PHYS 232 - UNIVERSITY PHYSICS
II - Calculation of electric forces
between point charges
Question Bank - Set 5
Liberty University
Question 1
Question
Two point charges, q1=−3µC and q2= 5 µC, 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 standard SI units.
Given that: q1=−3µC =−3×10−6C
q2= 5 µC = 5 ×10−6C
Step 2: Calculate the electric force using Coulomb’s Law.
The magnitude of the electric force between two point charges, q1and q2, sep-
arated by a distance r, is given by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the Coulomb constant, 8.9875×109Nm2/C2, and r= 10 cm = 0.1m.
Plugging in the values, we get:
F= (8.9875 ×109)·| − 3×10−6·5×10−6|
(0.1)2
Step 3: Calculate the magnitude of the electric force.
F= 8.9875 ×109·15 ×10−12
0.01 = 8.9875 ×109·1.5×10−10
F= 1.34813 ×10−1N
Therefore, the magnitude of the electric force between the charges q1and q2
is 1.34813 ×10−1N.
Question 2
Question
Two point charges, q1=−4µC and q2= 6 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
Solution
To find the electric force between two point charges, we can use Coulomb’s Law,
which states that the magnitude of the force between two point charges is given
by:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Step 1: Convert the charges to coulombs and the distance to meters. Given:
q1=−4µC=−4×10−6C, q2= 6 µC= 6 ×10−6C, r= 10 cm = 0.1m.
Step 2: Calculate the electric force. Plugging the given values into Coulomb’s
Law, we have:
F= (8.99 ×109)·| − 4×10−6×6×10−6|
(0.1)2
F= 8.99 ×109·24 ×10−12
0.01
F= 8.99 ×109·2.4×10−9
F= 21.576 ×10−9
F= 2.1576 ×10−8N
Therefore, the magnitude of the electric force between the charges is 2.1576×
10−8N.
Question 3
Question
Two point charges +3 µC and −5µC are placed 12 cm apart in air. Calculate
the magnitude of the electric force acting on each charge. Given: k= 8.99 ×
109N m2/C2.
2
Solution
Step 1: Convert the charges to coulombs.
+3 µC= 3 ×10−6C
−5µC=−5×10−6C
Step 2: Calculate the magnitude of the electric force exerted by the +3 µC
charge on the −5µC charge using Coulomb’s Law:
F=k· |q1|·|q2|
r2
where F= magnitude of the electric force, k= Coulomb’s constant (8.99 ×
109N m2/C2), |q1|= 3 ×10−6C, |q2|= 5 ×10−6C, and r= 0.12 m.
Substitute the values into the formula:
F=(8.99 ×109N m2/C2)·(3 ×10−6C)·(5 ×10−6C)
(0.12 m)2
Step 3: Calculate the magnitude of the electric force between the charges.
Question 4
Question
Two point charges, q1=−3µC and q2= 5µC, are located 6 cm apart. Calculate
the magnitude of the electric force between these charges.
Solution
Step 1: Calculate the distance between the charges in meters. Given: Distance,
r= 6 cm = 0.06 m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
Coulomb’s law states that the magnitude of the electric force between two point
charges is given by:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging in the given values:
F= 8.99 ×109| − 3×10−6×5×10−6|
0.062
Step 3: Calculate the magnitude of the electric force.
F= 8.99 ×109×15 ×10−12
0.0036
3
F= 8.99 ×109×4.1667 ×10−9
F= 37.45 N
Therefore, the magnitude of the electric force between the charges q1and q2
is 37.45 N.
Question 5
Question
Two point charges, q1and q2, are held 4 cm apart in air. If q1= 5µC and
q2=−3µC, calculate the magnitude of the electric force between them.
Solution
Step 1: Calculate the electric force using Coulomb’s Law formula:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2) and ris the distance
between the charges.
Step 2: Substitute the given values into the formula:
F=(8.99 ×109N m2/C2)· |5×10−6C·(−3×10−6C)|
(0.04 m)2
Step 3: Calculate the magnitude of the force:
F=(8.99 ×109)·(5 ×3) ×10−12
0.0016
Step 4: Simplify the expression:
F=44.95 ×10−3
0.0016
Step 5: Calculate the final answer:
F= 28.09 N
Thus, the magnitude of the electric force between the charges q1and q2is
28.09 N.
Question 6
Question
Two point charges, +4.0µC and −2.0µC, are placed 10.0 cm apart. Calculate
the magnitude of the electric force between these two charges.
4
Solution
Step 1: Convert the charges to Coulombs. Given: q1= +4.0µC = 4.0×10−6C
q2=−2.0µC =−2.0×10−6C
Step 2: Calculate the distance between the charges in meters. d= 10.0cm =
10.0×10−2m= 0.10 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law:
F=k
q1·q2
d2
where k= 8.99 ×109Nm2/C2is Coulomb’s constant.
Step 4: Substituting the given values into the formula:
F= 8.99 ×109×
4.0×10−6× −2.0×10−6
(0.10)2
Step 5: Calculate the magnitude of the electric force:
F= 8.99 ×109×8.0×10−12
0.01
F= 8.99 ×109×8.0×10−10
F= 7.19 ×10−1N
Therefore, the magnitude of the electric force between the two charges is
7.19 ×10−1N.
Question 7
Question
Two point charges, q1=−2µC and q2= 3 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between them.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−2µC and q2= 3 µC
Converting to coulombs: q1=−2×10−6C and q2= 3 ×10−6C.
Step 2: Calculate the magnitude of electric force using Coulomb’s Law. The
magnitude of the electric force between point charges is given by Coulomb’s
Law:
F=k
q1q2
r2
,
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
5
Plugging in the given values:
F= 8.99 ×109
(−2×10−6)(3 ×10−6)
(0.10)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= 5.394 N
Therefore, the magnitude of the electric force between the point charges q1
and q2is 5.394 N.
Question 8
Question
Two point charges, q1=−3.0µC and q2= 4.0µC, are placed 8.0 cm apart.
Calculate the magnitude and direction of the electric force between the two
charges.
Solution
Step 1: Convert the charges to coulombs and write down the given values.
Given: q1=−3.0µC = −3.0×10−6C (negative sign indicates charge is nega-
tive)
q2= 4.0µC = 4.0×10−6C
Distance r= 8.0 cm = 0.08 m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law
formula:
F=k·|q1·q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the given values:
F= 8.99 ×109·| − 3.0×10−6·4.0×10−6|
(0.08)2
Step 3: Calculate the magnitude of the electric force:
6
F= 8.99 ×109·12 ×10−12
0.0064
F= 8.99 ×109·1.875 ×10−6
F= 16.87 N
Step 4: Determine the direction of the force. Since the charges have opposite
signs, the force will be attractive. Thus, the force is directed from q1to q2.
Therefore, the magnitude of the electric force between the charges is 16.87
N, and it is directed from q1to q2.
Question 9
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 10.0 cm apart.
Calculate the magnitude and direction of the electric force that q1exerts on q2.
Solution
To find the electric force that q1exerts on q2, we can use Coulomb’s Law:
F=k·|q1|·|q2|
r2
where k= 8.99 ×109N·m2/C2is Coulomb’s constant, |q1|and |q2|are the
magnitudes of the charges, and ris the distance between the charges.
Step 1: Calculate the distance rin meters: Given: r= 10.0cm = 0.10 m.
Step 2: Substitute the values into Coulomb’s Law to find the electric force:
F= (8.99 ×109N·m2/C2)·(4.0×10−6C)·(6.0×10−6C)
(0.10 m)2
Step 3: Calculate the magnitude of the electric force:
F= (8.99 ×109)·24 ×10−12
0.01
F= 8.99 ×109×2.4×10−9
F= 21.576 ×100
F= 21.576 N
Step 4: Determine the direction of the force: Since q1is negative and q2is
positive, the electric force will be attractive, that is, the force will be directed
toward q1.
Therefore, the magnitude of the electric force that q1exerts on q2is 21.576 N
directed towards q1.
7
Question 10
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are fixed in place and
separated by a distance of 5.0cm. Calculate the magnitude of the electric force
that q1exerts on q2.
Solution
Step 1: Convert the charges to Coulombs. Given: q1=−4.0µC and q2=
6.0µC.
To convert from microCoulombs (µC) to Coulombs, we multiply by 10−6.q1=
−4.0×10−6C and q2= 6.0×10−6C.
Step 2: Calculate the magnitude of the electric force using the Coulomb’s law
formula. Coulomb’s law states that the magnitude of the electric force between
two charges is given by
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.
Plugging in the values gives us:
F= 8.99 ×109N m2/C2×| − 4.0×10−6C×6.0×10−6C|
(0.05 m)2.
Step 3: Substitute the values and calculate the electric force.
F= 8.99×109×4.0×6.0×10−6×10−6
0.052= 8.99×109×24 ×10−12
0.0025 = 8.99×109×9.6×10−10 = 8.62 N.
Therefore, the magnitude of the electric force that q1exerts on q2is 8.62 N.
Question 11
Question
In an equilateral triangle with side length a, three point charges are placed at
its vertices. The charges are +q,−2q, and +3q, respectively. Calculate the
magnitude and direction of the net electric force acting on the charge at the
first vertex (+q) due to the other two charges.
Solution
Given: Side length of equilateral triangle, a. Point charges at the vertices: +q,
−2q, and +3q.
8
Step 1: Calculate the electric force between the charge at the first vertex
(+q) and the charge at the second vertex (−2q). The electric force between two
charges q1and q2separated by a distance ris given by Coulomb’s law:
F=k·|q1q2|
r2
where kis Coulomb’s constant (8.9875 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
The distance between the first and second vertices of the equilateral triangle
is a. So, the magnitude of the force between +qand −2qis:
F1,2=k|+q(−2q)|
a2
Step 2: Determine the direction of the force between the charge at the first
vertex and the charge at the second vertex. Since the charges are opposite in
sign, the force between them is attractive. It acts along the line joining the
charges, pointing from the second vertex towards the first vertex.
Step 3: Calculate the electric force between the charge at the first vertex
(+q) and the charge at the third vertex (+3q). The distance between the first
and third vertices of the equilateral triangle is a. So, the magnitude of the force
between +qand +3qis:
F1,3=k|+q(+3q)|
a2
Step 4: Determine the direction of the force between the charge at the first
vertex and the charge at the third vertex. Since the charges are like in sign,
the force between them is repulsive. It acts along the line joining the charges,
pointing from the first vertex towards the third vertex.
Step 5: Find the net force on the charge at the first vertex. The net force
on the charge at the first vertex is the vector sum of the forces F1,2and F1,3.
Since these forces are along different directions, we need to consider their vector
nature.
The net force can be calculated using the parallelogram rule of vector addi-
tion:
Net force =q(F2
1,2+F2
1,3+ 2F1,2F1,3cos θ)
where θis the angle between the forces.
Solving this expression will give the magnitude and direction of the net
electric force acting on the charge at the first vertex due to the other two charges.
9
Question 12
Question
Three point charges are arranged in an equilateral triangle as shown below:
+q
↗
+q−2q
If each side of the triangle has a length a, calculate the net electric force on the
+qcharge.
Solution
Step 1: Calculate the net force on the +qcharge due to the +qcharge. The
electric force between two point charges q1and q2separated by distance ris
given by Coulomb’s Law:
F=kq1q2
r2
where kis Coulomb’s constant (8.99 ×109N·m2/C2). The magnitude of the
force on the +qcharge due to the other +qcharge will be:
F1=kq2
a2
Step 2: Calculate the direction of the force on the +qcharge due to the +q
charge. The force on the +qcharge due to the other +qcharge will be directed
along the line joining the two charges and will be repulsive.
Step 3: Calculate the net force on the +qcharge due to the −2qcharge.
The magnitude of the force on the +qcharge due to the −2qcharge will be:
F2=kq(2q)
a2√3
Step 4: Calculate the direction of the force on the +qcharge due to the −2q
charge. The force on the +qcharge due to the −2qcharge will be directed along
the line joining the two charges and will be attractive.
Step 5: Find the net force on the +qcharge. The net force Fnet on the
+qcharge will be the vector sum of the forces F1and F2, considering their
directions.
Step 6: Resolve the forces into xand ycomponents to find the net force in
terms of magnitude and direction.
Question 13
Question
Two point charges, q1= +3.0µC and q2=−4.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the charges.
10
Solution
Step 1: Determine the charges in coulombs. We convert µC to C by multiplying
by 10−6:q1= 3.0µC= 3.0×10−6Cq2=−4.0µC=−4.0×10−6C
Step 2: Calculate the distance rbetween the charges in meters. Given
r= 10.0cm = 0.10 m.
Step 3: Use Coulomb’s law to calculate the magnitude of the electric force
between the charges: F=k|q1q2|
r2
where k= 8.99 ×109Nm2/C2is the Coulomb constant.
Step 4: Substitute the given values into the formula to find the electric force:
F=(8.99 ×109)|(3.0×10−6)(−4.0×10−6)|
(0.10)2
Step 5: Calculate the magnitude of the electric force: F=(8.99 ×109)(12 ×10−12)
0.01
F=1.0788 ×10−2
0.01 F= 1.0788 N
Therefore, the magnitude of the electric force between the charges is 1.0788 N.
Question 14
Question
Two point charges, q1=−5µC and q2= 3 µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Recall the formula for the electric force between two point charges:
F=k·|q1·q2|
r2
where Fis the magnitude of the electric force, kis the Coulomb constant (8.99×
109N m2/C2), q1and q2are the magnitudes of the respective charges, and ris
the distance between the charges.
Step 2: Convert the charges to coulombs:
q1=−5µC =−5×10−6C
q2= 3 µC = 3 ×10−6C
Step 3: Plug in the given values into the formula to calculate the electric
force:
F= (8.99 ×109)·| − 5×10−6·3×10−6|
(0.10)2
Step 4: Perform the calculation:
F= (8.99 ×109)·15 ×10−12
0.01
11
F= 8.99 ×109·1.5×10−9
F= 13.485 ×100
F= 13.485 N
Step 5: Therefore, the magnitude of the electric force between the charges
q1and q2is 13.485 N.
Question 15
Question
Three point charges are placed at the corners of a right triangle as shown:
q1= +5 µC at point A,q2=−3µC at point B, and q3= +4 µC at point C.
Calculate the net electric force on q3due to q1and q2. The sides of the triangle
are of lengths 3m, 4m, and 5m.
Solution
Step 1: Calculate the distance between q1at point Aand q3at point C.
Distance between q1and q3=p32+ 42= 5 m
Step 2: Calculate the distance between q2at point Band q3at point C.
Distance between q2and q3= 4 m
Step 3: Calculate the electric force F13 on q3due to q1.
F13 =k|q1||q3|
r2=(9 ×109N m2/C2)(5 ×10−6C)(4 ×10−6C)
52= 144 mN
Step 4: Calculate the electric force F23 on q3due to q2.
F23 =k|q2||q3|
r2=(9 ×109N m2/C2)(3 ×10−6C)(4 ×10−6C)
42= 54 mN
Step 5: Calculate the net electric force on q3.
Fnet =F13 −F23 = 144 mN −54 mN = 90 mN
Therefore, the net electric force on q3due to q1and q2is 90 mN directed
towards q1.
Question 16
Question
Two point charges, q1= 3 µC and q2=−5µC, are placed 8 cm apart in air.
Calculate the magnitude and direction of the electric force on each charge.
12
Solution
Step 1: First calculate the magnitude of the electric force on charge q1.
Given: q1= 3 µC, q2=−5µC, r= 0.08 m
The formula for the magnitude of the electric force between two point charges
is given by Coulomb’s Law:
F=k·|q1q2|
r2
where kis the electrostatic constant approximately equal to 8.99×109N m2/C2.
Substitute the given values into the formula:
F= (8.99 ×109)·|3×10−6× −5×10−6|
0.082
F= 0.84375 N
Therefore, the magnitude of the electric force on charge q1is 0.84375 N.
Step 2: Next, determine the direction of the electric force on charge q1.
Since q1is positive, the electric force on q1will be repulsive. Thus, the
direction of the force on q1is away from q2.
Step 3: Now, calculate the magnitude of the electric force on charge q2.
The magnitude of the electric force on q2is the same as that on q1, but
acting in the opposite direction. Therefore, the magnitude of the electric force
on q2is also 0.84375 N.
Step 4: Determine the direction of the electric force on charge q2.
Since q2is negative, the electric force on q2will be attractive, pulling q2
towards q1. Therefore, the direction of the force on q2is towards q1.
Question 17
Question
Two point charges are placed on the x-axis. The first charge, q1=−3.0µC,
is located at x= 2.0m, and the second charge, q2= 6.0µC, is located at
x= 4.0m. Calculate the magnitude and direction of the electric force on q2due
to q1.
Solution
Step 1: Calculate the distance between the two charges. Given the positions of
the charges, the distance between them can be found using the formula:
r=|x2−x1|
r=|4.0m−2.0m|
r= 2.0m
13
Step 2: Calculate the electric force magnitude. The electric force between
two charges is given by Coulomb’s law:
F=k·|q1q2|
r2
where k= 8.99×109N m2/C2is the electrostatic constant. Substitute the given
values into the formula:
F= 8.99 ×109N m2/C2·|(−3.0µC)(6.0µC)|
(2.0m)2
F= 8.99 ×109N m2/C2·18.0×10−12 C2
4.0m2
F=8.99 N m2/C2·18.0×10−12 C2
4.0m2
F=161.82 ×10−12 N m
4.0m2
F= 40.455 ×10−12 N
F= 40.5pN
Step 3: Determine the direction of the force. The force will be attractive
since the charges have opposite signs. The force on q2is directed towards q1.
Question 18
Question
Two point charges, q1= +3.0µC and q2=−4.0µC, are placed 12.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Determine the distance between the charges in meters. Given: d=
12.0cm = 0.12 m.
Step 2: Calculate the magnitude of the electric force 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 Fis the magnitude of the electric force, kis the Coulomb constant 8.99×
109N m2/C2,|q1|and |q2|are the magnitudes of the charges, and ris the
distance between the charges.
Plugging in the values,
F=(8.99 ×109N m2/C2)·(3.0×10−6C)·(4.0×10−6C)
(0.12 m)2
14
Step 3: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(3.0·4.0) ×10−12
0.0144
F=107.88 ×10−3
0.0144
F=107.88
0.0144
F≈7491.67 N
Therefore, the magnitude of the electric force between the charges q1and q2
is approximately 7491.67 N.
Question 19
Question
Two point charges, q1= 4 µC and q2=−6µC, are separated by a distance of
5cm. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Write down the given values.
Given: q1= 4µC,q2=−6µC,r= 5 cm = 0.05 m.
Step 2: 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 kis the Coulomb constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
Step 3: Substitute the given values into Coulomb’s Law. Plugging in the
values, we get:
F=(9 ×109Nm2/C2)·(4 ×10−6C)·(6 ×10−6C)
(0.05 m)2
Step 4: Calculate the electric force.
F=216 ×10−15 Nm2/C2
0.0025 m2
F= 86.4×10−15 N
F= 8.64 ×10−14 N
Therefore, the magnitude of the electric force between the charges is 8.64 ×
10−14 N.
15
Question 20
Question
Two point charges, q1=−3nC and q2= 5 nC, are placed 10 cm apart. Calculate
the magnitude of the electric force between the two charges.
Solution
Step 1: Calculate the distance between the two charges in meters.
Distance = 10 cm = 10 ×10−2m= 0.1m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
The magnitude of the electric force (F) between two point charges is given by:
F=k·|q1·q2|
r2
where: - kis the electrostatic constant, 8.99 ×109N m2/C2, - q1and q2are the
two charges, and - ris the distance between the two charges.
Plugging in the given values:
F= (8.99 ×109N m2/C2)·| − 3×10−9C×5×10−9C|
(0.1m)2
Step 3: Simplify the expression to find the magnitude of the electric force.
F= (8.99 ×109)·15 ×10−18
0.01
F= 8.99 ×109·1.5×10−15
F= 13.485 ×10−6
F= 1.3485 ×10−5N
Therefore, the magnitude of the electric force between the two charges is
1.3485 ×10−5N.
Question 21
Question
Three point charges are arranged on the corners of an equilateral triangle with
side length a. The charges are +q,−q, and +2q. Calculate the magnitude of
the net electric force acting on the charge located at the corner with the +2q
charge.
16
Solution
Step 1: Calculate the magnitude of the individual electric forces between each
pair of charges:
The force between two charges q1and q2separated by a distance ris given
by Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the Coulomb’s constant (8.99 ×109N·m2/C2).
The distances between the corners of the equilateral triangle are all a.
Therefore, the magnitude of the forces for the given charges are: - +qand
−q:F+q,−q=k·|q|·|q|
a2=kq2
a2-−qand +2q:F−q,+2q=k·|q|·|2q|
a2=2kq2
a2-+2qand
+q:F+2q,+q=k·|2q|·|q|
a2=2kq2
a2
Step 2: Find the horizontal and vertical components of each force.
The forces can be divided into horizontal and vertical components based on
the symmetry of the equilateral triangle: - +qand −q: Horizontal and vertical
components are equal in magnitude but opposite in direction. - −qand +2q:
Horizontal component is twice the vertical component. - +2qand +q: Both
components are the same for this pair.
Step 3: Determine the net force acting on the +2qcharge.
Since the +2qcharge is at an angle of 60◦with the horizontal axis, the net
force can be found by resolving the components of the forces along the horizontal
and vertical directions and then adding them up vectorially.
The net force will depend on the angles and magnitudes of the different
components calculated in Steps 1 and 2.
Question 22
Question
Three point charges are arranged in a line as shown below:
+Q−2Q+Q
If the separation between the charges +Qand −2Qis a, and the separation
between the charges −2Qand +Qis 2a, calculate the magnitude and direction
of the net force on the charge −2Q.
Solution
Step 1: Calculate the electric force between +Qand −2Q. The electric force
between two point charges q1and q2separated by a distance ris given by
Coulomb’s law:
F=k· |q1|·|q2|
r2
17
where kis the electrostatic constant, |q1|and |q2|are the magnitudes of the
charges, and ris the distance between the charges.
For the charges +Qand −2Qwith separation abetween them, the force on
−2Qis attractive, so the force vector points towards the charge +Q.
Step 2: Calculate the electric force between −2Qand +Q. Using the same
formula as in Step 1, we find that the force between −2Qand +Qis attractive,
and the force vector points towards +Q.
Step 3: Calculate the net force on −2Q. Since the forces from +Qand +Q
on −2Qhave the same direction, we can simply add their magnitudes to find
the net force. The force between +Qand −2Qis F1and the force between −2Q
and +Qis F2.
Net force =F1+F2
Step 4: Substitute the electric force formulas into the net force equation.
F1=k·Q·2Q
a2=2kQ2
a2
F2=k·2Q·Q
(2a)2=kQ2
2a2
Substitute these into the net force equation:
Net force =2kQ2
a2+kQ2
2a2=5kQ2
2a2
Step 5: Determine the direction of the net force. Since both individual forces
were directed towards +Q, the net force on −2Qis also directed towards +Q.
Therefore, the magnitude of the net force on the charge −2Qis 5kQ2
2a2, and
the direction is towards +Q.
Question 23
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. The magnitude of each charge is q. Calculate the magnitude and
direction of the net electric force on the charge at the top vertex due to the
other two charges. Given that the side length of the equilateral triangle is a.
+q
/
−q−q
18
Solution
Step 1: Calculate the force due to the charge qat the right vertex.
The magnitude of the force F1is given by Coulomb’s Law:
F1=k|q|2
a2
where kis the Coulomb constant (8.99 ×109N m2/C2), qis the magnitude
of the charge, and ais the distance between the charges.
Step 2: Calculate the force due to the charge qat the left vertex.
Since the charges are equidistant from the charge at the top vertex and the
charges are equidistant from each other, the magnitude of the force F2due to
the charge qat the left vertex is also F1.
Step 3: Find the net force on the charge at the top vertex.
The net force is the vector sum of the forces F1and F2. Since they are both
acting in the vertical direction but in opposite directions, the net force will be
the difference between the two.
Thus, the net force on the charge at the top vertex is:
Net force =F1−F2
Now, substitute the expressions for F1and F2into the above equation and
simplify to find the net force.
Question 24
Question
Two point charges, q1=−2.5µC and q2= 3.0µC, are placed 8.0 cm apart in
air. Calculate the magnitude of the electric force between them.
Solution
Step 1: Convert the charges to standard units. Step 2: Calculate the distance
between the charges in meters. Step 3: Use Coulomb’s law to find the magnitude
of the electric force between the charges.
Step 1: Convert the charges to standard units. Given: q1=−2.5µC
and q2= 3.0µC. Converting to standard units: q1=−2.5×10−6C and
q2= 3.0×10−6C.
Step 2: Calculate the distance between the charges in meters. Given: dis-
tance = 8.0 cm = 0.08 m.
Step 3: Use Coulomb’s law to find the magnitude of the electric force
between the charges. Coulomb’s law:
F=k·|q1·q2|
r2
19
where: F= electric force, k= Coulomb’s constant (8.99 ×109N·m2/C2),q1
and q2= magnitudes of the charges, r= distance between the charges.
Plugging in the values:
F= (8.99 ×109)·|(−2.5×10−6)·(3.0×10−6)|
(0.08)2
F= (8.99 ×109)·7.5×10−12
0.0064
F= (8.99 ×109)·1.171875 ×10−9
F= 10.54 N
Therefore, the magnitude of the electric force between the charges is 10.54
N.
Question 25
Question
Two point charges, q1= +6.0µC and q2=−3.0µC, are placed 20 cm apart in
a vacuum. Calculate the magnitude of the electric force exerted by q1on q2.
Solution
Step 1: Convert the given charges from microcoulombs to coulombs. Step 2:
Determine the distance between the charges in meters. Step 3: Calculate the
magnitude of the electric force between the charges using Coulomb’s law.
Step 1: Convert the given charges from microcoulombs to coulombs.
q1= 6.0µC = 6.0×10−6C
q2=−3.0µC =−3.0×10−6C
Step 2: Determine the distance between the charges in meters.
r= 20 cm = 20 ×10−2m= 0.20 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law.
Coulomb’s law states that the magnitude of the electric force Fbetween two
point charges is given by
F=k|q1·q2|
r2
where - kis the Coulomb constant, 8.9875 ×109Nm2/C2, - q1, q2are the mag-
nitudes of the two point charges, and - ris the distance between the charges.
Plugging in the values:
F= 8.9875 ×109·|6.0×10−6· −3.0×10−6|
0.202
20
F= 4.49275 ×10−3N
Thus, the magnitude of the electric force exerted by q1on q2is 4.49275 ×
10−3N.
21
Question 2
Question
Two point charges, q1=−4µC and q2= 6 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
Solution
To find the electric force between two point charges, we can use Coulomb’s Law,
which states that the magnitude of the force between two point charges is given
by:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Step 1: Convert the charges to coulombs and the distance to meters. Given:
q1=−4µC=−4×10−6C, q2= 6 µC= 6 ×10−6C, r= 10 cm = 0.1m.
Step 2: Calculate the electric force. Plugging the given values into Coulomb’s
Law, we have:
F= (8.99 ×109)·| − 4×10−6×6×10−6|
(0.1)2
F= 8.99 ×109·24 ×10−12
0.01
F= 8.99 ×109·2.4×10−9
F= 21.576 ×10−9
F= 2.1576 ×10−8N
Therefore, the magnitude of the electric force between the charges is 2.1576×
10−8N.
Question 3
Question
Two point charges +3 µC and −5µC are placed 12 cm apart in air. Calculate
the magnitude of the electric force acting on each charge. Given: k= 8.99 ×
109N m2/C2.
2
Solution
Step 1: Convert the charges to coulombs.
+3 µC= 3 ×10−6C
−5µC=−5×10−6C
Step 2: Calculate the magnitude of the electric force exerted by the +3 µC
charge on the −5µC charge using Coulomb’s Law:
F=k· |q1|·|q2|
r2
where F= magnitude of the electric force, k= Coulomb’s constant (8.99 ×
109N m2/C2), |q1|= 3 ×10−6C, |q2|= 5 ×10−6C, and r= 0.12 m.
Substitute the values into the formula:
F=(8.99 ×109N m2/C2)·(3 ×10−6C)·(5 ×10−6C)
(0.12 m)2
Step 3: Calculate the magnitude of the electric force between the charges.
Question 4
Question
Two point charges, q1=−3µC and q2= 5µC, are located 6 cm apart. Calculate
the magnitude of the electric force between these charges.
Solution
Step 1: Calculate the distance between the charges in meters. Given: Distance,
r= 6 cm = 0.06 m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
Coulomb’s law states that the magnitude of the electric force between two point
charges is given by:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging in the given values:
F= 8.99 ×109| − 3×10−6×5×10−6|
0.062
Step 3: Calculate the magnitude of the electric force.
F= 8.99 ×109×15 ×10−12
0.0036
3
F= 8.99 ×109×4.1667 ×10−9
F= 37.45 N
Therefore, the magnitude of the electric force between the charges q1and q2
is 37.45 N.
Question 5
Question
Two point charges, q1and q2, are held 4 cm apart in air. If q1= 5µC and
q2=−3µC, calculate the magnitude of the electric force between them.
Solution
Step 1: Calculate the electric force using Coulomb’s Law formula:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2) and ris the distance
between the charges.
Step 2: Substitute the given values into the formula:
F=(8.99 ×109N m2/C2)· |5×10−6C·(−3×10−6C)|
(0.04 m)2
Step 3: Calculate the magnitude of the force:
F=(8.99 ×109)·(5 ×3) ×10−12
0.0016
Step 4: Simplify the expression:
F=44.95 ×10−3
0.0016
Step 5: Calculate the final answer:
F= 28.09 N
Thus, the magnitude of the electric force between the charges q1and q2is
28.09 N.
Question 6
Question
Two point charges, +4.0µC and −2.0µC, are placed 10.0 cm apart. Calculate
the magnitude of the electric force between these two charges.
4
Solution
Step 1: Convert the charges to Coulombs. Given: q1= +4.0µC = 4.0×10−6C
q2=−2.0µC =−2.0×10−6C
Step 2: Calculate the distance between the charges in meters. d= 10.0cm =
10.0×10−2m= 0.10 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law:
F=k
q1·q2
d2
where k= 8.99 ×109Nm2/C2is Coulomb’s constant.
Step 4: Substituting the given values into the formula:
F= 8.99 ×109×
4.0×10−6× −2.0×10−6
(0.10)2
Step 5: Calculate the magnitude of the electric force:
F= 8.99 ×109×8.0×10−12
0.01
F= 8.99 ×109×8.0×10−10
F= 7.19 ×10−1N
Therefore, the magnitude of the electric force between the two charges is
7.19 ×10−1N.
Question 7
Question
Two point charges, q1=−2µC and q2= 3 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between them.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−2µC and q2= 3 µC
Converting to coulombs: q1=−2×10−6C and q2= 3 ×10−6C.
Step 2: Calculate the magnitude of electric force using Coulomb’s Law. The
magnitude of the electric force between point charges is given by Coulomb’s
Law:
F=k
q1q2
r2
,
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
5
Plugging in the given values:
F= 8.99 ×109
(−2×10−6)(3 ×10−6)
(0.10)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= 5.394 N
Therefore, the magnitude of the electric force between the point charges q1
and q2is 5.394 N.
Question 8
Question
Two point charges, q1=−3.0µC and q2= 4.0µC, are placed 8.0 cm apart.
Calculate the magnitude and direction of the electric force between the two
charges.
Solution
Step 1: Convert the charges to coulombs and write down the given values.
Given: q1=−3.0µC = −3.0×10−6C (negative sign indicates charge is nega-
tive)
q2= 4.0µC = 4.0×10−6C
Distance r= 8.0 cm = 0.08 m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law
formula:
F=k·|q1·q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the given values:
F= 8.99 ×109·| − 3.0×10−6·4.0×10−6|
(0.08)2
Step 3: Calculate the magnitude of the electric force:
6
F= 8.99 ×109·12 ×10−12
0.0064
F= 8.99 ×109·1.875 ×10−6
F= 16.87 N
Step 4: Determine the direction of the force. Since the charges have opposite
signs, the force will be attractive. Thus, the force is directed from q1to q2.
Therefore, the magnitude of the electric force between the charges is 16.87
N, and it is directed from q1to q2.
Question 9
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 10.0 cm apart.
Calculate the magnitude and direction of the electric force that q1exerts on q2.
Solution
To find the electric force that q1exerts on q2, we can use Coulomb’s Law:
F=k·|q1|·|q2|
r2
where k= 8.99 ×109N·m2/C2is Coulomb’s constant, |q1|and |q2|are the
magnitudes of the charges, and ris the distance between the charges.
Step 1: Calculate the distance rin meters: Given: r= 10.0cm = 0.10 m.
Step 2: Substitute the values into Coulomb’s Law to find the electric force:
F= (8.99 ×109N·m2/C2)·(4.0×10−6C)·(6.0×10−6C)
(0.10 m)2
Step 3: Calculate the magnitude of the electric force:
F= (8.99 ×109)·24 ×10−12
0.01
F= 8.99 ×109×2.4×10−9
F= 21.576 ×100
F= 21.576 N
Step 4: Determine the direction of the force: Since q1is negative and q2is
positive, the electric force will be attractive, that is, the force will be directed
toward q1.
Therefore, the magnitude of the electric force that q1exerts on q2is 21.576 N
directed towards q1.
7
Question 10
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are fixed in place and
separated by a distance of 5.0cm. Calculate the magnitude of the electric force
that q1exerts on q2.
Solution
Step 1: Convert the charges to Coulombs. Given: q1=−4.0µC and q2=
6.0µC.
To convert from microCoulombs (µC) to Coulombs, we multiply by 10−6.q1=
−4.0×10−6C and q2= 6.0×10−6C.
Step 2: Calculate the magnitude of the electric force using the Coulomb’s law
formula. Coulomb’s law states that the magnitude of the electric force between
two charges is given by
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.
Plugging in the values gives us:
F= 8.99 ×109N m2/C2×| − 4.0×10−6C×6.0×10−6C|
(0.05 m)2.
Step 3: Substitute the values and calculate the electric force.
F= 8.99×109×4.0×6.0×10−6×10−6
0.052= 8.99×109×24 ×10−12
0.0025 = 8.99×109×9.6×10−10 = 8.62 N.
Therefore, the magnitude of the electric force that q1exerts on q2is 8.62 N.
Question 11
Question
In an equilateral triangle with side length a, three point charges are placed at
its vertices. The charges are +q,−2q, and +3q, respectively. Calculate the
magnitude and direction of the net electric force acting on the charge at the
first vertex (+q) due to the other two charges.
Solution
Given: Side length of equilateral triangle, a. Point charges at the vertices: +q,
−2q, and +3q.
8
Step 1: Calculate the electric force between the charge at the first vertex
(+q) and the charge at the second vertex (−2q). The electric force between two
charges q1and q2separated by a distance ris given by Coulomb’s law:
F=k·|q1q2|
r2
where kis Coulomb’s constant (8.9875 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
The distance between the first and second vertices of the equilateral triangle
is a. So, the magnitude of the force between +qand −2qis:
F1,2=k|+q(−2q)|
a2
Step 2: Determine the direction of the force between the charge at the first
vertex and the charge at the second vertex. Since the charges are opposite in
sign, the force between them is attractive. It acts along the line joining the
charges, pointing from the second vertex towards the first vertex.
Step 3: Calculate the electric force between the charge at the first vertex
(+q) and the charge at the third vertex (+3q). The distance between the first
and third vertices of the equilateral triangle is a. So, the magnitude of the force
between +qand +3qis:
F1,3=k|+q(+3q)|
a2
Step 4: Determine the direction of the force between the charge at the first
vertex and the charge at the third vertex. Since the charges are like in sign,
the force between them is repulsive. It acts along the line joining the charges,
pointing from the first vertex towards the third vertex.
Step 5: Find the net force on the charge at the first vertex. The net force
on the charge at the first vertex is the vector sum of the forces F1,2and F1,3.
Since these forces are along different directions, we need to consider their vector
nature.
The net force can be calculated using the parallelogram rule of vector addi-
tion:
Net force =q(F2
1,2+F2
1,3+ 2F1,2F1,3cos θ)
where θis the angle between the forces.
Solving this expression will give the magnitude and direction of the net
electric force acting on the charge at the first vertex due to the other two charges.
9
Question 12
Question
Three point charges are arranged in an equilateral triangle as shown below:
+q
↗
+q−2q
If each side of the triangle has a length a, calculate the net electric force on the
+qcharge.
Solution
Step 1: Calculate the net force on the +qcharge due to the +qcharge. The
electric force between two point charges q1and q2separated by distance ris
given by Coulomb’s Law:
F=kq1q2
r2
where kis Coulomb’s constant (8.99 ×109N·m2/C2). The magnitude of the
force on the +qcharge due to the other +qcharge will be:
F1=kq2
a2
Step 2: Calculate the direction of the force on the +qcharge due to the +q
charge. The force on the +qcharge due to the other +qcharge will be directed
along the line joining the two charges and will be repulsive.
Step 3: Calculate the net force on the +qcharge due to the −2qcharge.
The magnitude of the force on the +qcharge due to the −2qcharge will be:
F2=kq(2q)
a2√3
Step 4: Calculate the direction of the force on the +qcharge due to the −2q
charge. The force on the +qcharge due to the −2qcharge will be directed along
the line joining the two charges and will be attractive.
Step 5: Find the net force on the +qcharge. The net force Fnet on the
+qcharge will be the vector sum of the forces F1and F2, considering their
directions.
Step 6: Resolve the forces into xand ycomponents to find the net force in
terms of magnitude and direction.
Question 13
Question
Two point charges, q1= +3.0µC and q2=−4.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the charges.
10
Solution
Step 1: Determine the charges in coulombs. We convert µC to C by multiplying
by 10−6:q1= 3.0µC= 3.0×10−6Cq2=−4.0µC=−4.0×10−6C
Step 2: Calculate the distance rbetween the charges in meters. Given
r= 10.0cm = 0.10 m.
Step 3: Use Coulomb’s law to calculate the magnitude of the electric force
between the charges: F=k|q1q2|
r2
where k= 8.99 ×109Nm2/C2is the Coulomb constant.
Step 4: Substitute the given values into the formula to find the electric force:
F=(8.99 ×109)|(3.0×10−6)(−4.0×10−6)|
(0.10)2
Step 5: Calculate the magnitude of the electric force: F=(8.99 ×109)(12 ×10−12)
0.01
F=1.0788 ×10−2
0.01 F= 1.0788 N
Therefore, the magnitude of the electric force between the charges is 1.0788 N.
Question 14
Question
Two point charges, q1=−5µC and q2= 3 µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Recall the formula for the electric force between two point charges:
F=k·|q1·q2|
r2
where Fis the magnitude of the electric force, kis the Coulomb constant (8.99×
109N m2/C2), q1and q2are the magnitudes of the respective charges, and ris
the distance between the charges.
Step 2: Convert the charges to coulombs:
q1=−5µC =−5×10−6C
q2= 3 µC = 3 ×10−6C
Step 3: Plug in the given values into the formula to calculate the electric
force:
F= (8.99 ×109)·| − 5×10−6·3×10−6|
(0.10)2
Step 4: Perform the calculation:
F= (8.99 ×109)·15 ×10−12
0.01
11
F= 8.99 ×109·1.5×10−9
F= 13.485 ×100
F= 13.485 N
Step 5: Therefore, the magnitude of the electric force between the charges
q1and q2is 13.485 N.
Question 15
Question
Three point charges are placed at the corners of a right triangle as shown:
q1= +5 µC at point A,q2=−3µC at point B, and q3= +4 µC at point C.
Calculate the net electric force on q3due to q1and q2. The sides of the triangle
are of lengths 3m, 4m, and 5m.
Solution
Step 1: Calculate the distance between q1at point Aand q3at point C.
Distance between q1and q3=p32+ 42= 5 m
Step 2: Calculate the distance between q2at point Band q3at point C.
Distance between q2and q3= 4 m
Step 3: Calculate the electric force F13 on q3due to q1.
F13 =k|q1||q3|
r2=(9 ×109N m2/C2)(5 ×10−6C)(4 ×10−6C)
52= 144 mN
Step 4: Calculate the electric force F23 on q3due to q2.
F23 =k|q2||q3|
r2=(9 ×109N m2/C2)(3 ×10−6C)(4 ×10−6C)
42= 54 mN
Step 5: Calculate the net electric force on q3.
Fnet =F13 −F23 = 144 mN −54 mN = 90 mN
Therefore, the net electric force on q3due to q1and q2is 90 mN directed
towards q1.
Question 16
Question
Two point charges, q1= 3 µC and q2=−5µC, are placed 8 cm apart in air.
Calculate the magnitude and direction of the electric force on each charge.
12
Solution
Step 1: First calculate the magnitude of the electric force on charge q1.
Given: q1= 3 µC, q2=−5µC, r= 0.08 m
The formula for the magnitude of the electric force between two point charges
is given by Coulomb’s Law:
F=k·|q1q2|
r2
where kis the electrostatic constant approximately equal to 8.99×109N m2/C2.
Substitute the given values into the formula:
F= (8.99 ×109)·|3×10−6× −5×10−6|
0.082
F= 0.84375 N
Therefore, the magnitude of the electric force on charge q1is 0.84375 N.
Step 2: Next, determine the direction of the electric force on charge q1.
Since q1is positive, the electric force on q1will be repulsive. Thus, the
direction of the force on q1is away from q2.
Step 3: Now, calculate the magnitude of the electric force on charge q2.
The magnitude of the electric force on q2is the same as that on q1, but
acting in the opposite direction. Therefore, the magnitude of the electric force
on q2is also 0.84375 N.
Step 4: Determine the direction of the electric force on charge q2.
Since q2is negative, the electric force on q2will be attractive, pulling q2
towards q1. Therefore, the direction of the force on q2is towards q1.
Question 17
Question
Two point charges are placed on the x-axis. The first charge, q1=−3.0µC,
is located at x= 2.0m, and the second charge, q2= 6.0µC, is located at
x= 4.0m. Calculate the magnitude and direction of the electric force on q2due
to q1.
Solution
Step 1: Calculate the distance between the two charges. Given the positions of
the charges, the distance between them can be found using the formula:
r=|x2−x1|
r=|4.0m−2.0m|
r= 2.0m
13
Step 2: Calculate the electric force magnitude. The electric force between
two charges is given by Coulomb’s law:
F=k·|q1q2|
r2
where k= 8.99×109N m2/C2is the electrostatic constant. Substitute the given
values into the formula:
F= 8.99 ×109N m2/C2·|(−3.0µC)(6.0µC)|
(2.0m)2
F= 8.99 ×109N m2/C2·18.0×10−12 C2
4.0m2
F=8.99 N m2/C2·18.0×10−12 C2
4.0m2
F=161.82 ×10−12 N m
4.0m2
F= 40.455 ×10−12 N
F= 40.5pN
Step 3: Determine the direction of the force. The force will be attractive
since the charges have opposite signs. The force on q2is directed towards q1.
Question 18
Question
Two point charges, q1= +3.0µC and q2=−4.0µC, are placed 12.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Determine the distance between the charges in meters. Given: d=
12.0cm = 0.12 m.
Step 2: Calculate the magnitude of the electric force 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 Fis the magnitude of the electric force, kis the Coulomb constant 8.99×
109N m2/C2,|q1|and |q2|are the magnitudes of the charges, and ris the
distance between the charges.
Plugging in the values,
F=(8.99 ×109N m2/C2)·(3.0×10−6C)·(4.0×10−6C)
(0.12 m)2
14
Step 3: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(3.0·4.0) ×10−12
0.0144
F=107.88 ×10−3
0.0144
F=107.88
0.0144
F≈7491.67 N
Therefore, the magnitude of the electric force between the charges q1and q2
is approximately 7491.67 N.
Question 19
Question
Two point charges, q1= 4 µC and q2=−6µC, are separated by a distance of
5cm. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Write down the given values.
Given: q1= 4µC,q2=−6µC,r= 5 cm = 0.05 m.
Step 2: 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 kis the Coulomb constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
Step 3: Substitute the given values into Coulomb’s Law. Plugging in the
values, we get:
F=(9 ×109Nm2/C2)·(4 ×10−6C)·(6 ×10−6C)
(0.05 m)2
Step 4: Calculate the electric force.
F=216 ×10−15 Nm2/C2
0.0025 m2
F= 86.4×10−15 N
F= 8.64 ×10−14 N
Therefore, the magnitude of the electric force between the charges is 8.64 ×
10−14 N.
15
Question 20
Question
Two point charges, q1=−3nC and q2= 5 nC, are placed 10 cm apart. Calculate
the magnitude of the electric force between the two charges.
Solution
Step 1: Calculate the distance between the two charges in meters.
Distance = 10 cm = 10 ×10−2m= 0.1m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
The magnitude of the electric force (F) between two point charges is given by:
F=k·|q1·q2|
r2
where: - kis the electrostatic constant, 8.99 ×109N m2/C2, - q1and q2are the
two charges, and - ris the distance between the two charges.
Plugging in the given values:
F= (8.99 ×109N m2/C2)·| − 3×10−9C×5×10−9C|
(0.1m)2
Step 3: Simplify the expression to find the magnitude of the electric force.
F= (8.99 ×109)·15 ×10−18
0.01
F= 8.99 ×109·1.5×10−15
F= 13.485 ×10−6
F= 1.3485 ×10−5N
Therefore, the magnitude of the electric force between the two charges is
1.3485 ×10−5N.
Question 21
Question
Three point charges are arranged on the corners of an equilateral triangle with
side length a. The charges are +q,−q, and +2q. Calculate the magnitude of
the net electric force acting on the charge located at the corner with the +2q
charge.
16
Solution
Step 1: Calculate the magnitude of the individual electric forces between each
pair of charges:
The force between two charges q1and q2separated by a distance ris given
by Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the Coulomb’s constant (8.99 ×109N·m2/C2).
The distances between the corners of the equilateral triangle are all a.
Therefore, the magnitude of the forces for the given charges are: - +qand
−q:F+q,−q=k·|q|·|q|
a2=kq2
a2-−qand +2q:F−q,+2q=k·|q|·|2q|
a2=2kq2
a2-+2qand
+q:F+2q,+q=k·|2q|·|q|
a2=2kq2
a2
Step 2: Find the horizontal and vertical components of each force.
The forces can be divided into horizontal and vertical components based on
the symmetry of the equilateral triangle: - +qand −q: Horizontal and vertical
components are equal in magnitude but opposite in direction. - −qand +2q:
Horizontal component is twice the vertical component. - +2qand +q: Both
components are the same for this pair.
Step 3: Determine the net force acting on the +2qcharge.
Since the +2qcharge is at an angle of 60◦with the horizontal axis, the net
force can be found by resolving the components of the forces along the horizontal
and vertical directions and then adding them up vectorially.
The net force will depend on the angles and magnitudes of the different
components calculated in Steps 1 and 2.
Question 22
Question
Three point charges are arranged in a line as shown below:
+Q−2Q+Q
If the separation between the charges +Qand −2Qis a, and the separation
between the charges −2Qand +Qis 2a, calculate the magnitude and direction
of the net force on the charge −2Q.
Solution
Step 1: Calculate the electric force between +Qand −2Q. The electric force
between two point charges q1and q2separated by a distance ris given by
Coulomb’s law:
F=k· |q1|·|q2|
r2
17
where kis the electrostatic constant, |q1|and |q2|are the magnitudes of the
charges, and ris the distance between the charges.
For the charges +Qand −2Qwith separation abetween them, the force on
−2Qis attractive, so the force vector points towards the charge +Q.
Step 2: Calculate the electric force between −2Qand +Q. Using the same
formula as in Step 1, we find that the force between −2Qand +Qis attractive,
and the force vector points towards +Q.
Step 3: Calculate the net force on −2Q. Since the forces from +Qand +Q
on −2Qhave the same direction, we can simply add their magnitudes to find
the net force. The force between +Qand −2Qis F1and the force between −2Q
and +Qis F2.
Net force =F1+F2
Step 4: Substitute the electric force formulas into the net force equation.
F1=k·Q·2Q
a2=2kQ2
a2
F2=k·2Q·Q
(2a)2=kQ2
2a2
Substitute these into the net force equation:
Net force =2kQ2
a2+kQ2
2a2=5kQ2
2a2
Step 5: Determine the direction of the net force. Since both individual forces
were directed towards +Q, the net force on −2Qis also directed towards +Q.
Therefore, the magnitude of the net force on the charge −2Qis 5kQ2
2a2, and
the direction is towards +Q.
Question 23
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. The magnitude of each charge is q. Calculate the magnitude and
direction of the net electric force on the charge at the top vertex due to the
other two charges. Given that the side length of the equilateral triangle is a.
+q
/
−q−q
18
Solution
Step 1: Calculate the force due to the charge qat the right vertex.
The magnitude of the force F1is given by Coulomb’s Law:
F1=k|q|2
a2
where kis the Coulomb constant (8.99 ×109N m2/C2), qis the magnitude
of the charge, and ais the distance between the charges.
Step 2: Calculate the force due to the charge qat the left vertex.
Since the charges are equidistant from the charge at the top vertex and the
charges are equidistant from each other, the magnitude of the force F2due to
the charge qat the left vertex is also F1.
Step 3: Find the net force on the charge at the top vertex.
The net force is the vector sum of the forces F1and F2. Since they are both
acting in the vertical direction but in opposite directions, the net force will be
the difference between the two.
Thus, the net force on the charge at the top vertex is:
Net force =F1−F2
Now, substitute the expressions for F1and F2into the above equation and
simplify to find the net force.
Question 24
Question
Two point charges, q1=−2.5µC and q2= 3.0µC, are placed 8.0 cm apart in
air. Calculate the magnitude of the electric force between them.
Solution
Step 1: Convert the charges to standard units. Step 2: Calculate the distance
between the charges in meters. Step 3: Use Coulomb’s law to find the magnitude
of the electric force between the charges.
Step 1: Convert the charges to standard units. Given: q1=−2.5µC
and q2= 3.0µC. Converting to standard units: q1=−2.5×10−6C and
q2= 3.0×10−6C.
Step 2: Calculate the distance between the charges in meters. Given: dis-
tance = 8.0 cm = 0.08 m.
Step 3: Use Coulomb’s law to find the magnitude of the electric force
between the charges. Coulomb’s law:
F=k·|q1·q2|
r2
19
where: F= electric force, k= Coulomb’s constant (8.99 ×109N·m2/C2),q1
and q2= magnitudes of the charges, r= distance between the charges.
Plugging in the values:
F= (8.99 ×109)·|(−2.5×10−6)·(3.0×10−6)|
(0.08)2
F= (8.99 ×109)·7.5×10−12
0.0064
F= (8.99 ×109)·1.171875 ×10−9
F= 10.54 N
Therefore, the magnitude of the electric force between the charges is 10.54
N.
Question 25
Question
Two point charges, q1= +6.0µC and q2=−3.0µC, are placed 20 cm apart in
a vacuum. Calculate the magnitude of the electric force exerted by q1on q2.
Solution
Step 1: Convert the given charges from microcoulombs to coulombs. Step 2:
Determine the distance between the charges in meters. Step 3: Calculate the
magnitude of the electric force between the charges using Coulomb’s law.
Step 1: Convert the given charges from microcoulombs to coulombs.
q1= 6.0µC = 6.0×10−6C
q2=−3.0µC =−3.0×10−6C
Step 2: Determine the distance between the charges in meters.
r= 20 cm = 20 ×10−2m= 0.20 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law.
Coulomb’s law states that the magnitude of the electric force Fbetween two
point charges is given by
F=k|q1·q2|
r2
where - kis the Coulomb constant, 8.9875 ×109Nm2/C2, - q1, q2are the mag-
nitudes of the two point charges, and - ris the distance between the charges.
Plugging in the values:
F= 8.9875 ×109·|6.0×10−6· −3.0×10−6|
0.202
20
F= 4.49275 ×10−3N
Thus, the magnitude of the electric force exerted by q1on q2is 4.49275 ×
10−3N.
21
Question 2
Question
Two point charges, q1=−4µC and q2= 6 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
Solution
To find the electric force between two point charges, we can use Coulomb’s Law,
which states that the magnitude of the force between two point charges is given
by:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Step 1: Convert the charges to coulombs and the distance to meters. Given:
q1=−4µC=−4×10−6C, q2= 6 µC= 6 ×10−6C, r= 10 cm = 0.1m.
Step 2: Calculate the electric force. Plugging the given values into Coulomb’s
Law, we have:
F= (8.99 ×109)·| − 4×10−6×6×10−6|
(0.1)2
F= 8.99 ×109·24 ×10−12
0.01
F= 8.99 ×109·2.4×10−9
F= 21.576 ×10−9
F= 2.1576 ×10−8N
Therefore, the magnitude of the electric force between the charges is 2.1576×
10−8N.
Question 3
Question
Two point charges +3 µC and −5µC are placed 12 cm apart in air. Calculate
the magnitude of the electric force acting on each charge. Given: k= 8.99 ×
109N m2/C2.
2
Solution
Step 1: Convert the charges to coulombs.
+3 µC= 3 ×10−6C
−5µC=−5×10−6C
Step 2: Calculate the magnitude of the electric force exerted by the +3 µC
charge on the −5µC charge using Coulomb’s Law:
F=k· |q1|·|q2|
r2
where F= magnitude of the electric force, k= Coulomb’s constant (8.99 ×
109N m2/C2), |q1|= 3 ×10−6C, |q2|= 5 ×10−6C, and r= 0.12 m.
Substitute the values into the formula:
F=(8.99 ×109N m2/C2)·(3 ×10−6C)·(5 ×10−6C)
(0.12 m)2
Step 3: Calculate the magnitude of the electric force between the charges.
Question 4
Question
Two point charges, q1=−3µC and q2= 5µC, are located 6 cm apart. Calculate
the magnitude of the electric force between these charges.
Solution
Step 1: Calculate the distance between the charges in meters. Given: Distance,
r= 6 cm = 0.06 m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
Coulomb’s law states that the magnitude of the electric force between two point
charges is given by:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging in the given values:
F= 8.99 ×109| − 3×10−6×5×10−6|
0.062
Step 3: Calculate the magnitude of the electric force.
F= 8.99 ×109×15 ×10−12
0.0036
3
F= 8.99 ×109×4.1667 ×10−9
F= 37.45 N
Therefore, the magnitude of the electric force between the charges q1and q2
is 37.45 N.
Question 5
Question
Two point charges, q1and q2, are held 4 cm apart in air. If q1= 5µC and
q2=−3µC, calculate the magnitude of the electric force between them.
Solution
Step 1: Calculate the electric force using Coulomb’s Law formula:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2) and ris the distance
between the charges.
Step 2: Substitute the given values into the formula:
F=(8.99 ×109N m2/C2)· |5×10−6C·(−3×10−6C)|
(0.04 m)2
Step 3: Calculate the magnitude of the force:
F=(8.99 ×109)·(5 ×3) ×10−12
0.0016
Step 4: Simplify the expression:
F=44.95 ×10−3
0.0016
Step 5: Calculate the final answer:
F= 28.09 N
Thus, the magnitude of the electric force between the charges q1and q2is
28.09 N.
Question 6
Question
Two point charges, +4.0µC and −2.0µC, are placed 10.0 cm apart. Calculate
the magnitude of the electric force between these two charges.
4
Solution
Step 1: Convert the charges to Coulombs. Given: q1= +4.0µC = 4.0×10−6C
q2=−2.0µC =−2.0×10−6C
Step 2: Calculate the distance between the charges in meters. d= 10.0cm =
10.0×10−2m= 0.10 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law:
F=k
q1·q2
d2
where k= 8.99 ×109Nm2/C2is Coulomb’s constant.
Step 4: Substituting the given values into the formula:
F= 8.99 ×109×
4.0×10−6× −2.0×10−6
(0.10)2
Step 5: Calculate the magnitude of the electric force:
F= 8.99 ×109×8.0×10−12
0.01
F= 8.99 ×109×8.0×10−10
F= 7.19 ×10−1N
Therefore, the magnitude of the electric force between the two charges is
7.19 ×10−1N.
Question 7
Question
Two point charges, q1=−2µC and q2= 3 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between them.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−2µC and q2= 3 µC
Converting to coulombs: q1=−2×10−6C and q2= 3 ×10−6C.
Step 2: Calculate the magnitude of electric force using Coulomb’s Law. The
magnitude of the electric force between point charges is given by Coulomb’s
Law:
F=k
q1q2
r2
,
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
5
Plugging in the given values:
F= 8.99 ×109
(−2×10−6)(3 ×10−6)
(0.10)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= 5.394 N
Therefore, the magnitude of the electric force between the point charges q1
and q2is 5.394 N.
Question 8
Question
Two point charges, q1=−3.0µC and q2= 4.0µC, are placed 8.0 cm apart.
Calculate the magnitude and direction of the electric force between the two
charges.
Solution
Step 1: Convert the charges to coulombs and write down the given values.
Given: q1=−3.0µC = −3.0×10−6C (negative sign indicates charge is nega-
tive)
q2= 4.0µC = 4.0×10−6C
Distance r= 8.0 cm = 0.08 m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law
formula:
F=k·|q1·q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the given values:
F= 8.99 ×109·| − 3.0×10−6·4.0×10−6|
(0.08)2
Step 3: Calculate the magnitude of the electric force:
6
F= 8.99 ×109·12 ×10−12
0.0064
F= 8.99 ×109·1.875 ×10−6
F= 16.87 N
Step 4: Determine the direction of the force. Since the charges have opposite
signs, the force will be attractive. Thus, the force is directed from q1to q2.
Therefore, the magnitude of the electric force between the charges is 16.87
N, and it is directed from q1to q2.
Question 9
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 10.0 cm apart.
Calculate the magnitude and direction of the electric force that q1exerts on q2.
Solution
To find the electric force that q1exerts on q2, we can use Coulomb’s Law:
F=k·|q1|·|q2|
r2
where k= 8.99 ×109N·m2/C2is Coulomb’s constant, |q1|and |q2|are the
magnitudes of the charges, and ris the distance between the charges.
Step 1: Calculate the distance rin meters: Given: r= 10.0cm = 0.10 m.
Step 2: Substitute the values into Coulomb’s Law to find the electric force:
F= (8.99 ×109N·m2/C2)·(4.0×10−6C)·(6.0×10−6C)
(0.10 m)2
Step 3: Calculate the magnitude of the electric force:
F= (8.99 ×109)·24 ×10−12
0.01
F= 8.99 ×109×2.4×10−9
F= 21.576 ×100
F= 21.576 N
Step 4: Determine the direction of the force: Since q1is negative and q2is
positive, the electric force will be attractive, that is, the force will be directed
toward q1.
Therefore, the magnitude of the electric force that q1exerts on q2is 21.576 N
directed towards q1.
7
Question 10
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are fixed in place and
separated by a distance of 5.0cm. Calculate the magnitude of the electric force
that q1exerts on q2.
Solution
Step 1: Convert the charges to Coulombs. Given: q1=−4.0µC and q2=
6.0µC.
To convert from microCoulombs (µC) to Coulombs, we multiply by 10−6.q1=
−4.0×10−6C and q2= 6.0×10−6C.
Step 2: Calculate the magnitude of the electric force using the Coulomb’s law
formula. Coulomb’s law states that the magnitude of the electric force between
two charges is given by
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.
Plugging in the values gives us:
F= 8.99 ×109N m2/C2×| − 4.0×10−6C×6.0×10−6C|
(0.05 m)2.
Step 3: Substitute the values and calculate the electric force.
F= 8.99×109×4.0×6.0×10−6×10−6
0.052= 8.99×109×24 ×10−12
0.0025 = 8.99×109×9.6×10−10 = 8.62 N.
Therefore, the magnitude of the electric force that q1exerts on q2is 8.62 N.
Question 11
Question
In an equilateral triangle with side length a, three point charges are placed at
its vertices. The charges are +q,−2q, and +3q, respectively. Calculate the
magnitude and direction of the net electric force acting on the charge at the
first vertex (+q) due to the other two charges.
Solution
Given: Side length of equilateral triangle, a. Point charges at the vertices: +q,
−2q, and +3q.
8
Step 1: Calculate the electric force between the charge at the first vertex
(+q) and the charge at the second vertex (−2q). The electric force between two
charges q1and q2separated by a distance ris given by Coulomb’s law:
F=k·|q1q2|
r2
where kis Coulomb’s constant (8.9875 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
The distance between the first and second vertices of the equilateral triangle
is a. So, the magnitude of the force between +qand −2qis:
F1,2=k|+q(−2q)|
a2
Step 2: Determine the direction of the force between the charge at the first
vertex and the charge at the second vertex. Since the charges are opposite in
sign, the force between them is attractive. It acts along the line joining the
charges, pointing from the second vertex towards the first vertex.
Step 3: Calculate the electric force between the charge at the first vertex
(+q) and the charge at the third vertex (+3q). The distance between the first
and third vertices of the equilateral triangle is a. So, the magnitude of the force
between +qand +3qis:
F1,3=k|+q(+3q)|
a2
Step 4: Determine the direction of the force between the charge at the first
vertex and the charge at the third vertex. Since the charges are like in sign,
the force between them is repulsive. It acts along the line joining the charges,
pointing from the first vertex towards the third vertex.
Step 5: Find the net force on the charge at the first vertex. The net force
on the charge at the first vertex is the vector sum of the forces F1,2and F1,3.
Since these forces are along different directions, we need to consider their vector
nature.
The net force can be calculated using the parallelogram rule of vector addi-
tion:
Net force =q(F2
1,2+F2
1,3+ 2F1,2F1,3cos θ)
where θis the angle between the forces.
Solving this expression will give the magnitude and direction of the net
electric force acting on the charge at the first vertex due to the other two charges.
9
Question 12
Question
Three point charges are arranged in an equilateral triangle as shown below:
+q
↗
+q−2q
If each side of the triangle has a length a, calculate the net electric force on the
+qcharge.
Solution
Step 1: Calculate the net force on the +qcharge due to the +qcharge. The
electric force between two point charges q1and q2separated by distance ris
given by Coulomb’s Law:
F=kq1q2
r2
where kis Coulomb’s constant (8.99 ×109N·m2/C2). The magnitude of the
force on the +qcharge due to the other +qcharge will be:
F1=kq2
a2
Step 2: Calculate the direction of the force on the +qcharge due to the +q
charge. The force on the +qcharge due to the other +qcharge will be directed
along the line joining the two charges and will be repulsive.
Step 3: Calculate the net force on the +qcharge due to the −2qcharge.
The magnitude of the force on the +qcharge due to the −2qcharge will be:
F2=kq(2q)
a2√3
Step 4: Calculate the direction of the force on the +qcharge due to the −2q
charge. The force on the +qcharge due to the −2qcharge will be directed along
the line joining the two charges and will be attractive.
Step 5: Find the net force on the +qcharge. The net force Fnet on the
+qcharge will be the vector sum of the forces F1and F2, considering their
directions.
Step 6: Resolve the forces into xand ycomponents to find the net force in
terms of magnitude and direction.
Question 13
Question
Two point charges, q1= +3.0µC and q2=−4.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the charges.
10
Solution
Step 1: Determine the charges in coulombs. We convert µC to C by multiplying
by 10−6:q1= 3.0µC= 3.0×10−6Cq2=−4.0µC=−4.0×10−6C
Step 2: Calculate the distance rbetween the charges in meters. Given
r= 10.0cm = 0.10 m.
Step 3: Use Coulomb’s law to calculate the magnitude of the electric force
between the charges: F=k|q1q2|
r2
where k= 8.99 ×109Nm2/C2is the Coulomb constant.
Step 4: Substitute the given values into the formula to find the electric force:
F=(8.99 ×109)|(3.0×10−6)(−4.0×10−6)|
(0.10)2
Step 5: Calculate the magnitude of the electric force: F=(8.99 ×109)(12 ×10−12)
0.01
F=1.0788 ×10−2
0.01 F= 1.0788 N
Therefore, the magnitude of the electric force between the charges is 1.0788 N.
Question 14
Question
Two point charges, q1=−5µC and q2= 3 µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Recall the formula for the electric force between two point charges:
F=k·|q1·q2|
r2
where Fis the magnitude of the electric force, kis the Coulomb constant (8.99×
109N m2/C2), q1and q2are the magnitudes of the respective charges, and ris
the distance between the charges.
Step 2: Convert the charges to coulombs:
q1=−5µC =−5×10−6C
q2= 3 µC = 3 ×10−6C
Step 3: Plug in the given values into the formula to calculate the electric
force:
F= (8.99 ×109)·| − 5×10−6·3×10−6|
(0.10)2
Step 4: Perform the calculation:
F= (8.99 ×109)·15 ×10−12
0.01
11
F= 8.99 ×109·1.5×10−9
F= 13.485 ×100
F= 13.485 N
Step 5: Therefore, the magnitude of the electric force between the charges
q1and q2is 13.485 N.
Question 15
Question
Three point charges are placed at the corners of a right triangle as shown:
q1= +5 µC at point A,q2=−3µC at point B, and q3= +4 µC at point C.
Calculate the net electric force on q3due to q1and q2. The sides of the triangle
are of lengths 3m, 4m, and 5m.
Solution
Step 1: Calculate the distance between q1at point Aand q3at point C.
Distance between q1and q3=p32+ 42= 5 m
Step 2: Calculate the distance between q2at point Band q3at point C.
Distance between q2and q3= 4 m
Step 3: Calculate the electric force F13 on q3due to q1.
F13 =k|q1||q3|
r2=(9 ×109N m2/C2)(5 ×10−6C)(4 ×10−6C)
52= 144 mN
Step 4: Calculate the electric force F23 on q3due to q2.
F23 =k|q2||q3|
r2=(9 ×109N m2/C2)(3 ×10−6C)(4 ×10−6C)
42= 54 mN
Step 5: Calculate the net electric force on q3.
Fnet =F13 −F23 = 144 mN −54 mN = 90 mN
Therefore, the net electric force on q3due to q1and q2is 90 mN directed
towards q1.
Question 16
Question
Two point charges, q1= 3 µC and q2=−5µC, are placed 8 cm apart in air.
Calculate the magnitude and direction of the electric force on each charge.
12
Solution
Step 1: First calculate the magnitude of the electric force on charge q1.
Given: q1= 3 µC, q2=−5µC, r= 0.08 m
The formula for the magnitude of the electric force between two point charges
is given by Coulomb’s Law:
F=k·|q1q2|
r2
where kis the electrostatic constant approximately equal to 8.99×109N m2/C2.
Substitute the given values into the formula:
F= (8.99 ×109)·|3×10−6× −5×10−6|
0.082
F= 0.84375 N
Therefore, the magnitude of the electric force on charge q1is 0.84375 N.
Step 2: Next, determine the direction of the electric force on charge q1.
Since q1is positive, the electric force on q1will be repulsive. Thus, the
direction of the force on q1is away from q2.
Step 3: Now, calculate the magnitude of the electric force on charge q2.
The magnitude of the electric force on q2is the same as that on q1, but
acting in the opposite direction. Therefore, the magnitude of the electric force
on q2is also 0.84375 N.
Step 4: Determine the direction of the electric force on charge q2.
Since q2is negative, the electric force on q2will be attractive, pulling q2
towards q1. Therefore, the direction of the force on q2is towards q1.
Question 17
Question
Two point charges are placed on the x-axis. The first charge, q1=−3.0µC,
is located at x= 2.0m, and the second charge, q2= 6.0µC, is located at
x= 4.0m. Calculate the magnitude and direction of the electric force on q2due
to q1.
Solution
Step 1: Calculate the distance between the two charges. Given the positions of
the charges, the distance between them can be found using the formula:
r=|x2−x1|
r=|4.0m−2.0m|
r= 2.0m
13
Step 2: Calculate the electric force magnitude. The electric force between
two charges is given by Coulomb’s law:
F=k·|q1q2|
r2
where k= 8.99×109N m2/C2is the electrostatic constant. Substitute the given
values into the formula:
F= 8.99 ×109N m2/C2·|(−3.0µC)(6.0µC)|
(2.0m)2
F= 8.99 ×109N m2/C2·18.0×10−12 C2
4.0m2
F=8.99 N m2/C2·18.0×10−12 C2
4.0m2
F=161.82 ×10−12 N m
4.0m2
F= 40.455 ×10−12 N
F= 40.5pN
Step 3: Determine the direction of the force. The force will be attractive
since the charges have opposite signs. The force on q2is directed towards q1.
Question 18
Question
Two point charges, q1= +3.0µC and q2=−4.0µC, are placed 12.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Determine the distance between the charges in meters. Given: d=
12.0cm = 0.12 m.
Step 2: Calculate the magnitude of the electric force 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 Fis the magnitude of the electric force, kis the Coulomb constant 8.99×
109N m2/C2,|q1|and |q2|are the magnitudes of the charges, and ris the
distance between the charges.
Plugging in the values,
F=(8.99 ×109N m2/C2)·(3.0×10−6C)·(4.0×10−6C)
(0.12 m)2
14
Step 3: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(3.0·4.0) ×10−12
0.0144
F=107.88 ×10−3
0.0144
F=107.88
0.0144
F≈7491.67 N
Therefore, the magnitude of the electric force between the charges q1and q2
is approximately 7491.67 N.
Question 19
Question
Two point charges, q1= 4 µC and q2=−6µC, are separated by a distance of
5cm. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Write down the given values.
Given: q1= 4µC,q2=−6µC,r= 5 cm = 0.05 m.
Step 2: 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 kis the Coulomb constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
Step 3: Substitute the given values into Coulomb’s Law. Plugging in the
values, we get:
F=(9 ×109Nm2/C2)·(4 ×10−6C)·(6 ×10−6C)
(0.05 m)2
Step 4: Calculate the electric force.
F=216 ×10−15 Nm2/C2
0.0025 m2
F= 86.4×10−15 N
F= 8.64 ×10−14 N
Therefore, the magnitude of the electric force between the charges is 8.64 ×
10−14 N.
15
Question 20
Question
Two point charges, q1=−3nC and q2= 5 nC, are placed 10 cm apart. Calculate
the magnitude of the electric force between the two charges.
Solution
Step 1: Calculate the distance between the two charges in meters.
Distance = 10 cm = 10 ×10−2m= 0.1m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
The magnitude of the electric force (F) between two point charges is given by:
F=k·|q1·q2|
r2
where: - kis the electrostatic constant, 8.99 ×109N m2/C2, - q1and q2are the
two charges, and - ris the distance between the two charges.
Plugging in the given values:
F= (8.99 ×109N m2/C2)·| − 3×10−9C×5×10−9C|
(0.1m)2
Step 3: Simplify the expression to find the magnitude of the electric force.
F= (8.99 ×109)·15 ×10−18
0.01
F= 8.99 ×109·1.5×10−15
F= 13.485 ×10−6
F= 1.3485 ×10−5N
Therefore, the magnitude of the electric force between the two charges is
1.3485 ×10−5N.
Question 21
Question
Three point charges are arranged on the corners of an equilateral triangle with
side length a. The charges are +q,−q, and +2q. Calculate the magnitude of
the net electric force acting on the charge located at the corner with the +2q
charge.
16
Solution
Step 1: Calculate the magnitude of the individual electric forces between each
pair of charges:
The force between two charges q1and q2separated by a distance ris given
by Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the Coulomb’s constant (8.99 ×109N·m2/C2).
The distances between the corners of the equilateral triangle are all a.
Therefore, the magnitude of the forces for the given charges are: - +qand
−q:F+q,−q=k·|q|·|q|
a2=kq2
a2-−qand +2q:F−q,+2q=k·|q|·|2q|
a2=2kq2
a2-+2qand
+q:F+2q,+q=k·|2q|·|q|
a2=2kq2
a2
Step 2: Find the horizontal and vertical components of each force.
The forces can be divided into horizontal and vertical components based on
the symmetry of the equilateral triangle: - +qand −q: Horizontal and vertical
components are equal in magnitude but opposite in direction. - −qand +2q:
Horizontal component is twice the vertical component. - +2qand +q: Both
components are the same for this pair.
Step 3: Determine the net force acting on the +2qcharge.
Since the +2qcharge is at an angle of 60◦with the horizontal axis, the net
force can be found by resolving the components of the forces along the horizontal
and vertical directions and then adding them up vectorially.
The net force will depend on the angles and magnitudes of the different
components calculated in Steps 1 and 2.
Question 22
Question
Three point charges are arranged in a line as shown below:
+Q−2Q+Q
If the separation between the charges +Qand −2Qis a, and the separation
between the charges −2Qand +Qis 2a, calculate the magnitude and direction
of the net force on the charge −2Q.
Solution
Step 1: Calculate the electric force between +Qand −2Q. The electric force
between two point charges q1and q2separated by a distance ris given by
Coulomb’s law:
F=k· |q1|·|q2|
r2
17
where kis the electrostatic constant, |q1|and |q2|are the magnitudes of the
charges, and ris the distance between the charges.
For the charges +Qand −2Qwith separation abetween them, the force on
−2Qis attractive, so the force vector points towards the charge +Q.
Step 2: Calculate the electric force between −2Qand +Q. Using the same
formula as in Step 1, we find that the force between −2Qand +Qis attractive,
and the force vector points towards +Q.
Step 3: Calculate the net force on −2Q. Since the forces from +Qand +Q
on −2Qhave the same direction, we can simply add their magnitudes to find
the net force. The force between +Qand −2Qis F1and the force between −2Q
and +Qis F2.
Net force =F1+F2
Step 4: Substitute the electric force formulas into the net force equation.
F1=k·Q·2Q
a2=2kQ2
a2
F2=k·2Q·Q
(2a)2=kQ2
2a2
Substitute these into the net force equation:
Net force =2kQ2
a2+kQ2
2a2=5kQ2
2a2
Step 5: Determine the direction of the net force. Since both individual forces
were directed towards +Q, the net force on −2Qis also directed towards +Q.
Therefore, the magnitude of the net force on the charge −2Qis 5kQ2
2a2, and
the direction is towards +Q.
Question 23
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. The magnitude of each charge is q. Calculate the magnitude and
direction of the net electric force on the charge at the top vertex due to the
other two charges. Given that the side length of the equilateral triangle is a.
+q
/
−q−q
18
Solution
Step 1: Calculate the force due to the charge qat the right vertex.
The magnitude of the force F1is given by Coulomb’s Law:
F1=k|q|2
a2
where kis the Coulomb constant (8.99 ×109N m2/C2), qis the magnitude
of the charge, and ais the distance between the charges.
Step 2: Calculate the force due to the charge qat the left vertex.
Since the charges are equidistant from the charge at the top vertex and the
charges are equidistant from each other, the magnitude of the force F2due to
the charge qat the left vertex is also F1.
Step 3: Find the net force on the charge at the top vertex.
The net force is the vector sum of the forces F1and F2. Since they are both
acting in the vertical direction but in opposite directions, the net force will be
the difference between the two.
Thus, the net force on the charge at the top vertex is:
Net force =F1−F2
Now, substitute the expressions for F1and F2into the above equation and
simplify to find the net force.
Question 24
Question
Two point charges, q1=−2.5µC and q2= 3.0µC, are placed 8.0 cm apart in
air. Calculate the magnitude of the electric force between them.
Solution
Step 1: Convert the charges to standard units. Step 2: Calculate the distance
between the charges in meters. Step 3: Use Coulomb’s law to find the magnitude
of the electric force between the charges.
Step 1: Convert the charges to standard units. Given: q1=−2.5µC
and q2= 3.0µC. Converting to standard units: q1=−2.5×10−6C and
q2= 3.0×10−6C.
Step 2: Calculate the distance between the charges in meters. Given: dis-
tance = 8.0 cm = 0.08 m.
Step 3: Use Coulomb’s law to find the magnitude of the electric force
between the charges. Coulomb’s law:
F=k·|q1·q2|
r2
19
where: F= electric force, k= Coulomb’s constant (8.99 ×109N·m2/C2),q1
and q2= magnitudes of the charges, r= distance between the charges.
Plugging in the values:
F= (8.99 ×109)·|(−2.5×10−6)·(3.0×10−6)|
(0.08)2
F= (8.99 ×109)·7.5×10−12
0.0064
F= (8.99 ×109)·1.171875 ×10−9
F= 10.54 N
Therefore, the magnitude of the electric force between the charges is 10.54
N.
Question 25
Question
Two point charges, q1= +6.0µC and q2=−3.0µC, are placed 20 cm apart in
a vacuum. Calculate the magnitude of the electric force exerted by q1on q2.
Solution
Step 1: Convert the given charges from microcoulombs to coulombs. Step 2:
Determine the distance between the charges in meters. Step 3: Calculate the
magnitude of the electric force between the charges using Coulomb’s law.
Step 1: Convert the given charges from microcoulombs to coulombs.
q1= 6.0µC = 6.0×10−6C
q2=−3.0µC =−3.0×10−6C
Step 2: Determine the distance between the charges in meters.
r= 20 cm = 20 ×10−2m= 0.20 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law.
Coulomb’s law states that the magnitude of the electric force Fbetween two
point charges is given by
F=k|q1·q2|
r2
where - kis the Coulomb constant, 8.9875 ×109Nm2/C2, - q1, q2are the mag-
nitudes of the two point charges, and - ris the distance between the charges.
Plugging in the values:
F= 8.9875 ×109·|6.0×10−6· −3.0×10−6|
0.202
20
F= 4.49275 ×10−3N
Thus, the magnitude of the electric force exerted by q1on q2is 4.49275 ×
10−3N.
21
Question 2
Question
Two point charges, q1=−4µC and q2= 6 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
Solution
To find the electric force between two point charges, we can use Coulomb’s Law,
which states that the magnitude of the force between two point charges is given
by:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Step 1: Convert the charges to coulombs and the distance to meters. Given:
q1=−4µC=−4×10−6C, q2= 6 µC= 6 ×10−6C, r= 10 cm = 0.1m.
Step 2: Calculate the electric force. Plugging the given values into Coulomb’s
Law, we have:
F= (8.99 ×109)·| − 4×10−6×6×10−6|
(0.1)2
F= 8.99 ×109·24 ×10−12
0.01
F= 8.99 ×109·2.4×10−9
F= 21.576 ×10−9
F= 2.1576 ×10−8N
Therefore, the magnitude of the electric force between the charges is 2.1576×
10−8N.
Question 3
Question
Two point charges +3 µC and −5µC are placed 12 cm apart in air. Calculate
the magnitude of the electric force acting on each charge. Given: k= 8.99 ×
109N m2/C2.
2
Solution
Step 1: Convert the charges to coulombs.
+3 µC= 3 ×10−6C
−5µC=−5×10−6C
Step 2: Calculate the magnitude of the electric force exerted by the +3 µC
charge on the −5µC charge using Coulomb’s Law:
F=k· |q1|·|q2|
r2
where F= magnitude of the electric force, k= Coulomb’s constant (8.99 ×
109N m2/C2), |q1|= 3 ×10−6C, |q2|= 5 ×10−6C, and r= 0.12 m.
Substitute the values into the formula:
F=(8.99 ×109N m2/C2)·(3 ×10−6C)·(5 ×10−6C)
(0.12 m)2
Step 3: Calculate the magnitude of the electric force between the charges.
Question 4
Question
Two point charges, q1=−3µC and q2= 5µC, are located 6 cm apart. Calculate
the magnitude of the electric force between these charges.
Solution
Step 1: Calculate the distance between the charges in meters. Given: Distance,
r= 6 cm = 0.06 m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
Coulomb’s law states that the magnitude of the electric force between two point
charges is given by:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging in the given values:
F= 8.99 ×109| − 3×10−6×5×10−6|
0.062
Step 3: Calculate the magnitude of the electric force.
F= 8.99 ×109×15 ×10−12
0.0036
3
F= 8.99 ×109×4.1667 ×10−9
F= 37.45 N
Therefore, the magnitude of the electric force between the charges q1and q2
is 37.45 N.
Question 5
Question
Two point charges, q1and q2, are held 4 cm apart in air. If q1= 5µC and
q2=−3µC, calculate the magnitude of the electric force between them.
Solution
Step 1: Calculate the electric force using Coulomb’s Law formula:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2) and ris the distance
between the charges.
Step 2: Substitute the given values into the formula:
F=(8.99 ×109N m2/C2)· |5×10−6C·(−3×10−6C)|
(0.04 m)2
Step 3: Calculate the magnitude of the force:
F=(8.99 ×109)·(5 ×3) ×10−12
0.0016
Step 4: Simplify the expression:
F=44.95 ×10−3
0.0016
Step 5: Calculate the final answer:
F= 28.09 N
Thus, the magnitude of the electric force between the charges q1and q2is
28.09 N.
Question 6
Question
Two point charges, +4.0µC and −2.0µC, are placed 10.0 cm apart. Calculate
the magnitude of the electric force between these two charges.
4
Solution
Step 1: Convert the charges to Coulombs. Given: q1= +4.0µC = 4.0×10−6C
q2=−2.0µC =−2.0×10−6C
Step 2: Calculate the distance between the charges in meters. d= 10.0cm =
10.0×10−2m= 0.10 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law:
F=k
q1·q2
d2
where k= 8.99 ×109Nm2/C2is Coulomb’s constant.
Step 4: Substituting the given values into the formula:
F= 8.99 ×109×
4.0×10−6× −2.0×10−6
(0.10)2
Step 5: Calculate the magnitude of the electric force:
F= 8.99 ×109×8.0×10−12
0.01
F= 8.99 ×109×8.0×10−10
F= 7.19 ×10−1N
Therefore, the magnitude of the electric force between the two charges is
7.19 ×10−1N.
Question 7
Question
Two point charges, q1=−2µC and q2= 3 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between them.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−2µC and q2= 3 µC
Converting to coulombs: q1=−2×10−6C and q2= 3 ×10−6C.
Step 2: Calculate the magnitude of electric force using Coulomb’s Law. The
magnitude of the electric force between point charges is given by Coulomb’s
Law:
F=k
q1q2
r2
,
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
5
Plugging in the given values:
F= 8.99 ×109
(−2×10−6)(3 ×10−6)
(0.10)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= 5.394 N
Therefore, the magnitude of the electric force between the point charges q1
and q2is 5.394 N.
Question 8
Question
Two point charges, q1=−3.0µC and q2= 4.0µC, are placed 8.0 cm apart.
Calculate the magnitude and direction of the electric force between the two
charges.
Solution
Step 1: Convert the charges to coulombs and write down the given values.
Given: q1=−3.0µC = −3.0×10−6C (negative sign indicates charge is nega-
tive)
q2= 4.0µC = 4.0×10−6C
Distance r= 8.0 cm = 0.08 m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law
formula:
F=k·|q1·q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the given values:
F= 8.99 ×109·| − 3.0×10−6·4.0×10−6|
(0.08)2
Step 3: Calculate the magnitude of the electric force:
6
F= 8.99 ×109·12 ×10−12
0.0064
F= 8.99 ×109·1.875 ×10−6
F= 16.87 N
Step 4: Determine the direction of the force. Since the charges have opposite
signs, the force will be attractive. Thus, the force is directed from q1to q2.
Therefore, the magnitude of the electric force between the charges is 16.87
N, and it is directed from q1to q2.
Question 9
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 10.0 cm apart.
Calculate the magnitude and direction of the electric force that q1exerts on q2.
Solution
To find the electric force that q1exerts on q2, we can use Coulomb’s Law:
F=k·|q1|·|q2|
r2
where k= 8.99 ×109N·m2/C2is Coulomb’s constant, |q1|and |q2|are the
magnitudes of the charges, and ris the distance between the charges.
Step 1: Calculate the distance rin meters: Given: r= 10.0cm = 0.10 m.
Step 2: Substitute the values into Coulomb’s Law to find the electric force:
F= (8.99 ×109N·m2/C2)·(4.0×10−6C)·(6.0×10−6C)
(0.10 m)2
Step 3: Calculate the magnitude of the electric force:
F= (8.99 ×109)·24 ×10−12
0.01
F= 8.99 ×109×2.4×10−9
F= 21.576 ×100
F= 21.576 N
Step 4: Determine the direction of the force: Since q1is negative and q2is
positive, the electric force will be attractive, that is, the force will be directed
toward q1.
Therefore, the magnitude of the electric force that q1exerts on q2is 21.576 N
directed towards q1.
7
Question 10
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are fixed in place and
separated by a distance of 5.0cm. Calculate the magnitude of the electric force
that q1exerts on q2.
Solution
Step 1: Convert the charges to Coulombs. Given: q1=−4.0µC and q2=
6.0µC.
To convert from microCoulombs (µC) to Coulombs, we multiply by 10−6.q1=
−4.0×10−6C and q2= 6.0×10−6C.
Step 2: Calculate the magnitude of the electric force using the Coulomb’s law
formula. Coulomb’s law states that the magnitude of the electric force between
two charges is given by
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.
Plugging in the values gives us:
F= 8.99 ×109N m2/C2×| − 4.0×10−6C×6.0×10−6C|
(0.05 m)2.
Step 3: Substitute the values and calculate the electric force.
F= 8.99×109×4.0×6.0×10−6×10−6
0.052= 8.99×109×24 ×10−12
0.0025 = 8.99×109×9.6×10−10 = 8.62 N.
Therefore, the magnitude of the electric force that q1exerts on q2is 8.62 N.
Question 11
Question
In an equilateral triangle with side length a, three point charges are placed at
its vertices. The charges are +q,−2q, and +3q, respectively. Calculate the
magnitude and direction of the net electric force acting on the charge at the
first vertex (+q) due to the other two charges.
Solution
Given: Side length of equilateral triangle, a. Point charges at the vertices: +q,
−2q, and +3q.
8
Step 1: Calculate the electric force between the charge at the first vertex
(+q) and the charge at the second vertex (−2q). The electric force between two
charges q1and q2separated by a distance ris given by Coulomb’s law:
F=k·|q1q2|
r2
where kis Coulomb’s constant (8.9875 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
The distance between the first and second vertices of the equilateral triangle
is a. So, the magnitude of the force between +qand −2qis:
F1,2=k|+q(−2q)|
a2
Step 2: Determine the direction of the force between the charge at the first
vertex and the charge at the second vertex. Since the charges are opposite in
sign, the force between them is attractive. It acts along the line joining the
charges, pointing from the second vertex towards the first vertex.
Step 3: Calculate the electric force between the charge at the first vertex
(+q) and the charge at the third vertex (+3q). The distance between the first
and third vertices of the equilateral triangle is a. So, the magnitude of the force
between +qand +3qis:
F1,3=k|+q(+3q)|
a2
Step 4: Determine the direction of the force between the charge at the first
vertex and the charge at the third vertex. Since the charges are like in sign,
the force between them is repulsive. It acts along the line joining the charges,
pointing from the first vertex towards the third vertex.
Step 5: Find the net force on the charge at the first vertex. The net force
on the charge at the first vertex is the vector sum of the forces F1,2and F1,3.
Since these forces are along different directions, we need to consider their vector
nature.
The net force can be calculated using the parallelogram rule of vector addi-
tion:
Net force =q(F2
1,2+F2
1,3+ 2F1,2F1,3cos θ)
where θis the angle between the forces.
Solving this expression will give the magnitude and direction of the net
electric force acting on the charge at the first vertex due to the other two charges.
9
Question 12
Question
Three point charges are arranged in an equilateral triangle as shown below:
+q
↗
+q−2q
If each side of the triangle has a length a, calculate the net electric force on the
+qcharge.
Solution
Step 1: Calculate the net force on the +qcharge due to the +qcharge. The
electric force between two point charges q1and q2separated by distance ris
given by Coulomb’s Law:
F=kq1q2
r2
where kis Coulomb’s constant (8.99 ×109N·m2/C2). The magnitude of the
force on the +qcharge due to the other +qcharge will be:
F1=kq2
a2
Step 2: Calculate the direction of the force on the +qcharge due to the +q
charge. The force on the +qcharge due to the other +qcharge will be directed
along the line joining the two charges and will be repulsive.
Step 3: Calculate the net force on the +qcharge due to the −2qcharge.
The magnitude of the force on the +qcharge due to the −2qcharge will be:
F2=kq(2q)
a2√3
Step 4: Calculate the direction of the force on the +qcharge due to the −2q
charge. The force on the +qcharge due to the −2qcharge will be directed along
the line joining the two charges and will be attractive.
Step 5: Find the net force on the +qcharge. The net force Fnet on the
+qcharge will be the vector sum of the forces F1and F2, considering their
directions.
Step 6: Resolve the forces into xand ycomponents to find the net force in
terms of magnitude and direction.
Question 13
Question
Two point charges, q1= +3.0µC and q2=−4.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the charges.
10
Solution
Step 1: Determine the charges in coulombs. We convert µC to C by multiplying
by 10−6:q1= 3.0µC= 3.0×10−6Cq2=−4.0µC=−4.0×10−6C
Step 2: Calculate the distance rbetween the charges in meters. Given
r= 10.0cm = 0.10 m.
Step 3: Use Coulomb’s law to calculate the magnitude of the electric force
between the charges: F=k|q1q2|
r2
where k= 8.99 ×109Nm2/C2is the Coulomb constant.
Step 4: Substitute the given values into the formula to find the electric force:
F=(8.99 ×109)|(3.0×10−6)(−4.0×10−6)|
(0.10)2
Step 5: Calculate the magnitude of the electric force: F=(8.99 ×109)(12 ×10−12)
0.01
F=1.0788 ×10−2
0.01 F= 1.0788 N
Therefore, the magnitude of the electric force between the charges is 1.0788 N.
Question 14
Question
Two point charges, q1=−5µC and q2= 3 µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Recall the formula for the electric force between two point charges:
F=k·|q1·q2|
r2
where Fis the magnitude of the electric force, kis the Coulomb constant (8.99×
109N m2/C2), q1and q2are the magnitudes of the respective charges, and ris
the distance between the charges.
Step 2: Convert the charges to coulombs:
q1=−5µC =−5×10−6C
q2= 3 µC = 3 ×10−6C
Step 3: Plug in the given values into the formula to calculate the electric
force:
F= (8.99 ×109)·| − 5×10−6·3×10−6|
(0.10)2
Step 4: Perform the calculation:
F= (8.99 ×109)·15 ×10−12
0.01
11
F= 8.99 ×109·1.5×10−9
F= 13.485 ×100
F= 13.485 N
Step 5: Therefore, the magnitude of the electric force between the charges
q1and q2is 13.485 N.
Question 15
Question
Three point charges are placed at the corners of a right triangle as shown:
q1= +5 µC at point A,q2=−3µC at point B, and q3= +4 µC at point C.
Calculate the net electric force on q3due to q1and q2. The sides of the triangle
are of lengths 3m, 4m, and 5m.
Solution
Step 1: Calculate the distance between q1at point Aand q3at point C.
Distance between q1and q3=p32+ 42= 5 m
Step 2: Calculate the distance between q2at point Band q3at point C.
Distance between q2and q3= 4 m
Step 3: Calculate the electric force F13 on q3due to q1.
F13 =k|q1||q3|
r2=(9 ×109N m2/C2)(5 ×10−6C)(4 ×10−6C)
52= 144 mN
Step 4: Calculate the electric force F23 on q3due to q2.
F23 =k|q2||q3|
r2=(9 ×109N m2/C2)(3 ×10−6C)(4 ×10−6C)
42= 54 mN
Step 5: Calculate the net electric force on q3.
Fnet =F13 −F23 = 144 mN −54 mN = 90 mN
Therefore, the net electric force on q3due to q1and q2is 90 mN directed
towards q1.
Question 16
Question
Two point charges, q1= 3 µC and q2=−5µC, are placed 8 cm apart in air.
Calculate the magnitude and direction of the electric force on each charge.
12
Solution
Step 1: First calculate the magnitude of the electric force on charge q1.
Given: q1= 3 µC, q2=−5µC, r= 0.08 m
The formula for the magnitude of the electric force between two point charges
is given by Coulomb’s Law:
F=k·|q1q2|
r2
where kis the electrostatic constant approximately equal to 8.99×109N m2/C2.
Substitute the given values into the formula:
F= (8.99 ×109)·|3×10−6× −5×10−6|
0.082
F= 0.84375 N
Therefore, the magnitude of the electric force on charge q1is 0.84375 N.
Step 2: Next, determine the direction of the electric force on charge q1.
Since q1is positive, the electric force on q1will be repulsive. Thus, the
direction of the force on q1is away from q2.
Step 3: Now, calculate the magnitude of the electric force on charge q2.
The magnitude of the electric force on q2is the same as that on q1, but
acting in the opposite direction. Therefore, the magnitude of the electric force
on q2is also 0.84375 N.
Step 4: Determine the direction of the electric force on charge q2.
Since q2is negative, the electric force on q2will be attractive, pulling q2
towards q1. Therefore, the direction of the force on q2is towards q1.
Question 17
Question
Two point charges are placed on the x-axis. The first charge, q1=−3.0µC,
is located at x= 2.0m, and the second charge, q2= 6.0µC, is located at
x= 4.0m. Calculate the magnitude and direction of the electric force on q2due
to q1.
Solution
Step 1: Calculate the distance between the two charges. Given the positions of
the charges, the distance between them can be found using the formula:
r=|x2−x1|
r=|4.0m−2.0m|
r= 2.0m
13
Step 2: Calculate the electric force magnitude. The electric force between
two charges is given by Coulomb’s law:
F=k·|q1q2|
r2
where k= 8.99×109N m2/C2is the electrostatic constant. Substitute the given
values into the formula:
F= 8.99 ×109N m2/C2·|(−3.0µC)(6.0µC)|
(2.0m)2
F= 8.99 ×109N m2/C2·18.0×10−12 C2
4.0m2
F=8.99 N m2/C2·18.0×10−12 C2
4.0m2
F=161.82 ×10−12 N m
4.0m2
F= 40.455 ×10−12 N
F= 40.5pN
Step 3: Determine the direction of the force. The force will be attractive
since the charges have opposite signs. The force on q2is directed towards q1.
Question 18
Question
Two point charges, q1= +3.0µC and q2=−4.0µC, are placed 12.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Determine the distance between the charges in meters. Given: d=
12.0cm = 0.12 m.
Step 2: Calculate the magnitude of the electric force 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 Fis the magnitude of the electric force, kis the Coulomb constant 8.99×
109N m2/C2,|q1|and |q2|are the magnitudes of the charges, and ris the
distance between the charges.
Plugging in the values,
F=(8.99 ×109N m2/C2)·(3.0×10−6C)·(4.0×10−6C)
(0.12 m)2
14
Step 3: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(3.0·4.0) ×10−12
0.0144
F=107.88 ×10−3
0.0144
F=107.88
0.0144
F≈7491.67 N
Therefore, the magnitude of the electric force between the charges q1and q2
is approximately 7491.67 N.
Question 19
Question
Two point charges, q1= 4 µC and q2=−6µC, are separated by a distance of
5cm. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Write down the given values.
Given: q1= 4µC,q2=−6µC,r= 5 cm = 0.05 m.
Step 2: 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 kis the Coulomb constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
Step 3: Substitute the given values into Coulomb’s Law. Plugging in the
values, we get:
F=(9 ×109Nm2/C2)·(4 ×10−6C)·(6 ×10−6C)
(0.05 m)2
Step 4: Calculate the electric force.
F=216 ×10−15 Nm2/C2
0.0025 m2
F= 86.4×10−15 N
F= 8.64 ×10−14 N
Therefore, the magnitude of the electric force between the charges is 8.64 ×
10−14 N.
15
Question 20
Question
Two point charges, q1=−3nC and q2= 5 nC, are placed 10 cm apart. Calculate
the magnitude of the electric force between the two charges.
Solution
Step 1: Calculate the distance between the two charges in meters.
Distance = 10 cm = 10 ×10−2m= 0.1m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
The magnitude of the electric force (F) between two point charges is given by:
F=k·|q1·q2|
r2
where: - kis the electrostatic constant, 8.99 ×109N m2/C2, - q1and q2are the
two charges, and - ris the distance between the two charges.
Plugging in the given values:
F= (8.99 ×109N m2/C2)·| − 3×10−9C×5×10−9C|
(0.1m)2
Step 3: Simplify the expression to find the magnitude of the electric force.
F= (8.99 ×109)·15 ×10−18
0.01
F= 8.99 ×109·1.5×10−15
F= 13.485 ×10−6
F= 1.3485 ×10−5N
Therefore, the magnitude of the electric force between the two charges is
1.3485 ×10−5N.
Question 21
Question
Three point charges are arranged on the corners of an equilateral triangle with
side length a. The charges are +q,−q, and +2q. Calculate the magnitude of
the net electric force acting on the charge located at the corner with the +2q
charge.
16
Solution
Step 1: Calculate the magnitude of the individual electric forces between each
pair of charges:
The force between two charges q1and q2separated by a distance ris given
by Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the Coulomb’s constant (8.99 ×109N·m2/C2).
The distances between the corners of the equilateral triangle are all a.
Therefore, the magnitude of the forces for the given charges are: - +qand
−q:F+q,−q=k·|q|·|q|
a2=kq2
a2-−qand +2q:F−q,+2q=k·|q|·|2q|
a2=2kq2
a2-+2qand
+q:F+2q,+q=k·|2q|·|q|
a2=2kq2
a2
Step 2: Find the horizontal and vertical components of each force.
The forces can be divided into horizontal and vertical components based on
the symmetry of the equilateral triangle: - +qand −q: Horizontal and vertical
components are equal in magnitude but opposite in direction. - −qand +2q:
Horizontal component is twice the vertical component. - +2qand +q: Both
components are the same for this pair.
Step 3: Determine the net force acting on the +2qcharge.
Since the +2qcharge is at an angle of 60◦with the horizontal axis, the net
force can be found by resolving the components of the forces along the horizontal
and vertical directions and then adding them up vectorially.
The net force will depend on the angles and magnitudes of the different
components calculated in Steps 1 and 2.
Question 22
Question
Three point charges are arranged in a line as shown below:
+Q−2Q+Q
If the separation between the charges +Qand −2Qis a, and the separation
between the charges −2Qand +Qis 2a, calculate the magnitude and direction
of the net force on the charge −2Q.
Solution
Step 1: Calculate the electric force between +Qand −2Q. The electric force
between two point charges q1and q2separated by a distance ris given by
Coulomb’s law:
F=k· |q1|·|q2|
r2
17
where kis the electrostatic constant, |q1|and |q2|are the magnitudes of the
charges, and ris the distance between the charges.
For the charges +Qand −2Qwith separation abetween them, the force on
−2Qis attractive, so the force vector points towards the charge +Q.
Step 2: Calculate the electric force between −2Qand +Q. Using the same
formula as in Step 1, we find that the force between −2Qand +Qis attractive,
and the force vector points towards +Q.
Step 3: Calculate the net force on −2Q. Since the forces from +Qand +Q
on −2Qhave the same direction, we can simply add their magnitudes to find
the net force. The force between +Qand −2Qis F1and the force between −2Q
and +Qis F2.
Net force =F1+F2
Step 4: Substitute the electric force formulas into the net force equation.
F1=k·Q·2Q
a2=2kQ2
a2
F2=k·2Q·Q
(2a)2=kQ2
2a2
Substitute these into the net force equation:
Net force =2kQ2
a2+kQ2
2a2=5kQ2
2a2
Step 5: Determine the direction of the net force. Since both individual forces
were directed towards +Q, the net force on −2Qis also directed towards +Q.
Therefore, the magnitude of the net force on the charge −2Qis 5kQ2
2a2, and
the direction is towards +Q.
Question 23
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. The magnitude of each charge is q. Calculate the magnitude and
direction of the net electric force on the charge at the top vertex due to the
other two charges. Given that the side length of the equilateral triangle is a.
+q
/
−q−q
18
Solution
Step 1: Calculate the force due to the charge qat the right vertex.
The magnitude of the force F1is given by Coulomb’s Law:
F1=k|q|2
a2
where kis the Coulomb constant (8.99 ×109N m2/C2), qis the magnitude
of the charge, and ais the distance between the charges.
Step 2: Calculate the force due to the charge qat the left vertex.
Since the charges are equidistant from the charge at the top vertex and the
charges are equidistant from each other, the magnitude of the force F2due to
the charge qat the left vertex is also F1.
Step 3: Find the net force on the charge at the top vertex.
The net force is the vector sum of the forces F1and F2. Since they are both
acting in the vertical direction but in opposite directions, the net force will be
the difference between the two.
Thus, the net force on the charge at the top vertex is:
Net force =F1−F2
Now, substitute the expressions for F1and F2into the above equation and
simplify to find the net force.
Question 24
Question
Two point charges, q1=−2.5µC and q2= 3.0µC, are placed 8.0 cm apart in
air. Calculate the magnitude of the electric force between them.
Solution
Step 1: Convert the charges to standard units. Step 2: Calculate the distance
between the charges in meters. Step 3: Use Coulomb’s law to find the magnitude
of the electric force between the charges.
Step 1: Convert the charges to standard units. Given: q1=−2.5µC
and q2= 3.0µC. Converting to standard units: q1=−2.5×10−6C and
q2= 3.0×10−6C.
Step 2: Calculate the distance between the charges in meters. Given: dis-
tance = 8.0 cm = 0.08 m.
Step 3: Use Coulomb’s law to find the magnitude of the electric force
between the charges. Coulomb’s law:
F=k·|q1·q2|
r2
19
where: F= electric force, k= Coulomb’s constant (8.99 ×109N·m2/C2),q1
and q2= magnitudes of the charges, r= distance between the charges.
Plugging in the values:
F= (8.99 ×109)·|(−2.5×10−6)·(3.0×10−6)|
(0.08)2
F= (8.99 ×109)·7.5×10−12
0.0064
F= (8.99 ×109)·1.171875 ×10−9
F= 10.54 N
Therefore, the magnitude of the electric force between the charges is 10.54
N.
Question 25
Question
Two point charges, q1= +6.0µC and q2=−3.0µC, are placed 20 cm apart in
a vacuum. Calculate the magnitude of the electric force exerted by q1on q2.
Solution
Step 1: Convert the given charges from microcoulombs to coulombs. Step 2:
Determine the distance between the charges in meters. Step 3: Calculate the
magnitude of the electric force between the charges using Coulomb’s law.
Step 1: Convert the given charges from microcoulombs to coulombs.
q1= 6.0µC = 6.0×10−6C
q2=−3.0µC =−3.0×10−6C
Step 2: Determine the distance between the charges in meters.
r= 20 cm = 20 ×10−2m= 0.20 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law.
Coulomb’s law states that the magnitude of the electric force Fbetween two
point charges is given by
F=k|q1·q2|
r2
where - kis the Coulomb constant, 8.9875 ×109Nm2/C2, - q1, q2are the mag-
nitudes of the two point charges, and - ris the distance between the charges.
Plugging in the values:
F= 8.9875 ×109·|6.0×10−6· −3.0×10−6|
0.202
20
F= 4.49275 ×10−3N
Thus, the magnitude of the electric force exerted by q1on q2is 4.49275 ×
10−3N.
21
Question 2
Question
Two point charges, q1=−4µC and q2= 6 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
Solution
To find the electric force between two point charges, we can use Coulomb’s Law,
which states that the magnitude of the force between two point charges is given
by:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Step 1: Convert the charges to coulombs and the distance to meters. Given:
q1=−4µC=−4×10−6C, q2= 6 µC= 6 ×10−6C, r= 10 cm = 0.1m.
Step 2: Calculate the electric force. Plugging the given values into Coulomb’s
Law, we have:
F= (8.99 ×109)·| − 4×10−6×6×10−6|
(0.1)2
F= 8.99 ×109·24 ×10−12
0.01
F= 8.99 ×109·2.4×10−9
F= 21.576 ×10−9
F= 2.1576 ×10−8N
Therefore, the magnitude of the electric force between the charges is 2.1576×
10−8N.
Question 3
Question
Two point charges +3 µC and −5µC are placed 12 cm apart in air. Calculate
the magnitude of the electric force acting on each charge. Given: k= 8.99 ×
109N m2/C2.
2
Solution
Step 1: Convert the charges to coulombs.
+3 µC= 3 ×10−6C
−5µC=−5×10−6C
Step 2: Calculate the magnitude of the electric force exerted by the +3 µC
charge on the −5µC charge using Coulomb’s Law:
F=k· |q1|·|q2|
r2
where F= magnitude of the electric force, k= Coulomb’s constant (8.99 ×
109N m2/C2), |q1|= 3 ×10−6C, |q2|= 5 ×10−6C, and r= 0.12 m.
Substitute the values into the formula:
F=(8.99 ×109N m2/C2)·(3 ×10−6C)·(5 ×10−6C)
(0.12 m)2
Step 3: Calculate the magnitude of the electric force between the charges.
Question 4
Question
Two point charges, q1=−3µC and q2= 5µC, are located 6 cm apart. Calculate
the magnitude of the electric force between these charges.
Solution
Step 1: Calculate the distance between the charges in meters. Given: Distance,
r= 6 cm = 0.06 m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
Coulomb’s law states that the magnitude of the electric force between two point
charges is given by:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging in the given values:
F= 8.99 ×109| − 3×10−6×5×10−6|
0.062
Step 3: Calculate the magnitude of the electric force.
F= 8.99 ×109×15 ×10−12
0.0036
3
F= 8.99 ×109×4.1667 ×10−9
F= 37.45 N
Therefore, the magnitude of the electric force between the charges q1and q2
is 37.45 N.
Question 5
Question
Two point charges, q1and q2, are held 4 cm apart in air. If q1= 5µC and
q2=−3µC, calculate the magnitude of the electric force between them.
Solution
Step 1: Calculate the electric force using Coulomb’s Law formula:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2) and ris the distance
between the charges.
Step 2: Substitute the given values into the formula:
F=(8.99 ×109N m2/C2)· |5×10−6C·(−3×10−6C)|
(0.04 m)2
Step 3: Calculate the magnitude of the force:
F=(8.99 ×109)·(5 ×3) ×10−12
0.0016
Step 4: Simplify the expression:
F=44.95 ×10−3
0.0016
Step 5: Calculate the final answer:
F= 28.09 N
Thus, the magnitude of the electric force between the charges q1and q2is
28.09 N.
Question 6
Question
Two point charges, +4.0µC and −2.0µC, are placed 10.0 cm apart. Calculate
the magnitude of the electric force between these two charges.
4
Solution
Step 1: Convert the charges to Coulombs. Given: q1= +4.0µC = 4.0×10−6C
q2=−2.0µC =−2.0×10−6C
Step 2: Calculate the distance between the charges in meters. d= 10.0cm =
10.0×10−2m= 0.10 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law:
F=k
q1·q2
d2
where k= 8.99 ×109Nm2/C2is Coulomb’s constant.
Step 4: Substituting the given values into the formula:
F= 8.99 ×109×
4.0×10−6× −2.0×10−6
(0.10)2
Step 5: Calculate the magnitude of the electric force:
F= 8.99 ×109×8.0×10−12
0.01
F= 8.99 ×109×8.0×10−10
F= 7.19 ×10−1N
Therefore, the magnitude of the electric force between the two charges is
7.19 ×10−1N.
Question 7
Question
Two point charges, q1=−2µC and q2= 3 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between them.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−2µC and q2= 3 µC
Converting to coulombs: q1=−2×10−6C and q2= 3 ×10−6C.
Step 2: Calculate the magnitude of electric force using Coulomb’s Law. The
magnitude of the electric force between point charges is given by Coulomb’s
Law:
F=k
q1q2
r2
,
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
5
Plugging in the given values:
F= 8.99 ×109
(−2×10−6)(3 ×10−6)
(0.10)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= 5.394 N
Therefore, the magnitude of the electric force between the point charges q1
and q2is 5.394 N.
Question 8
Question
Two point charges, q1=−3.0µC and q2= 4.0µC, are placed 8.0 cm apart.
Calculate the magnitude and direction of the electric force between the two
charges.
Solution
Step 1: Convert the charges to coulombs and write down the given values.
Given: q1=−3.0µC = −3.0×10−6C (negative sign indicates charge is nega-
tive)
q2= 4.0µC = 4.0×10−6C
Distance r= 8.0 cm = 0.08 m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law
formula:
F=k·|q1·q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the given values:
F= 8.99 ×109·| − 3.0×10−6·4.0×10−6|
(0.08)2
Step 3: Calculate the magnitude of the electric force:
6
F= 8.99 ×109·12 ×10−12
0.0064
F= 8.99 ×109·1.875 ×10−6
F= 16.87 N
Step 4: Determine the direction of the force. Since the charges have opposite
signs, the force will be attractive. Thus, the force is directed from q1to q2.
Therefore, the magnitude of the electric force between the charges is 16.87
N, and it is directed from q1to q2.
Question 9
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 10.0 cm apart.
Calculate the magnitude and direction of the electric force that q1exerts on q2.
Solution
To find the electric force that q1exerts on q2, we can use Coulomb’s Law:
F=k·|q1|·|q2|
r2
where k= 8.99 ×109N·m2/C2is Coulomb’s constant, |q1|and |q2|are the
magnitudes of the charges, and ris the distance between the charges.
Step 1: Calculate the distance rin meters: Given: r= 10.0cm = 0.10 m.
Step 2: Substitute the values into Coulomb’s Law to find the electric force:
F= (8.99 ×109N·m2/C2)·(4.0×10−6C)·(6.0×10−6C)
(0.10 m)2
Step 3: Calculate the magnitude of the electric force:
F= (8.99 ×109)·24 ×10−12
0.01
F= 8.99 ×109×2.4×10−9
F= 21.576 ×100
F= 21.576 N
Step 4: Determine the direction of the force: Since q1is negative and q2is
positive, the electric force will be attractive, that is, the force will be directed
toward q1.
Therefore, the magnitude of the electric force that q1exerts on q2is 21.576 N
directed towards q1.
7
Question 10
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are fixed in place and
separated by a distance of 5.0cm. Calculate the magnitude of the electric force
that q1exerts on q2.
Solution
Step 1: Convert the charges to Coulombs. Given: q1=−4.0µC and q2=
6.0µC.
To convert from microCoulombs (µC) to Coulombs, we multiply by 10−6.q1=
−4.0×10−6C and q2= 6.0×10−6C.
Step 2: Calculate the magnitude of the electric force using the Coulomb’s law
formula. Coulomb’s law states that the magnitude of the electric force between
two charges is given by
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.
Plugging in the values gives us:
F= 8.99 ×109N m2/C2×| − 4.0×10−6C×6.0×10−6C|
(0.05 m)2.
Step 3: Substitute the values and calculate the electric force.
F= 8.99×109×4.0×6.0×10−6×10−6
0.052= 8.99×109×24 ×10−12
0.0025 = 8.99×109×9.6×10−10 = 8.62 N.
Therefore, the magnitude of the electric force that q1exerts on q2is 8.62 N.
Question 11
Question
In an equilateral triangle with side length a, three point charges are placed at
its vertices. The charges are +q,−2q, and +3q, respectively. Calculate the
magnitude and direction of the net electric force acting on the charge at the
first vertex (+q) due to the other two charges.
Solution
Given: Side length of equilateral triangle, a. Point charges at the vertices: +q,
−2q, and +3q.
8
Step 1: Calculate the electric force between the charge at the first vertex
(+q) and the charge at the second vertex (−2q). The electric force between two
charges q1and q2separated by a distance ris given by Coulomb’s law:
F=k·|q1q2|
r2
where kis Coulomb’s constant (8.9875 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
The distance between the first and second vertices of the equilateral triangle
is a. So, the magnitude of the force between +qand −2qis:
F1,2=k|+q(−2q)|
a2
Step 2: Determine the direction of the force between the charge at the first
vertex and the charge at the second vertex. Since the charges are opposite in
sign, the force between them is attractive. It acts along the line joining the
charges, pointing from the second vertex towards the first vertex.
Step 3: Calculate the electric force between the charge at the first vertex
(+q) and the charge at the third vertex (+3q). The distance between the first
and third vertices of the equilateral triangle is a. So, the magnitude of the force
between +qand +3qis:
F1,3=k|+q(+3q)|
a2
Step 4: Determine the direction of the force between the charge at the first
vertex and the charge at the third vertex. Since the charges are like in sign,
the force between them is repulsive. It acts along the line joining the charges,
pointing from the first vertex towards the third vertex.
Step 5: Find the net force on the charge at the first vertex. The net force
on the charge at the first vertex is the vector sum of the forces F1,2and F1,3.
Since these forces are along different directions, we need to consider their vector
nature.
The net force can be calculated using the parallelogram rule of vector addi-
tion:
Net force =q(F2
1,2+F2
1,3+ 2F1,2F1,3cos θ)
where θis the angle between the forces.
Solving this expression will give the magnitude and direction of the net
electric force acting on the charge at the first vertex due to the other two charges.
9
Question 12
Question
Three point charges are arranged in an equilateral triangle as shown below:
+q
↗
+q−2q
If each side of the triangle has a length a, calculate the net electric force on the
+qcharge.
Solution
Step 1: Calculate the net force on the +qcharge due to the +qcharge. The
electric force between two point charges q1and q2separated by distance ris
given by Coulomb’s Law:
F=kq1q2
r2
where kis Coulomb’s constant (8.99 ×109N·m2/C2). The magnitude of the
force on the +qcharge due to the other +qcharge will be:
F1=kq2
a2
Step 2: Calculate the direction of the force on the +qcharge due to the +q
charge. The force on the +qcharge due to the other +qcharge will be directed
along the line joining the two charges and will be repulsive.
Step 3: Calculate the net force on the +qcharge due to the −2qcharge.
The magnitude of the force on the +qcharge due to the −2qcharge will be:
F2=kq(2q)
a2√3
Step 4: Calculate the direction of the force on the +qcharge due to the −2q
charge. The force on the +qcharge due to the −2qcharge will be directed along
the line joining the two charges and will be attractive.
Step 5: Find the net force on the +qcharge. The net force Fnet on the
+qcharge will be the vector sum of the forces F1and F2, considering their
directions.
Step 6: Resolve the forces into xand ycomponents to find the net force in
terms of magnitude and direction.
Question 13
Question
Two point charges, q1= +3.0µC and q2=−4.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the charges.
10
Solution
Step 1: Determine the charges in coulombs. We convert µC to C by multiplying
by 10−6:q1= 3.0µC= 3.0×10−6Cq2=−4.0µC=−4.0×10−6C
Step 2: Calculate the distance rbetween the charges in meters. Given
r= 10.0cm = 0.10 m.
Step 3: Use Coulomb’s law to calculate the magnitude of the electric force
between the charges: F=k|q1q2|
r2
where k= 8.99 ×109Nm2/C2is the Coulomb constant.
Step 4: Substitute the given values into the formula to find the electric force:
F=(8.99 ×109)|(3.0×10−6)(−4.0×10−6)|
(0.10)2
Step 5: Calculate the magnitude of the electric force: F=(8.99 ×109)(12 ×10−12)
0.01
F=1.0788 ×10−2
0.01 F= 1.0788 N
Therefore, the magnitude of the electric force between the charges is 1.0788 N.
Question 14
Question
Two point charges, q1=−5µC and q2= 3 µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Recall the formula for the electric force between two point charges:
F=k·|q1·q2|
r2
where Fis the magnitude of the electric force, kis the Coulomb constant (8.99×
109N m2/C2), q1and q2are the magnitudes of the respective charges, and ris
the distance between the charges.
Step 2: Convert the charges to coulombs:
q1=−5µC =−5×10−6C
q2= 3 µC = 3 ×10−6C
Step 3: Plug in the given values into the formula to calculate the electric
force:
F= (8.99 ×109)·| − 5×10−6·3×10−6|
(0.10)2
Step 4: Perform the calculation:
F= (8.99 ×109)·15 ×10−12
0.01
11
F= 8.99 ×109·1.5×10−9
F= 13.485 ×100
F= 13.485 N
Step 5: Therefore, the magnitude of the electric force between the charges
q1and q2is 13.485 N.
Question 15
Question
Three point charges are placed at the corners of a right triangle as shown:
q1= +5 µC at point A,q2=−3µC at point B, and q3= +4 µC at point C.
Calculate the net electric force on q3due to q1and q2. The sides of the triangle
are of lengths 3m, 4m, and 5m.
Solution
Step 1: Calculate the distance between q1at point Aand q3at point C.
Distance between q1and q3=p32+ 42= 5 m
Step 2: Calculate the distance between q2at point Band q3at point C.
Distance between q2and q3= 4 m
Step 3: Calculate the electric force F13 on q3due to q1.
F13 =k|q1||q3|
r2=(9 ×109N m2/C2)(5 ×10−6C)(4 ×10−6C)
52= 144 mN
Step 4: Calculate the electric force F23 on q3due to q2.
F23 =k|q2||q3|
r2=(9 ×109N m2/C2)(3 ×10−6C)(4 ×10−6C)
42= 54 mN
Step 5: Calculate the net electric force on q3.
Fnet =F13 −F23 = 144 mN −54 mN = 90 mN
Therefore, the net electric force on q3due to q1and q2is 90 mN directed
towards q1.
Question 16
Question
Two point charges, q1= 3 µC and q2=−5µC, are placed 8 cm apart in air.
Calculate the magnitude and direction of the electric force on each charge.
12
Solution
Step 1: First calculate the magnitude of the electric force on charge q1.
Given: q1= 3 µC, q2=−5µC, r= 0.08 m
The formula for the magnitude of the electric force between two point charges
is given by Coulomb’s Law:
F=k·|q1q2|
r2
where kis the electrostatic constant approximately equal to 8.99×109N m2/C2.
Substitute the given values into the formula:
F= (8.99 ×109)·|3×10−6× −5×10−6|
0.082
F= 0.84375 N
Therefore, the magnitude of the electric force on charge q1is 0.84375 N.
Step 2: Next, determine the direction of the electric force on charge q1.
Since q1is positive, the electric force on q1will be repulsive. Thus, the
direction of the force on q1is away from q2.
Step 3: Now, calculate the magnitude of the electric force on charge q2.
The magnitude of the electric force on q2is the same as that on q1, but
acting in the opposite direction. Therefore, the magnitude of the electric force
on q2is also 0.84375 N.
Step 4: Determine the direction of the electric force on charge q2.
Since q2is negative, the electric force on q2will be attractive, pulling q2
towards q1. Therefore, the direction of the force on q2is towards q1.
Question 17
Question
Two point charges are placed on the x-axis. The first charge, q1=−3.0µC,
is located at x= 2.0m, and the second charge, q2= 6.0µC, is located at
x= 4.0m. Calculate the magnitude and direction of the electric force on q2due
to q1.
Solution
Step 1: Calculate the distance between the two charges. Given the positions of
the charges, the distance between them can be found using the formula:
r=|x2−x1|
r=|4.0m−2.0m|
r= 2.0m
13
Step 2: Calculate the electric force magnitude. The electric force between
two charges is given by Coulomb’s law:
F=k·|q1q2|
r2
where k= 8.99×109N m2/C2is the electrostatic constant. Substitute the given
values into the formula:
F= 8.99 ×109N m2/C2·|(−3.0µC)(6.0µC)|
(2.0m)2
F= 8.99 ×109N m2/C2·18.0×10−12 C2
4.0m2
F=8.99 N m2/C2·18.0×10−12 C2
4.0m2
F=161.82 ×10−12 N m
4.0m2
F= 40.455 ×10−12 N
F= 40.5pN
Step 3: Determine the direction of the force. The force will be attractive
since the charges have opposite signs. The force on q2is directed towards q1.
Question 18
Question
Two point charges, q1= +3.0µC and q2=−4.0µC, are placed 12.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Determine the distance between the charges in meters. Given: d=
12.0cm = 0.12 m.
Step 2: Calculate the magnitude of the electric force 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 Fis the magnitude of the electric force, kis the Coulomb constant 8.99×
109N m2/C2,|q1|and |q2|are the magnitudes of the charges, and ris the
distance between the charges.
Plugging in the values,
F=(8.99 ×109N m2/C2)·(3.0×10−6C)·(4.0×10−6C)
(0.12 m)2
14
Step 3: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(3.0·4.0) ×10−12
0.0144
F=107.88 ×10−3
0.0144
F=107.88
0.0144
F≈7491.67 N
Therefore, the magnitude of the electric force between the charges q1and q2
is approximately 7491.67 N.
Question 19
Question
Two point charges, q1= 4 µC and q2=−6µC, are separated by a distance of
5cm. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Write down the given values.
Given: q1= 4µC,q2=−6µC,r= 5 cm = 0.05 m.
Step 2: 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 kis the Coulomb constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
Step 3: Substitute the given values into Coulomb’s Law. Plugging in the
values, we get:
F=(9 ×109Nm2/C2)·(4 ×10−6C)·(6 ×10−6C)
(0.05 m)2
Step 4: Calculate the electric force.
F=216 ×10−15 Nm2/C2
0.0025 m2
F= 86.4×10−15 N
F= 8.64 ×10−14 N
Therefore, the magnitude of the electric force between the charges is 8.64 ×
10−14 N.
15
Question 20
Question
Two point charges, q1=−3nC and q2= 5 nC, are placed 10 cm apart. Calculate
the magnitude of the electric force between the two charges.
Solution
Step 1: Calculate the distance between the two charges in meters.
Distance = 10 cm = 10 ×10−2m= 0.1m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
The magnitude of the electric force (F) between two point charges is given by:
F=k·|q1·q2|
r2
where: - kis the electrostatic constant, 8.99 ×109N m2/C2, - q1and q2are the
two charges, and - ris the distance between the two charges.
Plugging in the given values:
F= (8.99 ×109N m2/C2)·| − 3×10−9C×5×10−9C|
(0.1m)2
Step 3: Simplify the expression to find the magnitude of the electric force.
F= (8.99 ×109)·15 ×10−18
0.01
F= 8.99 ×109·1.5×10−15
F= 13.485 ×10−6
F= 1.3485 ×10−5N
Therefore, the magnitude of the electric force between the two charges is
1.3485 ×10−5N.
Question 21
Question
Three point charges are arranged on the corners of an equilateral triangle with
side length a. The charges are +q,−q, and +2q. Calculate the magnitude of
the net electric force acting on the charge located at the corner with the +2q
charge.
16
Solution
Step 1: Calculate the magnitude of the individual electric forces between each
pair of charges:
The force between two charges q1and q2separated by a distance ris given
by Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the Coulomb’s constant (8.99 ×109N·m2/C2).
The distances between the corners of the equilateral triangle are all a.
Therefore, the magnitude of the forces for the given charges are: - +qand
−q:F+q,−q=k·|q|·|q|
a2=kq2
a2-−qand +2q:F−q,+2q=k·|q|·|2q|
a2=2kq2
a2-+2qand
+q:F+2q,+q=k·|2q|·|q|
a2=2kq2
a2
Step 2: Find the horizontal and vertical components of each force.
The forces can be divided into horizontal and vertical components based on
the symmetry of the equilateral triangle: - +qand −q: Horizontal and vertical
components are equal in magnitude but opposite in direction. - −qand +2q:
Horizontal component is twice the vertical component. - +2qand +q: Both
components are the same for this pair.
Step 3: Determine the net force acting on the +2qcharge.
Since the +2qcharge is at an angle of 60◦with the horizontal axis, the net
force can be found by resolving the components of the forces along the horizontal
and vertical directions and then adding them up vectorially.
The net force will depend on the angles and magnitudes of the different
components calculated in Steps 1 and 2.
Question 22
Question
Three point charges are arranged in a line as shown below:
+Q−2Q+Q
If the separation between the charges +Qand −2Qis a, and the separation
between the charges −2Qand +Qis 2a, calculate the magnitude and direction
of the net force on the charge −2Q.
Solution
Step 1: Calculate the electric force between +Qand −2Q. The electric force
between two point charges q1and q2separated by a distance ris given by
Coulomb’s law:
F=k· |q1|·|q2|
r2
17
where kis the electrostatic constant, |q1|and |q2|are the magnitudes of the
charges, and ris the distance between the charges.
For the charges +Qand −2Qwith separation abetween them, the force on
−2Qis attractive, so the force vector points towards the charge +Q.
Step 2: Calculate the electric force between −2Qand +Q. Using the same
formula as in Step 1, we find that the force between −2Qand +Qis attractive,
and the force vector points towards +Q.
Step 3: Calculate the net force on −2Q. Since the forces from +Qand +Q
on −2Qhave the same direction, we can simply add their magnitudes to find
the net force. The force between +Qand −2Qis F1and the force between −2Q
and +Qis F2.
Net force =F1+F2
Step 4: Substitute the electric force formulas into the net force equation.
F1=k·Q·2Q
a2=2kQ2
a2
F2=k·2Q·Q
(2a)2=kQ2
2a2
Substitute these into the net force equation:
Net force =2kQ2
a2+kQ2
2a2=5kQ2
2a2
Step 5: Determine the direction of the net force. Since both individual forces
were directed towards +Q, the net force on −2Qis also directed towards +Q.
Therefore, the magnitude of the net force on the charge −2Qis 5kQ2
2a2, and
the direction is towards +Q.
Question 23
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. The magnitude of each charge is q. Calculate the magnitude and
direction of the net electric force on the charge at the top vertex due to the
other two charges. Given that the side length of the equilateral triangle is a.
+q
/
−q−q
18
Solution
Step 1: Calculate the force due to the charge qat the right vertex.
The magnitude of the force F1is given by Coulomb’s Law:
F1=k|q|2
a2
where kis the Coulomb constant (8.99 ×109N m2/C2), qis the magnitude
of the charge, and ais the distance between the charges.
Step 2: Calculate the force due to the charge qat the left vertex.
Since the charges are equidistant from the charge at the top vertex and the
charges are equidistant from each other, the magnitude of the force F2due to
the charge qat the left vertex is also F1.
Step 3: Find the net force on the charge at the top vertex.
The net force is the vector sum of the forces F1and F2. Since they are both
acting in the vertical direction but in opposite directions, the net force will be
the difference between the two.
Thus, the net force on the charge at the top vertex is:
Net force =F1−F2
Now, substitute the expressions for F1and F2into the above equation and
simplify to find the net force.
Question 24
Question
Two point charges, q1=−2.5µC and q2= 3.0µC, are placed 8.0 cm apart in
air. Calculate the magnitude of the electric force between them.
Solution
Step 1: Convert the charges to standard units. Step 2: Calculate the distance
between the charges in meters. Step 3: Use Coulomb’s law to find the magnitude
of the electric force between the charges.
Step 1: Convert the charges to standard units. Given: q1=−2.5µC
and q2= 3.0µC. Converting to standard units: q1=−2.5×10−6C and
q2= 3.0×10−6C.
Step 2: Calculate the distance between the charges in meters. Given: dis-
tance = 8.0 cm = 0.08 m.
Step 3: Use Coulomb’s law to find the magnitude of the electric force
between the charges. Coulomb’s law:
F=k·|q1·q2|
r2
19
where: F= electric force, k= Coulomb’s constant (8.99 ×109N·m2/C2),q1
and q2= magnitudes of the charges, r= distance between the charges.
Plugging in the values:
F= (8.99 ×109)·|(−2.5×10−6)·(3.0×10−6)|
(0.08)2
F= (8.99 ×109)·7.5×10−12
0.0064
F= (8.99 ×109)·1.171875 ×10−9
F= 10.54 N
Therefore, the magnitude of the electric force between the charges is 10.54
N.
Question 25
Question
Two point charges, q1= +6.0µC and q2=−3.0µC, are placed 20 cm apart in
a vacuum. Calculate the magnitude of the electric force exerted by q1on q2.
Solution
Step 1: Convert the given charges from microcoulombs to coulombs. Step 2:
Determine the distance between the charges in meters. Step 3: Calculate the
magnitude of the electric force between the charges using Coulomb’s law.
Step 1: Convert the given charges from microcoulombs to coulombs.
q1= 6.0µC = 6.0×10−6C
q2=−3.0µC =−3.0×10−6C
Step 2: Determine the distance between the charges in meters.
r= 20 cm = 20 ×10−2m= 0.20 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law.
Coulomb’s law states that the magnitude of the electric force Fbetween two
point charges is given by
F=k|q1·q2|
r2
where - kis the Coulomb constant, 8.9875 ×109Nm2/C2, - q1, q2are the mag-
nitudes of the two point charges, and - ris the distance between the charges.
Plugging in the values:
F= 8.9875 ×109·|6.0×10−6· −3.0×10−6|
0.202
20
F= 4.49275 ×10−3N
Thus, the magnitude of the electric force exerted by q1on q2is 4.49275 ×
10−3N.
21
Question 2
Question
Two point charges, q1=−4µC and q2= 6 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
Solution
To find the electric force between two point charges, we can use Coulomb’s Law,
which states that the magnitude of the force between two point charges is given
by:
F=k·|q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Step 1: Convert the charges to coulombs and the distance to meters. Given:
q1=−4µC=−4×10−6C, q2= 6 µC= 6 ×10−6C, r= 10 cm = 0.1m.
Step 2: Calculate the electric force. Plugging the given values into Coulomb’s
Law, we have:
F= (8.99 ×109)·| − 4×10−6×6×10−6|
(0.1)2
F= 8.99 ×109·24 ×10−12
0.01
F= 8.99 ×109·2.4×10−9
F= 21.576 ×10−9
F= 2.1576 ×10−8N
Therefore, the magnitude of the electric force between the charges is 2.1576×
10−8N.
Question 3
Question
Two point charges +3 µC and −5µC are placed 12 cm apart in air. Calculate
the magnitude of the electric force acting on each charge. Given: k= 8.99 ×
109N m2/C2.
2
Solution
Step 1: Convert the charges to coulombs.
+3 µC= 3 ×10−6C
−5µC=−5×10−6C
Step 2: Calculate the magnitude of the electric force exerted by the +3 µC
charge on the −5µC charge using Coulomb’s Law:
F=k· |q1|·|q2|
r2
where F= magnitude of the electric force, k= Coulomb’s constant (8.99 ×
109N m2/C2), |q1|= 3 ×10−6C, |q2|= 5 ×10−6C, and r= 0.12 m.
Substitute the values into the formula:
F=(8.99 ×109N m2/C2)·(3 ×10−6C)·(5 ×10−6C)
(0.12 m)2
Step 3: Calculate the magnitude of the electric force between the charges.
Question 4
Question
Two point charges, q1=−3µC and q2= 5µC, are located 6 cm apart. Calculate
the magnitude of the electric force between these charges.
Solution
Step 1: Calculate the distance between the charges in meters. Given: Distance,
r= 6 cm = 0.06 m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
Coulomb’s law states that the magnitude of the electric force between two point
charges is given by:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging in the given values:
F= 8.99 ×109| − 3×10−6×5×10−6|
0.062
Step 3: Calculate the magnitude of the electric force.
F= 8.99 ×109×15 ×10−12
0.0036
3
F= 8.99 ×109×4.1667 ×10−9
F= 37.45 N
Therefore, the magnitude of the electric force between the charges q1and q2
is 37.45 N.
Question 5
Question
Two point charges, q1and q2, are held 4 cm apart in air. If q1= 5µC and
q2=−3µC, calculate the magnitude of the electric force between them.
Solution
Step 1: Calculate the electric force using Coulomb’s Law formula:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N m2/C2) and ris the distance
between the charges.
Step 2: Substitute the given values into the formula:
F=(8.99 ×109N m2/C2)· |5×10−6C·(−3×10−6C)|
(0.04 m)2
Step 3: Calculate the magnitude of the force:
F=(8.99 ×109)·(5 ×3) ×10−12
0.0016
Step 4: Simplify the expression:
F=44.95 ×10−3
0.0016
Step 5: Calculate the final answer:
F= 28.09 N
Thus, the magnitude of the electric force between the charges q1and q2is
28.09 N.
Question 6
Question
Two point charges, +4.0µC and −2.0µC, are placed 10.0 cm apart. Calculate
the magnitude of the electric force between these two charges.
4
Solution
Step 1: Convert the charges to Coulombs. Given: q1= +4.0µC = 4.0×10−6C
q2=−2.0µC =−2.0×10−6C
Step 2: Calculate the distance between the charges in meters. d= 10.0cm =
10.0×10−2m= 0.10 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law:
F=k
q1·q2
d2
where k= 8.99 ×109Nm2/C2is Coulomb’s constant.
Step 4: Substituting the given values into the formula:
F= 8.99 ×109×
4.0×10−6× −2.0×10−6
(0.10)2
Step 5: Calculate the magnitude of the electric force:
F= 8.99 ×109×8.0×10−12
0.01
F= 8.99 ×109×8.0×10−10
F= 7.19 ×10−1N
Therefore, the magnitude of the electric force between the two charges is
7.19 ×10−1N.
Question 7
Question
Two point charges, q1=−2µC and q2= 3 µC, are placed 10 cm apart in a
vacuum. Calculate the magnitude of the electric force between them.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−2µC and q2= 3 µC
Converting to coulombs: q1=−2×10−6C and q2= 3 ×10−6C.
Step 2: Calculate the magnitude of electric force using Coulomb’s Law. The
magnitude of the electric force between point charges is given by Coulomb’s
Law:
F=k
q1q2
r2
,
where kis the electrostatic constant (8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
5
Plugging in the given values:
F= 8.99 ×109
(−2×10−6)(3 ×10−6)
(0.10)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= 5.394 N
Therefore, the magnitude of the electric force between the point charges q1
and q2is 5.394 N.
Question 8
Question
Two point charges, q1=−3.0µC and q2= 4.0µC, are placed 8.0 cm apart.
Calculate the magnitude and direction of the electric force between the two
charges.
Solution
Step 1: Convert the charges to coulombs and write down the given values.
Given: q1=−3.0µC = −3.0×10−6C (negative sign indicates charge is nega-
tive)
q2= 4.0µC = 4.0×10−6C
Distance r= 8.0 cm = 0.08 m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law
formula:
F=k·|q1·q2|
r2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the given values:
F= 8.99 ×109·| − 3.0×10−6·4.0×10−6|
(0.08)2
Step 3: Calculate the magnitude of the electric force:
6
F= 8.99 ×109·12 ×10−12
0.0064
F= 8.99 ×109·1.875 ×10−6
F= 16.87 N
Step 4: Determine the direction of the force. Since the charges have opposite
signs, the force will be attractive. Thus, the force is directed from q1to q2.
Therefore, the magnitude of the electric force between the charges is 16.87
N, and it is directed from q1to q2.
Question 9
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 10.0 cm apart.
Calculate the magnitude and direction of the electric force that q1exerts on q2.
Solution
To find the electric force that q1exerts on q2, we can use Coulomb’s Law:
F=k·|q1|·|q2|
r2
where k= 8.99 ×109N·m2/C2is Coulomb’s constant, |q1|and |q2|are the
magnitudes of the charges, and ris the distance between the charges.
Step 1: Calculate the distance rin meters: Given: r= 10.0cm = 0.10 m.
Step 2: Substitute the values into Coulomb’s Law to find the electric force:
F= (8.99 ×109N·m2/C2)·(4.0×10−6C)·(6.0×10−6C)
(0.10 m)2
Step 3: Calculate the magnitude of the electric force:
F= (8.99 ×109)·24 ×10−12
0.01
F= 8.99 ×109×2.4×10−9
F= 21.576 ×100
F= 21.576 N
Step 4: Determine the direction of the force: Since q1is negative and q2is
positive, the electric force will be attractive, that is, the force will be directed
toward q1.
Therefore, the magnitude of the electric force that q1exerts on q2is 21.576 N
directed towards q1.
7
Question 10
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are fixed in place and
separated by a distance of 5.0cm. Calculate the magnitude of the electric force
that q1exerts on q2.
Solution
Step 1: Convert the charges to Coulombs. Given: q1=−4.0µC and q2=
6.0µC.
To convert from microCoulombs (µC) to Coulombs, we multiply by 10−6.q1=
−4.0×10−6C and q2= 6.0×10−6C.
Step 2: Calculate the magnitude of the electric force using the Coulomb’s law
formula. Coulomb’s law states that the magnitude of the electric force between
two charges is given by
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.
Plugging in the values gives us:
F= 8.99 ×109N m2/C2×| − 4.0×10−6C×6.0×10−6C|
(0.05 m)2.
Step 3: Substitute the values and calculate the electric force.
F= 8.99×109×4.0×6.0×10−6×10−6
0.052= 8.99×109×24 ×10−12
0.0025 = 8.99×109×9.6×10−10 = 8.62 N.
Therefore, the magnitude of the electric force that q1exerts on q2is 8.62 N.
Question 11
Question
In an equilateral triangle with side length a, three point charges are placed at
its vertices. The charges are +q,−2q, and +3q, respectively. Calculate the
magnitude and direction of the net electric force acting on the charge at the
first vertex (+q) due to the other two charges.
Solution
Given: Side length of equilateral triangle, a. Point charges at the vertices: +q,
−2q, and +3q.
8
Step 1: Calculate the electric force between the charge at the first vertex
(+q) and the charge at the second vertex (−2q). The electric force between two
charges q1and q2separated by a distance ris given by Coulomb’s law:
F=k·|q1q2|
r2
where kis Coulomb’s constant (8.9875 ×109N m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
The distance between the first and second vertices of the equilateral triangle
is a. So, the magnitude of the force between +qand −2qis:
F1,2=k|+q(−2q)|
a2
Step 2: Determine the direction of the force between the charge at the first
vertex and the charge at the second vertex. Since the charges are opposite in
sign, the force between them is attractive. It acts along the line joining the
charges, pointing from the second vertex towards the first vertex.
Step 3: Calculate the electric force between the charge at the first vertex
(+q) and the charge at the third vertex (+3q). The distance between the first
and third vertices of the equilateral triangle is a. So, the magnitude of the force
between +qand +3qis:
F1,3=k|+q(+3q)|
a2
Step 4: Determine the direction of the force between the charge at the first
vertex and the charge at the third vertex. Since the charges are like in sign,
the force between them is repulsive. It acts along the line joining the charges,
pointing from the first vertex towards the third vertex.
Step 5: Find the net force on the charge at the first vertex. The net force
on the charge at the first vertex is the vector sum of the forces F1,2and F1,3.
Since these forces are along different directions, we need to consider their vector
nature.
The net force can be calculated using the parallelogram rule of vector addi-
tion:
Net force =q(F2
1,2+F2
1,3+ 2F1,2F1,3cos θ)
where θis the angle between the forces.
Solving this expression will give the magnitude and direction of the net
electric force acting on the charge at the first vertex due to the other two charges.
9
Question 12
Question
Three point charges are arranged in an equilateral triangle as shown below:
+q
↗
+q−2q
If each side of the triangle has a length a, calculate the net electric force on the
+qcharge.
Solution
Step 1: Calculate the net force on the +qcharge due to the +qcharge. The
electric force between two point charges q1and q2separated by distance ris
given by Coulomb’s Law:
F=kq1q2
r2
where kis Coulomb’s constant (8.99 ×109N·m2/C2). The magnitude of the
force on the +qcharge due to the other +qcharge will be:
F1=kq2
a2
Step 2: Calculate the direction of the force on the +qcharge due to the +q
charge. The force on the +qcharge due to the other +qcharge will be directed
along the line joining the two charges and will be repulsive.
Step 3: Calculate the net force on the +qcharge due to the −2qcharge.
The magnitude of the force on the +qcharge due to the −2qcharge will be:
F2=kq(2q)
a2√3
Step 4: Calculate the direction of the force on the +qcharge due to the −2q
charge. The force on the +qcharge due to the −2qcharge will be directed along
the line joining the two charges and will be attractive.
Step 5: Find the net force on the +qcharge. The net force Fnet on the
+qcharge will be the vector sum of the forces F1and F2, considering their
directions.
Step 6: Resolve the forces into xand ycomponents to find the net force in
terms of magnitude and direction.
Question 13
Question
Two point charges, q1= +3.0µC and q2=−4.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the charges.
10
Solution
Step 1: Determine the charges in coulombs. We convert µC to C by multiplying
by 10−6:q1= 3.0µC= 3.0×10−6Cq2=−4.0µC=−4.0×10−6C
Step 2: Calculate the distance rbetween the charges in meters. Given
r= 10.0cm = 0.10 m.
Step 3: Use Coulomb’s law to calculate the magnitude of the electric force
between the charges: F=k|q1q2|
r2
where k= 8.99 ×109Nm2/C2is the Coulomb constant.
Step 4: Substitute the given values into the formula to find the electric force:
F=(8.99 ×109)|(3.0×10−6)(−4.0×10−6)|
(0.10)2
Step 5: Calculate the magnitude of the electric force: F=(8.99 ×109)(12 ×10−12)
0.01
F=1.0788 ×10−2
0.01 F= 1.0788 N
Therefore, the magnitude of the electric force between the charges is 1.0788 N.
Question 14
Question
Two point charges, q1=−5µC and q2= 3 µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Recall the formula for the electric force between two point charges:
F=k·|q1·q2|
r2
where Fis the magnitude of the electric force, kis the Coulomb constant (8.99×
109N m2/C2), q1and q2are the magnitudes of the respective charges, and ris
the distance between the charges.
Step 2: Convert the charges to coulombs:
q1=−5µC =−5×10−6C
q2= 3 µC = 3 ×10−6C
Step 3: Plug in the given values into the formula to calculate the electric
force:
F= (8.99 ×109)·| − 5×10−6·3×10−6|
(0.10)2
Step 4: Perform the calculation:
F= (8.99 ×109)·15 ×10−12
0.01
11
F= 8.99 ×109·1.5×10−9
F= 13.485 ×100
F= 13.485 N
Step 5: Therefore, the magnitude of the electric force between the charges
q1and q2is 13.485 N.
Question 15
Question
Three point charges are placed at the corners of a right triangle as shown:
q1= +5 µC at point A,q2=−3µC at point B, and q3= +4 µC at point C.
Calculate the net electric force on q3due to q1and q2. The sides of the triangle
are of lengths 3m, 4m, and 5m.
Solution
Step 1: Calculate the distance between q1at point Aand q3at point C.
Distance between q1and q3=p32+ 42= 5 m
Step 2: Calculate the distance between q2at point Band q3at point C.
Distance between q2and q3= 4 m
Step 3: Calculate the electric force F13 on q3due to q1.
F13 =k|q1||q3|
r2=(9 ×109N m2/C2)(5 ×10−6C)(4 ×10−6C)
52= 144 mN
Step 4: Calculate the electric force F23 on q3due to q2.
F23 =k|q2||q3|
r2=(9 ×109N m2/C2)(3 ×10−6C)(4 ×10−6C)
42= 54 mN
Step 5: Calculate the net electric force on q3.
Fnet =F13 −F23 = 144 mN −54 mN = 90 mN
Therefore, the net electric force on q3due to q1and q2is 90 mN directed
towards q1.
Question 16
Question
Two point charges, q1= 3 µC and q2=−5µC, are placed 8 cm apart in air.
Calculate the magnitude and direction of the electric force on each charge.
12
Solution
Step 1: First calculate the magnitude of the electric force on charge q1.
Given: q1= 3 µC, q2=−5µC, r= 0.08 m
The formula for the magnitude of the electric force between two point charges
is given by Coulomb’s Law:
F=k·|q1q2|
r2
where kis the electrostatic constant approximately equal to 8.99×109N m2/C2.
Substitute the given values into the formula:
F= (8.99 ×109)·|3×10−6× −5×10−6|
0.082
F= 0.84375 N
Therefore, the magnitude of the electric force on charge q1is 0.84375 N.
Step 2: Next, determine the direction of the electric force on charge q1.
Since q1is positive, the electric force on q1will be repulsive. Thus, the
direction of the force on q1is away from q2.
Step 3: Now, calculate the magnitude of the electric force on charge q2.
The magnitude of the electric force on q2is the same as that on q1, but
acting in the opposite direction. Therefore, the magnitude of the electric force
on q2is also 0.84375 N.
Step 4: Determine the direction of the electric force on charge q2.
Since q2is negative, the electric force on q2will be attractive, pulling q2
towards q1. Therefore, the direction of the force on q2is towards q1.
Question 17
Question
Two point charges are placed on the x-axis. The first charge, q1=−3.0µC,
is located at x= 2.0m, and the second charge, q2= 6.0µC, is located at
x= 4.0m. Calculate the magnitude and direction of the electric force on q2due
to q1.
Solution
Step 1: Calculate the distance between the two charges. Given the positions of
the charges, the distance between them can be found using the formula:
r=|x2−x1|
r=|4.0m−2.0m|
r= 2.0m
13
Step 2: Calculate the electric force magnitude. The electric force between
two charges is given by Coulomb’s law:
F=k·|q1q2|
r2
where k= 8.99×109N m2/C2is the electrostatic constant. Substitute the given
values into the formula:
F= 8.99 ×109N m2/C2·|(−3.0µC)(6.0µC)|
(2.0m)2
F= 8.99 ×109N m2/C2·18.0×10−12 C2
4.0m2
F=8.99 N m2/C2·18.0×10−12 C2
4.0m2
F=161.82 ×10−12 N m
4.0m2
F= 40.455 ×10−12 N
F= 40.5pN
Step 3: Determine the direction of the force. The force will be attractive
since the charges have opposite signs. The force on q2is directed towards q1.
Question 18
Question
Two point charges, q1= +3.0µC and q2=−4.0µC, are placed 12.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Determine the distance between the charges in meters. Given: d=
12.0cm = 0.12 m.
Step 2: Calculate the magnitude of the electric force 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 Fis the magnitude of the electric force, kis the Coulomb constant 8.99×
109N m2/C2,|q1|and |q2|are the magnitudes of the charges, and ris the
distance between the charges.
Plugging in the values,
F=(8.99 ×109N m2/C2)·(3.0×10−6C)·(4.0×10−6C)
(0.12 m)2
14
Step 3: Calculate the magnitude of the electric force.
F=(8.99 ×109)·(3.0·4.0) ×10−12
0.0144
F=107.88 ×10−3
0.0144
F=107.88
0.0144
F≈7491.67 N
Therefore, the magnitude of the electric force between the charges q1and q2
is approximately 7491.67 N.
Question 19
Question
Two point charges, q1= 4 µC and q2=−6µC, are separated by a distance of
5cm. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Write down the given values.
Given: q1= 4µC,q2=−6µC,r= 5 cm = 0.05 m.
Step 2: 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 kis the Coulomb constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
Step 3: Substitute the given values into Coulomb’s Law. Plugging in the
values, we get:
F=(9 ×109Nm2/C2)·(4 ×10−6C)·(6 ×10−6C)
(0.05 m)2
Step 4: Calculate the electric force.
F=216 ×10−15 Nm2/C2
0.0025 m2
F= 86.4×10−15 N
F= 8.64 ×10−14 N
Therefore, the magnitude of the electric force between the charges is 8.64 ×
10−14 N.
15
Question 20
Question
Two point charges, q1=−3nC and q2= 5 nC, are placed 10 cm apart. Calculate
the magnitude of the electric force between the two charges.
Solution
Step 1: Calculate the distance between the two charges in meters.
Distance = 10 cm = 10 ×10−2m= 0.1m
Step 2: Calculate the magnitude of the electric force using Coulomb’s law.
The magnitude of the electric force (F) between two point charges is given by:
F=k·|q1·q2|
r2
where: - kis the electrostatic constant, 8.99 ×109N m2/C2, - q1and q2are the
two charges, and - ris the distance between the two charges.
Plugging in the given values:
F= (8.99 ×109N m2/C2)·| − 3×10−9C×5×10−9C|
(0.1m)2
Step 3: Simplify the expression to find the magnitude of the electric force.
F= (8.99 ×109)·15 ×10−18
0.01
F= 8.99 ×109·1.5×10−15
F= 13.485 ×10−6
F= 1.3485 ×10−5N
Therefore, the magnitude of the electric force between the two charges is
1.3485 ×10−5N.
Question 21
Question
Three point charges are arranged on the corners of an equilateral triangle with
side length a. The charges are +q,−q, and +2q. Calculate the magnitude of
the net electric force acting on the charge located at the corner with the +2q
charge.
16
Solution
Step 1: Calculate the magnitude of the individual electric forces between each
pair of charges:
The force between two charges q1and q2separated by a distance ris given
by Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the Coulomb’s constant (8.99 ×109N·m2/C2).
The distances between the corners of the equilateral triangle are all a.
Therefore, the magnitude of the forces for the given charges are: - +qand
−q:F+q,−q=k·|q|·|q|
a2=kq2
a2-−qand +2q:F−q,+2q=k·|q|·|2q|
a2=2kq2
a2-+2qand
+q:F+2q,+q=k·|2q|·|q|
a2=2kq2
a2
Step 2: Find the horizontal and vertical components of each force.
The forces can be divided into horizontal and vertical components based on
the symmetry of the equilateral triangle: - +qand −q: Horizontal and vertical
components are equal in magnitude but opposite in direction. - −qand +2q:
Horizontal component is twice the vertical component. - +2qand +q: Both
components are the same for this pair.
Step 3: Determine the net force acting on the +2qcharge.
Since the +2qcharge is at an angle of 60◦with the horizontal axis, the net
force can be found by resolving the components of the forces along the horizontal
and vertical directions and then adding them up vectorially.
The net force will depend on the angles and magnitudes of the different
components calculated in Steps 1 and 2.
Question 22
Question
Three point charges are arranged in a line as shown below:
+Q−2Q+Q
If the separation between the charges +Qand −2Qis a, and the separation
between the charges −2Qand +Qis 2a, calculate the magnitude and direction
of the net force on the charge −2Q.
Solution
Step 1: Calculate the electric force between +Qand −2Q. The electric force
between two point charges q1and q2separated by a distance ris given by
Coulomb’s law:
F=k· |q1|·|q2|
r2
17
where kis the electrostatic constant, |q1|and |q2|are the magnitudes of the
charges, and ris the distance between the charges.
For the charges +Qand −2Qwith separation abetween them, the force on
−2Qis attractive, so the force vector points towards the charge +Q.
Step 2: Calculate the electric force between −2Qand +Q. Using the same
formula as in Step 1, we find that the force between −2Qand +Qis attractive,
and the force vector points towards +Q.
Step 3: Calculate the net force on −2Q. Since the forces from +Qand +Q
on −2Qhave the same direction, we can simply add their magnitudes to find
the net force. The force between +Qand −2Qis F1and the force between −2Q
and +Qis F2.
Net force =F1+F2
Step 4: Substitute the electric force formulas into the net force equation.
F1=k·Q·2Q
a2=2kQ2
a2
F2=k·2Q·Q
(2a)2=kQ2
2a2
Substitute these into the net force equation:
Net force =2kQ2
a2+kQ2
2a2=5kQ2
2a2
Step 5: Determine the direction of the net force. Since both individual forces
were directed towards +Q, the net force on −2Qis also directed towards +Q.
Therefore, the magnitude of the net force on the charge −2Qis 5kQ2
2a2, and
the direction is towards +Q.
Question 23
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. The magnitude of each charge is q. Calculate the magnitude and
direction of the net electric force on the charge at the top vertex due to the
other two charges. Given that the side length of the equilateral triangle is a.
+q
/
−q−q
18
Solution
Step 1: Calculate the force due to the charge qat the right vertex.
The magnitude of the force F1is given by Coulomb’s Law:
F1=k|q|2
a2
where kis the Coulomb constant (8.99 ×109N m2/C2), qis the magnitude
of the charge, and ais the distance between the charges.
Step 2: Calculate the force due to the charge qat the left vertex.
Since the charges are equidistant from the charge at the top vertex and the
charges are equidistant from each other, the magnitude of the force F2due to
the charge qat the left vertex is also F1.
Step 3: Find the net force on the charge at the top vertex.
The net force is the vector sum of the forces F1and F2. Since they are both
acting in the vertical direction but in opposite directions, the net force will be
the difference between the two.
Thus, the net force on the charge at the top vertex is:
Net force =F1−F2
Now, substitute the expressions for F1and F2into the above equation and
simplify to find the net force.
Question 24
Question
Two point charges, q1=−2.5µC and q2= 3.0µC, are placed 8.0 cm apart in
air. Calculate the magnitude of the electric force between them.
Solution
Step 1: Convert the charges to standard units. Step 2: Calculate the distance
between the charges in meters. Step 3: Use Coulomb’s law to find the magnitude
of the electric force between the charges.
Step 1: Convert the charges to standard units. Given: q1=−2.5µC
and q2= 3.0µC. Converting to standard units: q1=−2.5×10−6C and
q2= 3.0×10−6C.
Step 2: Calculate the distance between the charges in meters. Given: dis-
tance = 8.0 cm = 0.08 m.
Step 3: Use Coulomb’s law to find the magnitude of the electric force
between the charges. Coulomb’s law:
F=k·|q1·q2|
r2
19
where: F= electric force, k= Coulomb’s constant (8.99 ×109N·m2/C2),q1
and q2= magnitudes of the charges, r= distance between the charges.
Plugging in the values:
F= (8.99 ×109)·|(−2.5×10−6)·(3.0×10−6)|
(0.08)2
F= (8.99 ×109)·7.5×10−12
0.0064
F= (8.99 ×109)·1.171875 ×10−9
F= 10.54 N
Therefore, the magnitude of the electric force between the charges is 10.54
N.
Question 25
Question
Two point charges, q1= +6.0µC and q2=−3.0µC, are placed 20 cm apart in
a vacuum. Calculate the magnitude of the electric force exerted by q1on q2.
Solution
Step 1: Convert the given charges from microcoulombs to coulombs. Step 2:
Determine the distance between the charges in meters. Step 3: Calculate the
magnitude of the electric force between the charges using Coulomb’s law.
Step 1: Convert the given charges from microcoulombs to coulombs.
q1= 6.0µC = 6.0×10−6C
q2=−3.0µC =−3.0×10−6C
Step 2: Determine the distance between the charges in meters.
r= 20 cm = 20 ×10−2m= 0.20 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law.
Coulomb’s law states that the magnitude of the electric force Fbetween two
point charges is given by
F=k|q1·q2|
r2
where - kis the Coulomb constant, 8.9875 ×109Nm2/C2, - q1, q2are the mag-
nitudes of the two point charges, and - ris the distance between the charges.
Plugging in the values:
F= 8.9875 ×109·|6.0×10−6· −3.0×10−6|
0.202
20
F= 4.49275 ×10−3N
Thus, the magnitude of the electric force exerted by q1on q2is 4.49275 ×
10−3N.
21
