges.
Substitute the known values into the formula:
F=(8.99 ×109N m2/C2)· |(5 ×10−6C)·(−3×10−6C)|
(0.10 m)2
Step 4: Calculate the electric force.
F=(8.99 ×109)·(5 ×10−6)·(3 ×10−6)
0.01
F=134850
100
F= 1348.50 N
Therefore, the magnitude of the electric force between the two charges is
1348.50 N.
Question 25
Question
Two point charges, q1=−3.00 µC and q2= 5.00 µC, are placed 10.0 cm apart
in vacuum.
Calculate the magnitude of the electric force between the charges, and de-
termine the direction of the force exerted on each charge.
Solution
Step 1: Convert the charges to coulombs. The charges are given in micro-
coulombs (µC), so we need to convert them to coulombs. q1=−3.00 µC=
−3.00 ×10−6Cq2= 5.00 µC= 5.00 ×10−6C
20
Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force between two point charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant and r= 0.10 m is
the separation distance between the charges. Substituting in the given values:
F=8.99 ×109·3.00 ×10−6·5.00 ×10−6
(0.10)2
Step 3: Calculate the numerical value of the electric force.
F=8.99 ×3×5
103=134.85
103= 0.13485 N
Step 4: Determine the direction of the force. The force between the charges
is attractive since the charges have opposite signs.
Therefore, the magnitude of the electric force between the charges is 0.13485 N,
and the force exerted on q1is attractive towards q2, while the force exerted on
q2is attractive towards q1.
21
F= 8.99 ×109N m2/C2·8×10−16 C2
F= 7.192 ×10−6N
Therefore, the magnitude of the electric force between the charges is 7.192 ×
10−6N.
Question 2
Question
Two point charges, q1=−4µC and q2= 8 µC, are separated by a distance of
0.5m. Calculate the magnitude of the electric force that q1exerts on q2.
Solution
Step 1: Determine the electric force between the two charges using Coulomb’s
Law:
F=k·|q1·q2|
r2,
where k= 8.99 ×109N m2/C2is the Coulomb’s constant, q1and q2are the
charges, and ris the distance between the charges.
Step 2: Calculate the force by plugging in the given values:
F= (8.99 ×109)·|(−4×10−6)·(8 ×10−6)|
(0.5)2.
Step 3: Simplify the expression:
F= (8.99 ×109)·32 ×10−12
0.25 .
F= (8.99 ×109)·128 ×10−12.
Step 4: Perform the multiplication:
F= 1150.72 ×10−3.
Step 5: Express the result in Newtons by converting the magnitude to sci-
entific notation:
F= 1.15072 N.
Therefore, the magnitude of the electric force that q1exerts on q2is 1.15072 N.
Question 3
Question
Two point charges are fixed in place, with charge q1=−2.0µC at coordinates
(0,0) and charge q2= 4.0µC at coordinates (2.0m,0). Calculate the magnitude
and direction of the electric force that q1exerts on q2.
2
Solution
Step 1: Calculate the distance between the two charges.
Let rbe the distance between charge q1and charge q2. Using the distance
formula in 2D space:
r=√(2.0m)2+ (0 −0)2=√4.0m2= 2.0m
Step 2: Calculate the magnitude of the electric force.
The electric force between two point charges is given by Coulomb’s law:
F=k|q1q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the equation:
F= (8.99 ×109)|(−2.0×10−6)(4.0×10−6)|
(2.0)2
F= (8.99 ×109)8.0×10−12
4.0= (8.99 ×109)×2.0×10−12 = 1.80 ×10−2N
The magnitude of the electric force is 1.80 ×10−2N.
Step 3: Determine the direction of the electric force.
The direction of the electric force is along the line joining the two charges. Since
charge q1has a negative charge, the force on q2will be attractive towards q1in
the negative xdirection.
Therefore, the electric force that q1exerts on q2is 1.80 ×10−2N in the
negative xdirection.
Question 4
Question
Two point charges, q1=−5µC and q2= 8 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges in
Newtons.
Solution
Step 1: Convert the charges to coulombs (µC to C). Given charges: q1=−5µC
and q2= 8 µC
Converting to coulombs: q1=−5×10−6C and q2= 8 ×10−6C
Step 2: Calculate the electric force using Coulomb’s Law: The magnitude of
the electric force between two point charges is given by Coulomb’s Law:
F=k· |q1·q2|
r2
3
where: F= electric force between the charges, k= Coulomb’s constant (8.99 ×
109N m2/C2), q1and q2= magnitudes of the point charges, r= distance be-
tween the charges.
In this case: k= 8.99 ×109N m2/C2,q1=−5×10−6C, q2= 8 ×10−6C,
r= 0.1m (since 10 cm is 0.1 m).
Step 3: Plug the values into Coulomb’s Law and calculate the electric force:
F=8.99 ×109×|−5×10−6×8×10−6|
(0.1)2
F=8.99 ×109×5×8×10−12
0.01
F=8.99 ×5×8
10 ×10−3
F= 8.99 ×5×0.8×10−3
F= 35.96 ×10−3
F= 0.03596 N
Therefore, the magnitude of the electric force between the charges is 0.03596
N.
Question 5
Question
Three point charges are placed on the vertices of an equilateral triangle. The
charges are +2µC,−4µC, and +3µC. Calculate the net electric force on the
charge at the vertex with the +2µC charge.
Solution
Step 1: We first need to calculate the distance between the charges. Since
the charges are placed on the vertices of an equilateral triangle, the distances
between them are equal. Let’s denote this distance as d.
Step 2: The electric force between two charges q1and q2separated by dis-
tance ris given by Coulomb’s law:
F=k·|q1·q2|
r2,
where k≈8.99 ×109N m2/C2is the Coulomb’s constant.
4
Step 3: Let’s first calculate the net electric force on the +2µC charge due
to the −4µC and +3µC charges. The force on +2µC due to −4µC can be
calculated using Coulomb’s law, and the direction of this force will be attractive.
Step 4: The force on +2µC due to +3µC can be calculated using Coulomb’s
law as well, and the direction of this force will be repulsive.
Step 5: Once we have the magnitudes and directions of the forces due to
each charge, we can find the net force on the +2µC charge by vector addition
of individual forces.
Step 6: Calculate the net electric force on the charge at the vertex with the
+2µC charge.
Question 7
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below:
+Q
+Q
−2Q
The charges are +Qat each of the two upper vertices and −2Qat the lower
vertex. Find the net electrostatic force on the charge at the lower vertex due to
the other two charges.
Solution
Step 1: Calculate the magnitude of the force between the charge at the lower
vertex (−2Q) and the charge at the upper left vertex (+Q).
F1=k· | + 2Q||Q|
r2
where kis the Coulomb’s constant (8.99×109N·m2/C2), |+2Q|= 2Q,|Q|=Q,
and ris the distance between the charges. Since the charges are at the vertices
of an equilateral triangle, the distance between the charges is equal to the side
length of the triangle.
5
Step 2: Calculate the magnitude of the force between the charge at the lower
vertex and the charge at the upper right vertex (+Q) using the same formula
as in Step 1.
F2=k· | + 2Q||Q|
r2
Step 3: Determine the direction of the forces. The force between the charge
at the lower vertex and the charge at the upper left vertex is repulsive, and the
force between the charge at the lower vertex and the charge at the upper right
vertex is also repulsive. Thus, both forces act away from the charge at the lower
vertex.
Step 4: Calculate the net force on the charge at the lower vertex by summing
the individual forces vectorially. The net force will be directed away from the
charge at the lower vertex and can be found using the Pythagorean theorem:
Fnet =√F2
1+F2
2
Question 8
Question
Three point charges are arranged along a straight line as shown below:
+3µC−2µC+5µC
The charges are spaced 10 cm apart with the +3 µC charge at the origin,
the -2 µC charge at 10 cm, and the +5 µC charge at 20 cm. Calculate the net
electric force on the -2 µC charge due to the other two charges.
Solution
Let’s denote the charges as follows: - Charge at 0cm: q1= +3µC - Charge at
10 cm: q2=−2µC - Charge at 20 cm: q3= +5µC
The net electric force on the -2 µC charge (q2) is the sum of the forces due to
the other two charges. We will first calculate the force between q1and q2, and
then the force between q3and q2, and finally sum them to find the net force.
Step 1: Calculate the force between q1and q2. The formula for the magni-
tude of the force between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
Where: - kis the Coulomb’s constant (8.99 ×109N m2/C2), - q1and q2are
the magnitudes of the charges, and - ris the distance between the charges.
Given q1= +3µC, q2=−2µC, and r= 10 cm = 0.1m, we have:
6
F12 =8.99 ×109· |3×10−6·(−2×10−6)|
(0.1)2
F12 =8.99 ×109·6×10−12
0.01
F12 =53.94 ×10−3
0.01
F12 = 5.394 N
So, the force between q1and q2is 5.394 N (pointing to the left).
Step 2: Calculate the force between q2and q3. This calculation follows the
same process as in Step 1.
Step 3: Sum the forces to find the net force on q2.
Thus, the net force on the -2 µC charge due to the other two charges is 5.394
N to the left.
Question 9
Question
Two point charges, +2µC and −3µC, are placed 10 cm apart in air. Calculate
the magnitude of the electric force between them.
Solution
Step 1: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k|q1·q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
charges (+2µC and −3µC), and ris the distance between the charges (0.1 m).
Step 2: Substitute the given values into the equation:
F= (8.99 ×109)|2×10−6· −3×10−6|
(0.1)2
Step 3: Calculate the magnitude of the electric force:
F= (8.99 ×109)6×10−12
0.01
F= 8.99 ×109×6×10−10
F= 53.94 ×10−1
F= 5.394N
Therefore, the magnitude of the electric force between the charges +2µC
and −3µC is 5.394N.
7
Question 10
Question
Three point charges are arranged in the following configuration: a charge of
+4.0µC at the origin, a charge of −2.0µC at coordinates (0,0,3m), and a
charge of +3.0µC at coordinates (4 m,0,0). Calculate the net electric force on
the charge at the origin due to the other two charges.
Solution
The electric force
Felec between two charges q1and q2separated by a distance
ris given by Coulomb’s law:
Felec =kq1q2
r2ˆr
where k≈8.99×109N·m2/C2is the Coulomb constant and ˆris the unit vector
pointing from q2to q1.
Step 1: Find the force on the charge at the origin due to the charge at
coordinates (0,0,3m). The distance between these two charges is r1= 3 m.
The electric force can be calculated as:
|
F1|=k|(+4.0µC)(−2.0µC)|
(3 m)2=8.99 ×109×4.0×2.0×10−12
9= 8.0×10−3N
Step 2: Find the force on the charge at the origin due to the charge at
coordinates (4 m,0,0). The distance between these two charges is r2= 4 m.
The electric force can be calculated as:
|
F2|=k|(+4.0µC)(+3.0µC)|
(4 m)2=8.99 ×109×4.0×3.0×10−12
16 = 6.7×10−3N
Step 3: Find the net force on the charge at the origin. The net force is the
vector sum of the individual forces. Since one force acts in the z-direction and
the other in the x-direction, the net force will have both xand zcomponents.
Fnet =
F1+
F2Calculating the magnitudes of the components:
Fnet,x =F2= 6.7×10−3N
Fnet,z =F1= 8.0×10−3N
Therefore, the net electric force on the charge at the origin is 6.7×10−3N
in the x-direction and 8.0×10−3N in the z-direction.
Question 11
Question
Three point charges are arranged in an equilateral triangle with sides of length d.
The charges are +qat point A,−2qat point B, and +3qat point C. Calculate
the net electric force on the charge at point Adue to the other two charges.
8
Solution
To find the net electric force on the charge at point A, we need to calculate the
electric forces between Aand B, and between Aand C, and then add these
forces as vectors to get the net force.
Step 1: Find the force between Aand B.
The electric force between two point charges q1and q2separated by a dis-
tance ris given by Coulomb’s Law:
F=kq1q2
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
The distance between Aand Bis also dbecause they are at opposite ends of
the side of length dof the equilateral triangle. Therefore, the net electric force
between Aand Bis:
FAB =k|+q|·|−2q|
d2=k2q2
d2
Step 2: Find the force between Aand C.
The distance between Aand Cis das well due to the equilateral triangle
arrangement. Therefore, the net electric force between Aand Cis:
FAC =k|+q|·|3q|
d2=k3q2
d2
Step 3: Find the net force on A.
The net force on the charge at point Ais the vector sum of FAB and FAC .
Since these forces are in opposite directions and have the same magnitude, the
net force is simply given by their difference:
Net Force on A =|FAC |−|FAB |=k3q2
d2−k2q2
d2=kq2
d2
Therefore, the net electric force on the charge at point Adue to the other
two charges is kq2
d2.
Question 12
Question
Two point charges, q1=−3.0µC and q2= 5.0µC, are placed 20 cm apart in a
vacuum. Calculate the magnitude of the electric force between them.
9
Solution
Step 1: Convert the given charges to coulombs.
q1=−3.0µC =−3.0×10−6C
q2= 5.0µC = 5.0×10−6C
Step 2: Write down the formula for the electric force between two point
charges. The magnitude of the electric force between two point charges is given
by Coulomb’s law:
F=k· |q1·q2|
r2
where k= 8.99 ×109N·m2/C2is the Coulomb constant, q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 3: Substitute the given values into the formula.
F=(8.99 ×109N·m2/C2)· |(−3.0×10−6C)·(5.0×10−6C)|
(0.20 m)2
Step 4: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(3.0×5.0) ×10−12
0.04
=1.35 ×10−11 ×8.99 ×109
0.04
=1.21365 ×10−1
0.04
= 3.034 ×10−10 N
Step 5: Therefore, the magnitude of the electric force between the two
charges is 3.034 ×10−10 N.
Question 13
Question
Three point charges are arranged along the x-axis as follows: +3.0 nC at x = 0
m, -6.0 nC at x = 2.0 m, and +4.0 nC at x = 3.0 m. Calculate the net electric
force on the +4.0 nC charge due to the other two charges.
10
Solution
Step 1: Calculate the electric force on the +4.0 nC charge due to the +3.0 nC
charge. The electric force between two charges is given by Coulomb’s Law:
F=kq1q2
r2
where: F= electric force k= Coulomb’s constant = 8.9875×109Nm²/C² q1, q2
= magnitudes of the charges r= distance between the charges
Plugging in the values: k= 8.9875 ×109Nm²/C² q1= +3.0×10−9C
q2= +4.0×10−9Cr= 3.0m
F1= 8.9875 ×109×(3.0×10−9)(4.0×10−9)
3.02
F1= 8.9875 ×109×12 ×10−18
9
F1= 8.9875 ×109×1.3333 ×10−18
F1= 11.98 ×10−9
F1= 1.198 ×10−8N
Step 2: Calculate the electric force on the +4.0 nC charge due to the -6.0
nC charge. Following the same steps as above with q1=−6.0×10−9C:
F2= 8.9875 ×109×(6.0×10−9)(4.0×10−9)
1
F2= 8.9875 ×109×24.0×10−18
F2= 215.7×10−9
F2= 2.157 ×10−7N
Step 3: Calculate the net electric force on the +4.0 nC charge. Since the
forces are in opposite directions (one repelling and the other attracting), we
need to find the difference between them:
Net force =F1−F2= 1.198 ×10−8−2.157 ×10−7
Net force =−1.037 ×10−7N
Therefore, the net electric force on the +4.0 nC charge due to the other two
charges is −1.037 ×10−7N, directed towards the -6.0 nC charge.
Question 14
Question
Two point charges are placed at the two opposite corners of a square with sides
of length a. The charges have magnitudes qand 2q. Calculate the magnitude of
the electric force between the charges along one of the diagonals of the square.
11
Solution
Let’s denote the magnitude of the charge qas q1=qand the magnitude of the
charge 2qas q2= 2q.
Step 1: Calculate the electric force between q1and q2along one of the
diagonals of the square. Let’s choose the diagonal from q1to q2.
Using the formula for the magnitude of the electric force between two point
charges, we have:
F=k|q1q2|
r2
where - kis the Coulomb constant, - q1and q2are the magnitudes of the charges,
and - ris the distance between the charges.
First, let’s find the distance ralong the diagonal of the square: Since the
charges are at opposite corners of a square with sides of length a, the diagonal
length can be calculated using the Pythagorean theorem:
r=√a2+a2=√2a
Step 2: Substitute the values of q1,q2, and rinto the formula for the electric
force:
F=k|q1q2|
r2=k|q(2q)|
(√2a)2=k2q2
2a2=kq2
a2
Therefore, the magnitude of the electric force between the charges qand 2q
along one of the diagonals of the square is kq2
a2.
Question 15
Question
Two point charges, q1=−3µC and q2= 5 µC, are placed 10 cm apart along
the x-axis. Calculate the magnitude and direction of the electric force that q2
exerts on q1.
Solution
Step 1: First, calculate the electric force between the two charges using Coulomb’s
Law:
F=k·|q1·q2|
r2
where Fis the magnitude of the electric force, kis the electrostatic constant
(8.99 ×109N m2/C2), q1and q2are the charges, and ris the distance between
the charges.
Step 2: Substitute the given values into the formula:
F= (8.99 ×109)·|(−3×10−6)(5 ×10−6)|
(0.1)2
12
Step 3: Calculate the electric force:
F= (8.99 ×109)·15 ×10−12
0.01
F= (8.99 ×109)·1.5×10−9
F= 13.485 ×100N
F= 13.485 N
Step 4: Now, determine the direction of the force. The force will be attractive
since the charges are of opposite signs. Therefore, the force will act along the
line joining the charges, i.e., along the negative x-axis.
Therefore, the magnitude of the electric force that q2exerts on q1is 13.485
N along the negative x-axis.
Question 16
Question
Two point charges, q1=−2.5µC and q2= 3.0µC, are placed 10 cm apart in
air. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to standard units. Step 2: Calculate the distance
between the charges. Step 3: Use Coulomb’s Law to calculate the electric force.
Step 4: Provide the final answer with appropriate units.
Step 1: Convert the charges to standard units. q1=−2.5µC =−2.5×
10−6C
q2= 3.0µC = 3.0×10−6C
Step 2: Calculate the distance between the charges. Given that the charges
are placed 10 cm apart, the distance rbetween them is r= 10 cm = 0.10 m.
Step 3: Use Coulomb’s Law to calculate the electric force. The magnitude
of the electric force between two charges is given by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109Nm2/C2).
Plugging in the values:
F= 8.99 ×109Nm2/C2·| − 2.5×10−6C·3.0×10−6C|
(0.10 m)2
Calculating the force will give us the magnitude of the electric force between
the charges.
Step 4: Provide the final answer with appropriate units.
F= 53.7×10−6N= 53.7µN
Therefore, the magnitude of the electric force between the charges is 53.7µN .
13
Question 17
Question
Two point charges, q1= 4.0×10−6C and q2=−6.0×10−6C, are placed 0.30
m apart. Calculate the magnitude and direction of the electric force that q2
exerts on q1.
Solution
Step 1: Calculate the electric force magnitude using Coulomb’s Law.
F=k|q1||q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Step 2: Substitute the given values into the formula.
F=(8.99 ×109N m2/C2)(4.0×10−6C)(6.0×10−6C)
(0.30 m)2
Step 3: Calculate the electric force magnitude.
F=(8.99 ×109)(4.0×10−6)(6.0×10−6)
(0.30)2
F=215.76
0.09
F= 2397.33 N
Step 4: Determine the direction of the force. The force is attractive in this
case because the charges q1and q2have opposite signs. Thus, the force points
from q2to q1.
Therefore, the magnitude of the electric force that q2exerts on q1is 2397.33 N
and the direction is from q2to q1.
Question 18
Question
Two point charges, q1= 3.0µC and q2=−2.0µC, are located 10.0 cm apart.
Calculate the magnitude of the electric force between these charges.
14
Solution
Step 1: Convert the charges to Coulombs. Given: q1= 3.0µC,q2=−2.0µC
1 microCoulomb (µC) = 1×10−6Coulombs So, q1= 3.0×10−6C and q2=
−2.0×10−6C.
Step 2: Calculate the magnitude of the electric force using Coulomb’s Law.
The electric force Fbetween two point charges q1and q2separated by a distance
ris given by Coulomb’s Law:
F=k·|q1·q2|
r2
where k= 8.99 ×109Nm2/C2is Coulomb’s constant.
Plugging in the given values: F= (8.99 ×109)·|3.0×10−6·−2.0×10−6|
(0.10)2
Step 3: Calculate the magnitude of the electric force.
F= 8.99 ×109·6.0×10−12
0.01
F= 8.99 ×109·6.0×10−10
F= 53.94 ×10−1
F= 5.39 ×100N
Therefore, the magnitude of the electric force between these charges is 5.39
N.
Question 19
Question
Three point charges are arranged along the x-axis as follows: +3.00 µC at the
origin, −4.00 µC at x= 50.0cm, and +5.00 µC at x= 70.0cm. What is the net
electric force on the −4.00 µC charge?
Solution
1. Calculate the electric force on the −4.00 µC charge due to the +3.00 µC
charge at the origin. The electric force F1between two point charges is given
by Coulomb’s Law:
F1=kq1q2
r2
where - kis the Coulomb constant (8.99 ×109N·m2/C2), - q1and q2are
the magnitudes of the point charges (3.00 µC and −4.00 µC, respectively), - r
is the distance between the charges (origin to x= 50.0cm = 0.50 m).
Plugging in the values, we get:
F1= (8.99 ×109)(3.00 ×10−6)·(4.00 ×10−6)
(0.50)2
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Calculating the result gives F1=−4.79 N, directed towards the origin.
2. Calculate the electric force on the −4.00 µC charge due to the +5.00 µC
charge at x= 70.0cm. Using Coulomb’s Law again, this time with q2=
+5.00 µC and r= 0.20 m, we have:
F2= (8.99 ×109)(5.00 ×10−6)·(4.00 ×10−6)
(0.20)2
Calculating the result gives F2= 44.94 N, directed towards the positive
charge.
3. The net force is the vector sum of F1and F2.
Fnet =F2−F1= 44.94 N−4.79 N= 40.15 N
Therefore, the net electric force on the −4.00 µC charge is 40.15 N directed
towards the +5.00 µC charge at x= 70.0cm.
Question 20
Question
Two point charges, q1=−2.5µC and q2= 4.0µC, are located at points
A(−2.0m,0) and B(3.0m,0) respectively on the x-axis. Calculate the electric
force exerted on q2due to q1.
Solution
Step 1: First, we need to calculate the distance between the two charges. Given:
xA=−2.0m
xB= 3.0m
Let dbe the distance between the charges. d=xB−xA= 3.0m−(−2.0m) =
5.0m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law:
F=k|q1·q2|
d2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Step 3: Substituting the given values into the formula:
F= 8.99 ×109N m2/C2·| − 2.5×10−6C·4.0×10−6C|
(5.0m)2
F= 8.99 ×109·10 ×10−12
25
F= 8.99 ×109·4×10−12
F= 35.96 ×10−3
16
F= 3.596 mN
Therefore, the electric force exerted on q2due to q1is 3.596 mN directed
towards q1.
Question 21
Question
Three point charges are fixed on the x-axis. The charges are located at x=−2
m, x= 0 m, and x= 3 m, with magnitudes q1= 4 µC, q2=−6µC, and q3= 2
µC, respectively. Calculate the net electrostatic force acting on q1due to the
other two charges.
Solution
Step 1: Calculate the force on q1due to q2. The magnitude of the force F12 can
be calculated using Coulomb’s Law:
F12 =k|q1||q2|
r2
12
where kis the Coulomb’s constant, r12 is the distance between q1and q2, and
|q1|and |q2|are the magnitudes of the charges.
Substitute the given values into the equation:
F12 =(9 ×109N·m2/C2)(4 ×10−6C)(6 ×10−6C)
(−2m)2
F12 =216 ×10−12
4= 54 ×10−12 = 5.4×10−11 N
Step 2: Calculate the force on q1due to q3. Similarly, calculate the force
F13 on q1due to q3using Coulomb’s Law:
F13 =k|q1||q3|
r2
13
where r13 is the distance between q1and q3.
Substitute the given values into the equation:
F13 =(9 ×109N·m2/C2)(4 ×10−6C)(2 ×10−6C)
(3 m−(−2m))2
F13 =72 ×10−12
25 = 2.88 ×10−12 N
Step 3: Calculate the net force on q1. The net force on q1is the vector
sum of F12 and F13. Since F12 and F13 are in opposite directions, we need to
subtract them:
Fnet =F12 −F13 = 5.4×10−11 N−2.88 ×10−12 N
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Fnet = 5.112 ×10−11 N
Therefore, the net electrostatic force acting on q1due to the other two
charges is 5.112 ×10−11 N.
Question 22
Question
Two point charges, q1=−4µC and q2= 2 µC, are placed 10 cm apart along the
x-axis. Calculate the magnitude and direction of the electric force experienced
by each charge.
Solution
Step 1: Determine the distance between the charges Given that the charges are
10 cm apart along the x-axis, the distance between them is r= 10 cm = 0.10
m.
Step 2: Calculate the magnitude of the electric force experienced by q1The
magnitude of the electric force between two point charges q1and q2is given by
Coulomb’s Law:
F=k|q1q2|
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant.
Substitute q1=−4µC, q2= 2 µC, and r= 0.10 m into the formula:
F=(8.99 ×109N·m2/C2)|(−4×10−6C)(2 ×10−6C)|
(0.10 m)2
Calculating this expression gives the magnitude of the electric force experi-
enced by q1.
Step 3: Determine the direction of the electric force experienced by q1The
direction of the electric force experienced by q1will be attractive towards q2.
Step 4: Calculate the magnitude of the electric force experienced by q2Using
the same formula as in Step 2, but switching the signs of q1and q2, calculate
the electric force experienced by q2.
Step 5: Determine the direction of the electric force experienced by q2The
direction of the electric force experienced by q2will be repulsive away from q1.
Question 23
Question
A charge of −2.0µC is located at the origin, and a charge of 4.0µC is located
at the point (0,3.0) m. Calculate the magnitude and direction of the electric
force on the positive charge due to the negative charge.
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Solution
To find the electric force, we will use Coulomb’s Law, given by:
Electric force =k·|q1·q2|
r2
where kis Coulomb’s constant (8.99 ×109Nm2/C2), q1and q2are the two
charges, and ris the distance between the charges.
Step 1: Calculate the distance rbetween the charges
The distance rbetween the charges can be found using the distance formula
in two dimensions:
r=√(0 −0)2+ (3.0−0)2
r=√0+9.0
r=√9.0
r= 3.0m
Step 2: Calculate the magnitude of the electric force
Now, we can substitute the given values into Coulomb’s Law:
Electric force = (8.99 ×109)·| − 2.0×10−6×4.0×10−6|
(3.0)2
Electric force = 8.99 ×109·8.0×10−12
9.0
Electric force = 8.99 ×109·8.89 ×10−13
Electric force = 7.99 ×10−3N
Step 3: Determine the direction of the electric force
The electric force will be attractive since the charges are of opposite signs.
Therefore, the direction of the electric force is towards the negative charge (from
the positive charge to the negative charge) along the line connecting the charges.
Question 24
Question
Two point charges, q1= +5 µC and q2=−3µC, are placed 10 cm apart. What
is the magnitude of the electric force between these two charges?
19
Solution
Step 1: Convert the charges to Coulombs. The charges given are q1= +5 µC
and q2=−3µC. To convert microcoulombs to coulombs, we use the following
conversion factor: 1µC= 10−6C. Therefore, q1= +5 ×10−6C= 5 ×10−6C
and q2=−3×10−6C=−3×10−6C.
Step 2: Calculate the distance between the charges in meters. The charges
are placed 10 cm apart, which is 0.10 meters.
Step 3: Calculate the electric force between the charges using Coulomb’s
Law. Coulomb’s Law states that the magnitude of the electric force between
two point charges is given by:
F=k· |q1·q2|
r2
where F= electric force, k= Coulomb’s constant (8.99 ×109N m2/C2), q1, q2
= magnitudes of the two charges, and r= distance between the charges.
Substitute the known values into the formula:
F=(8.99 ×109N m2/C2)· |(5 ×10−6C)·(−3×10−6C)|
(0.10 m)2
Step 4: Calculate the electric force.
F=(8.99 ×109)·(5 ×10−6)·(3 ×10−6)
0.01
F=134850
100
F= 1348.50 N
Therefore, the magnitude of the electric force between the two charges is
1348.50 N.
Question 25
Question
Two point charges, q1=−3.00 µC and q2= 5.00 µC, are placed 10.0 cm apart
in vacuum.
Calculate the magnitude of the electric force between the charges, and de-
termine the direction of the force exerted on each charge.
Solution
Step 1: Convert the charges to coulombs. The charges are given in micro-
coulombs (µC), so we need to convert them to coulombs. q1=−3.00 µC=
−3.00 ×10−6Cq2= 5.00 µC= 5.00 ×10−6C
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Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force between two point charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2
where k= 8.99 ×109N·m2/C2is the electrostatic constant and r= 0.10 m is
the separation distance between the charges. Substituting in the given values:
F=8.99 ×109·3.00 ×10−6·5.00 ×10−6
(0.10)2
Step 3: Calculate the numerical value of the electric force.
F=8.99 ×3×5
103=134.85
103= 0.13485 N
Step 4: Determine the direction of the force. The force between the charges
is attractive since the charges have opposite signs.
Therefore, the magnitude of the electric force between the charges is 0.13485 N,
and the force exerted on q1is attractive towards q2, while the force exerted on
q2is attractive towards q1.
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