PHYS 232 - UNIVERSITY PHYSICS
II - Calculation of electric forces
between point charges
Question Bank - Set 6
Liberty University
Question 1
Question
Two point charges are placed on the x-axis. Charge q1= 3 µC is located at
x= 1 m, and charge q2=−5µC is located at x= 4 m. Calculate the electric
force that charge q1exerts on charge q2.
Solution
Step 1: Find the distance between the two charges. The distance rbetween the
two charges is given by the difference in their positions along the x-axis:
r=|x2−x1|=|4m−1m|= 3 m
Step 2: Calculate the magnitude of the electric force using Coulomb’s Law.
The magnitude of the electric force Fbetween the two charges is given by
Coulomb’s Law:
F=k|q1q2|
r2
where kis the electrostatic constant 8.99 ×109N m2/C2.
Substitute the given values:
F= 8.99 ×109(3 ×10−6C)(5 ×10−6C)
(3 m)2
F= 8.99 ×10915 ×10−12 C2
9m2
F= 8.99 ×109×1.67 ×10−12 N
F= 14.98 ×10−3N
F= 0.01498 N
Therefore, the electric force that charge q1exerts on charge q2is 0.01498 N.
Question 2
Question
Three point charges are arranged at the vertices of an equilateral triangle as
shown below:
Q
−Q Q
If each charge magnitude is Q= 2 mC, what is the magnitude of the electric
force experienced by the charge at the top vertex due to the other two charges?
Solution
Step 1: Begin by labeling the charges and distances in the triangle. Let the
distance between adjacent charges be a.
Q
a
−Q Q
a
Step 2: Calculate the electric force experienced by the charge at the top
vertex due to the charge on the bottom left.
F1=kQ2
a2
Step 3: Calculate the electric force experienced by the charge at the top
vertex due to the charge on the bottom right.
F2=kQ2
(2a)2
Step 4: Determine the net force experienced by the charge at the top vertex.
Since the two forces are acting in opposite directions along the same line, the
net force is the difference between them.
2
Fnet =F1−F2
=kQ2
a2−kQ2
(2a)2
=kQ21
a2−1
4a2
=kQ23
4a2
Step 5: Substitute the known values k= 8.99×109N·m2/C2and Q= 2 mC,
and afor an equilateral triangle (a=a√3/3).
Fnet = (8.99 ×109)·(2 ×10−3)2·
3
4a√3
32
= (8.99 ×109)·4×10−6·
3
a√3
32
= 3√3×104N
Therefore, the magnitude of the electric force experienced by the charge at
the top vertex due to the other two charges is 3√3×104N.
Question 3
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are placed 20.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges from microcoulombs to coulombs.
q1=−4.0µC=−4.0×10−6C
q2= 8.0µC= 8.0×10−6C
Step 2: Determine the distance between the charges in meters.
r= 20.0cm = 20.0×10−2m= 0.20 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k·|q1·q2|
r2
3
where k= 8.99 ×109N·m2/C2is the Coulomb’s constant.
Step 4: Substitute the values into the equation to find the electric force.
F= (8.99 ×109)·|(−4.0×10−6)·(8.0×10−6)|
(0.20)2
= 8.99 ×109·32 ×10−12
0.04
= 8.99 ×109·8×10−11
= 71.92 N
Therefore, the magnitude of the electric force between the charges is 71.92
N.
Question 5
Question
Three point charges are arranged as shown in the diagram below: a +2.0 µC
charge at the origin, a -3.0 µC charge at (4.0 m, 0), and a +4.0 µC charge at
(0, 3.0 m). Calculate the magnitude of the net electric force acting on the +4.0
µC charge due to the other two charges.
Solution
1. Let’s denote the charges at the origin, (4.0 m, 0), and (0, 3.0 m) respectively
as q1= +2.0µC, q2=−3.0µC, and q3= +4.0µC.
2. We will calculate the net electric force on the +4.0µC charge, q3, due to
charges q1and q2using the principle of superposition. The electric force exerted
by each point charge can be calculated using the formula:
Felec =k|q1q2|
r2
3. First, we will calculate the force on q3due to q1:
F1=k|q1q3|
r2
13
4. Given that k= 8.9875×109Nm2
C2and r13 = 3.0m, we can substitute these
values into the formula.
5. Solving for F1:
F1= 8.9875 ×109Nm2
C2·(2.0µC)(4.0µC)
(3.0m)2
6. Next, we will calculate the force on q3due to q2:
F2=k|q2q3|
r2
23
4
7. Given that r23 = 4.0m, we can substitute the known values into the
formula.
8. Solving for F2:
F2= 8.9875 ×109Nm2
C2·(3.0µC)(4.0µC)
(4.0m)2
9. The net electric force on q3would be the vector sum of F1and F2.
10. Calculate the magnitudes and directions of F1and F2, then sum them
to find the net force acting on q3.
Question 7
Question
Three point charges are arranged along the x-axis. Charge q1=−2.0µC is
located at the origin, charge q2= 3.0µC is located at x= 4.0m, and charge
q3= 1.0µC is located at x= 8.0m. Calculate the net force on charge q3due
to charges q1and q2.
Solution
Step 1: Calculate the force F31 on charge q3due to charge q1.
Since the charges are on the x-axis, we can use Coulomb’s Law to calculate
the force:
F31 =k|q1||q3|
r2
31
where k= 8.99 ×109N·m2/C2is the Coulomb’s constant, q1=−2.0µC =
−2.0×10−6Cis the charge at the origin, and q3= 1.0µC = 1.0×10−6Cis the
charge at x= 8.0m.
The distance r31 between charges q3(at x= 8.0m) and q1(at the origin) is:
r31 = 8.0m
Therefore, the force F31 is:
F31 = (8.99 ×109)| − 2.0×10−6×1.0×10−6|
(8.0)2
Calculating the force, we get:
F31 = 4.49 ×10−4N
Step 2: Calculate the force F32 on charge q3due to charge q2.
Using Coulomb’s Law again, the force F32 is:
F32 =k|q2||q3|
r2
32
5
where q2= 3.0µC = 3.0×10−6Cis the charge at x= 4.0m.
The distance r32 between charges q3(at x= 8.0m) and q2(at x= 4.0m) is:
r32 = 8.0−4.0 = 4.0m
Plugging in the values, we get:
F32 = (8.99 ×109)3.0×10−6×1.0×10−6
(4.0)2
Solving for the force F32, we find:
F32 = 6.74 ×10−4N
Step 3: Calculate the net force Fnet on charge q3.
The net force Fnet is the vector sum of forces F31 and F32. Since F31 is to
the left (negative x direction) and F32 is to the right (positive x direction), the
net force is:
Fnet =F32 −F31
Fnet = 6.74 ×10−4−4.49 ×10−4
Fnet = 2.25 ×10−4N
Therefore, the net force on charge q3due to charges q1and q2is 2.25×
Question 8
Question
Two point charges, q1=−3µC and q2= 5 µC, are placed 10 cm apart in air.
Calculate the magnitude and direction of the electric force that each charge
exerts on the other.
Solution
Step 1: Calculate the magnitude of the electric force each charge exerts on the
other 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 k= 8.99 ×109Nm2/C2is Coulomb’s constant, q1and q2are the magni-
tudes of the charges, and ris the distance between the charges.
Plugging in the values:
F=(8.99 ×109)·3×10−6·5×10−6
0.12
6
F=4.495 ×104
0.01
F= 4495 N
Therefore, the magnitude of the electric force that each charge exerts on the
other is 4495 N.
Step 2: Determine the direction of the electric force on each charge. Since
the charges are opposite in sign, there will be an attractive force between them.
The force on q1due to q2will be directed toward q2, and the force on q2due to
q1will be directed toward q1.
Question 9
Question
Two point charges, q1=−4µC and q2= 8 µC, are located 10 cm apart on
the x-axis. Calculate the magnitude and direction of the electric force on each
charge.
Solution
Step 1: Calculate the distance between the two charges. Given that the charges
are located 10 cm (or 0.1 m) apart on the x-axis, the distance (r) between them
is 0.1 m.
Step 2: Calculate the magnitude of the electric force on q1. The magnitude
of the electric force between two point charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the electrostatic constant (8.99 ×109N·m2/C2).
Substitute the given values into the formula:
F=(8.99 ×109N·m2/C2)·(4 ×10−6C)·(8 ×10−6C)
(0.1m)2
F=287.68 ×10−6
0.01
F= 28.768 N
So, the magnitude of the electric force acting on q1is 28.768 N.
Step 3: Determine the direction of the force on q1. Since q1is negative, the
force on it will be attractive and directed towards q2.
Step 4: Calculate the magnitude of the electric force on q2. The magnitude
of the electric force on q2will be the same as that on q1, i.e., 28.768 N.
Step 5: Determine the direction of the force on q2. Since q2is positive, the
force on it will be repulsive and directed away from q1.
7
Question 10
Question
Three point charges are arranged in an equilateral triangle with sides of length
a. The charges are +q,−2q, and +3q. Calculate the net electric force on the
+qcharge due to the other two charges.
Solution
To find the net electric force on the +qcharge, we need to calculate the in-
dividual forces on it due to the other two charges and then sum these forces
vectorially.
Step 1: Calculate the force on the +qcharge due to the −2qcharge. The
magnitude of the electric force between two point charges q1and q2separated
by a distance ris given by Coulomb’s Law:
F=k·|q1q2|
r2,
where kis the Coulomb constant (8.99 ×109N m2/C2).
The distance between the +qand −2qcharges is a. Therefore, the magnitude
of the force on the +qcharge due to the −2qcharge is:
F+qdue to −2q=k·|+q(−2q)|
a2.
Step 2: Calculate the force on the +qcharge due to the +3qcharge. The
distance between the +qand +3qcharges is also a. The magnitude of the force
on the +qcharge due to the +3qcharge is:
F+qdue to +3q=k·|+q(3q)|
a2.
Step 3: Find the net force on the +qcharge. The net force on the +q
charge is the vector sum of the forces from the −2qand +3qcharges. Since the
forces are acting along the sides of an equilateral triangle, they will have the
same magnitude but will be in opposite directions.
Therefore, the net force is:
Fnet =F+qdue to +3q−F+qdue to −2q.
Substitute the expressions for the forces into the equation above and simplify
to find the net force on the +qcharge.
Question 11
Question
Two point charges are fixed in place along the x-axis. Charge q1=−4.0µC is
located at x= 0 m and charge q2= 7.0µC is located at x= 2.0m. Find the
8
electric force on q2due to q1.
Solution
Step 1: Calculate the distance between the two charges. Given that q2is located
at x= 2.0m and q1is located at x= 0 m, the distance rbetween the two charges
is:
r=x2−x1= 2.0m−0m= 2.0m
Step 2: Calculate the electric force. The magnitude of the electric force F
between two point charges can be calculated using Coulomb’s Law:
F=k· |q1|·|q2|
r2
where
•k≈8.99 ×109N m2/C2is the Coulomb constant,
•|q1|= 4.0×10−6C,
•|q2|= 7.0×10−6C, and
•r= 2.0m.
Plugging in the values, we get:
F=(8.99 ×109N m2/C2)·(4.0×10−6C)·(7.0×10−6C)
(2.0m)2
F=(8.99 ×103)·(4.0) ·(7.0)
4.0
F= 8.99 ×4×7 = 251.32 N
Therefore, the electric force on q2due to q1is 251.32 N directed towards q1.
Question 12
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between these two
charges.
9
Solution
Step 1: Convert the charges to coulombs: The charges in microcoulombs can be
converted to coulombs by multiplying by 10−6.q1=−4.0µC = −4.0×10−6C
=−4.0µCq2= 6.0µC = 6.0×10−6C = 6.0µC
Step 2: Find the distance between the charges in meters: Given that the
charges are placed 10.0 cm apart, convert the distance to meters by dividing by
100. r= 10.0cm = 10.0×10−2m = 0.10 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
The electric force between two point charges q1and q2separated by a distance
ris given by Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
Substitute the values into the formula:
F=(8.99 ×109N m2/C2)·|−4.0×10−6C·6.0×10−6C|
(0.10 m)2
Step 4: Calculate the magnitude of the electric force:
F=(8.99 ×109)·(4.0×6.0) ×10−6
0.01 N
F=8.99 ×109·24
0.01 N
F= 2.16 ×10−2N
Therefore, the magnitude of the electric force between the charges q1=−4.0
µC and q2= 6.0µC placed 10.0 cm apart is 2.16 ×10−2N.
Question 13
Question
Three charges are arranged along the x-axis. Charge q1= 4 nC is at the origin,
charge q2=−6nC is located at x= 5 m, and charge q3= 8 nC is placed at
x= 10 m. Calculate the electric force acting on charge q3due to the other two
charges.
Solution
Step 1: Calculate the electric force on charge q3due to charge q1using Coulomb’s
law:
F13 =k|q1q3|
r2
13
10
Step 2: Calculate the distance r13 between charge q1and charge q3:
r13 = 10 m
Step 3: Substitute the given values into Coulomb’s law:
F13 =(8.99 ×109N m2/C2)(4 ×10−9C)(8 ×10−9C)
(10 m)2
Step 4: Calculate the electric force on charge q3due to charge q1:
F13 =(35.96 ×101)(8 ×10−9)
100
F13 =287.68 ×10−8
100
F13 = 2.8768 ×10−6N
Step 5: Calculate the electric force on charge q3due to charge q2using the
same process:
F23 =k|q2q3|
r2
23
Step 6: Calculate the distance r23 between charge q2and charge q3:
r23 = 5 m
Step 7: Substitute the given values into Coulomb’s law:
F23 =(8.99 ×109)(−6×10−9)(8 ×10−9)
(5)2
Step 8: Calculate the electric force on charge q3due to charge q2:
F23 =−43.92 ×101
25
F23 =−1.7568 ×10−6N
Step 9: Finally, calculate the total electric force on charge q3by summing
the forces F13 and F23:
Ftotal =F13 +F23
Ftotal = 2.8768 ×10−6−1.7568 ×10−6
Ftotal = 1.12 ×10−6N
Therefore, the total electric force on charge q3is 1.12 ×10−6N, directed
towards charge q1.
11
Question 14
Question
Two point charges, q1=−8nC and q2= 4 nC, are placed 6cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs using the conversion factor 1nC =
10−9C.
q1=−8×10−9C
q2= 4 ×10−9C
Step 2: Calculate the distance between the charges in meters.
6cm = 6 ×10−2m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law:
F=k·|q1·q2|
r2where k= 8.99 ×109N m2/C2is the electrostatic constant.
F=(8.99 ×109)·|−8×10−9·4×10−9|
(6 ×10−2)2
=8.99 ×109·32 ×10−18
36 ×10−4
=287.68 ×10−9
36 ×10−4
= 7.99 ×10−5N
Therefore, the magnitude of the electric force between the charges is 7.99 ×
10−5N.
Question 15
Question
Three point charges are arranged on the vertices of a right-angled triangle as
shown:
Q1(+3 C)
Q2(−2C)
Q3(+4 C)
The side lengths are 3 m, 4 m, and 5 m. Calculate the electric force on Q1
due to Q2and Q3.
12
Solution
Step 1: Calculate the electric force on Q1due to Q2. Given: Q1= +3 C,
Q2=−2C, r= 4 m.
The electric force between two point charges is given by Coulomb’s Law:
F=k·|Q1|·|Q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the formula to find the electric force on Q1
due to Q2:
FQ1Q2= 8.99 ×109·3·2
42= 1.3485 ×109N
Step 2: Calculate the electric force on Q1due to Q3. Given: Q1= +3 C,
Q3= +4 C, r= 5 m.
Using Coulomb’s Law again, the electric force on Q1due to Q3is:
FQ1Q3= 8.99 ×109·3·4
52= 2.8792 ×109N
Step 3: Calculate the net electric force on Q1. The electric force on Q1due
to Q2is attractive (since Q2is negative) and the force due to Q3is repulsive
(since Q3is positive).
Therefore, the net force on Q1is:
Fnet =FQ1Q3−FQ1Q2= 2.8792 ×109−1.3485 ×109= 1.5307 ×109N
So, the net electric force on Q1is 1.5307 ×109N.
Question 16
Question
Two point charges, q1=−2µC and q2= 4 µC, are placed 10 cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
To calculate the magnitude of the electric force between the charges, we can use
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 Coulomb’s constant
(8.99 ×109N·m2/C2), - q1and q2are the magnitudes of the point charges
involved, - ris the distance between the charges.
13
Given: - q1=−2µC, - q2= 4 µC, - r= 0.10 m.
Step 1: Calculate the electric force using Coulomb’s Law.
Plugging in the values:
F= (8.99 ×109)·|(−2×10−6)·(4 ×10−6)|
(0.10)2
F= 8.99 ×109·8×10−12
0.01
F= 8.99 ×109·8×10−10
F= 71.92 ×10−1
F= 7.192 N
Therefore, the magnitude of the electric force between the charges is 7.192
N.
Question 17
Question
Two point charges, q1=−3µC and q2= 6µC, are placed 10 cm apart. Calculate
the magnitude and direction of the electric force that q1exerts on q2.
Solution
Step 1: Determine the distance between the charges. Given that the charges
are 10 cm apart, we convert this distance to meters:
d= 10 cm = 0.10 m
Step 2: Calculate the magnitude of the electric force between the charges
using Coulomb’s law:
F=k·
q1·q2
d2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the values, we have:
F= 8.99 ×109N m2/C2·| − 3×10−6C·6×10−6C|
(0.10 m)2
F= 8.99 ×109N m2/C2·1.8×10−5C2/(0.01 m2)
F= 8.99 ×109N·1.8×10−3m/0.01 m2
14
F= 8.99 ×109·1.8×10−2N= 1.6182 ×108N
Step 3: Determine the direction of the force. The force is repulsive since the
charges have opposite signs, therefore the force points away from q1and towards
q2.
So, the magnitude of the electric force that q1exerts on q2is 1.6182 ×108
N, directed away from q1and towards q2.
Question 18
Question
Two point charges, +4.0µC and −2.0µC, are placed 10.0 cm apart in a vacuum.
Calculate the magnitude of the electric force between these two charges.
Solution
To calculate the magnitude of the electric force between the two charges, we
will use Coulomb’s Law, which states that the magnitude of the electric force
between two point charges is given by:
F=k·|q1·q2|
r2
where: - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the two charges, and -
ris the distance between the charges.
Given: - q1= +4.0µC= 4.0×10−6C, - q2=−2.0µC=−2.0×10−6C, and
-r= 10.0cm = 0.10 m.
Step 1: Substitute the given values into Coulomb’s Law formula:
F= (8.99 ×109)·|4.0×10−6· −2.0×10−6|
(0.10)2
F= (8.99 ×109)·8.0×10−12
0.01
Step 2: Calculate the magnitude of the electric force:
F= (8.99 ×109)·8.0×10−10
F= 7.192 ×10−1N
Therefore, the magnitude of the electric force between the two charges is
0.7192 N.
15
Question 19
Question
Three point charges are placed at the vertices of an equilateral triangle as shown
below:
−q
q
−q
If the charges qare positive and the charges −qare negative, calculate the
magnitude and direction of the electric force experienced by the charge at point
B.
Given: Each charge q= 2 µC, the side length of the equilateral triangle is 2
m, and the Coulomb’s constant k= 9 ×109N m2/C2.
Solution
Step 1: Calculate the distance between the charges at point B and point A or C.
Given that the side length of the equilateral triangle is 2 m, the distance between
the charges at points A and B (or C) can be calculated using the Pythagorean
theorem:
d=q22+ (2√3)2=√52 = 2√13 m
Step 2: Calculate the electric force between the charges qand −qat points B
and A (or C). The electric force between two point charges is given by Coulomb’s
Law:
F=k|q1q2|
r2
where kis the Coulomb’s constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
Substitute the given values into the equation to calculate the force:
F= 9 ×109×(2 ×10−6)(2 ×10−6)
(2√13)2
F= 36 ×10−12 ×4×10−12
52
F=144
52 ×10−24
16
F≈2.769 ×10−23 N
The force between the charges at point B and point A (or C) is approximately
2.769 ×10−23 N.
Step 3: Determine the direction of the electric force. Since the charges at
points B and A (or C) have opposite signs (attraction between positive and
negative charges), the electric force experienced by the charge at point B is
directed towards the charge at point A (or C).
Question 20
Question
Two point charges, q1= 4 µC and q2=−6µC, are placed 1 meter apart.
Determine the magnitude and direction of the electric force experienced by
each charge due to the other charge.
Solution
Step 1: Calculate the magnitude of the electric force on q1due to q2. The
magnitude of the electric force between two point charges is given by Coulomb’s
law:
F=k|q1q2|
r2
where kis Coulomb’s constant (8.99 ×109N·m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Plugging in the values, we get:
F1,2= (8.99 ×109)|4×10−6× −6×10−6|
(1)2
F1,2= 4314 N
The force on q1due to q2is attractive and points towards q2.
Step 2: Calculate the magnitude of the electric force on q2due to q1. Since
the forces between charges are equal in magnitude and opposite in direction by
Newton’s third law, the magnitude of the force on q2due to q1is also 4314 N,
but in the opposite direction (towards q1).
Therefore, the magnitude of the electric force experienced by each charge
due to the other charge is 4314 N, with the direction of each force being towards
the other charge.
Question 21
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. The charges are +2 µC,−3µC, and +4 µC. Calculate the mag-
17
nitude and direction of the net electric force on the charge located at the top
vertex.
+2 µC
−3µC
+4 µC
Solution
Step 1: Calculate the electric force between all pairs of charges using Coulomb’s
law: The electric force between two charges q1and q2separated by a distance
ris given by
F=k·|q1q2|
r2,
where k= 8.99 ×109N m2/C2is Coulomb’s constant.
We denote the charge at the top vertex as q1= +2 µC, the charge at the
right vertex as q2=−3µC, and the charge at the left vertex as q3= +4 µC.
The distance between any two charges in an equilateral triangle is the length
of a side, denoted as d. For an equilateral triangle, d= 2r√3where ris the
length from a vertex to the centroid. Given that the side of the triangle is
d= 4 cm, we have r=d
2√3=4
2√3=2√3
3cm. So, r=2√3
3cm.
Now, we calculate the distances between charges:
r12 =r13 =r23 =2√3
3cm.
The electric forces between each pair of charges are:
F12 =k·|+ 2 ×(−3)|
(2√3/3)2=18k√3
4,
F13 =k·|+ 2 ×4|
(2√3/3)2=8k√3
4,
F23 =k·| − 3×4|
(2√3/3)2=12k√3
4.
18
Question 22
Question
Two point charges, q1=−3.00 µC and q2= 1.50 µC, are placed 20.0cm apart.
Calculate the magnitude of the electric force between them.
Solution
To calculate the magnitude of the electric force between the two charges, we
can use Coulomb’s law. Coulomb’s law states that the magnitude of the electric
force between two point charges is given by the equation:
F=k·|q1|·|q2|
r2,
where: - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the two charges, and -
ris the distance between the charges.
Let’s substitute the given values into the equation and solve for F.
Step 1: Calculate the distance between the charges in meters Given:
r= 20.0cm = 0.20 m
Step 2: Calculate the electric force using Coulomb’s law Given:
q1=−3.00 µC=−3.00 ×10−6C Given: q2= 1.50 µC= 1.50 ×10−6C
Plugging the values into Coulomb’s law:
F=k·|q1|·|q2|
r2
F= (8.99 ×109N m2/C2)·3.00 ×10−6C·1.50 ×10−6C
(0.20 m)2
Step 3: Calculate the electric force
F= 8.99 ×109×3.00 ×1.50
0.202N
F= 8.99 ×109×4.50
0.04 N
F= 8.99 ×109×112.5N
F= 1.013 ×1012 N
Therefore, the magnitude of the electric force between the two charges is
1.013 ×1012 N.
Question 23
Question
Two point charges, Q1= 3µC and Q2=−4µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
19
Solution
Step 1: Convert the given charges to standard units.
Given: Q1= 3µC and Q2=−4µC.
To convert from µC to C, we use the conversion factor 1µC = 10−6C.
So, Q1= 3µC = 3 ×10−6Cand Q2=−4µC =−4×10−6C.
Step 2: Determine the distance between the charges.
Given that the charges are placed 10 cm apart, we convert 10 cm to meters
using the conversion factor 1cm = 10−2m.
So, the distance between the charges is 10 cm = 10 ×10−2m= 0.1m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s law.
The magnitude of the electric force between point charges is given by Coulomb’s
law:
F=k·|Q1·Q2|
r2,
where kis the electrostatic constant (k≈8.9875 ×109N·m2/C2), Q1and Q2
are the magnitudes of the charges, and ris the distance between the charges.
Substitute the given values:
F= 8.9875 ×109N·m2/C2×|3×10−6· −4×10−6|
(0.1)2
F= 8.9875 ×109×12 ×10−12
0.01
F= 8.9875 ×109×1.2×10−9
F= 10.785 ×100N
F= 10.785 N
Therefore, the magnitude of the electric force between the charges is 10.785
N.
Question 24
Question
Three point charges are arranged in a line. Charge q1= +5.0µC is at the
origin, charge q2=−3.0µC is located at x= 0.30 m, and charge q3= +2.0µC
is located at x= 0.45 m. Calculate the net electric force on charge q2.
Solution
Step 1: Calculate the electric force on charge q2due to charge q1.
F12 =k· |q1|·|q2|
r2
12
20
Step 2: Calculate the direction of the electric force on charge q2due to
charge q1. The force will be repulsive because the charges have opposite signs,
so the force vector points in the positive x-direction.
Step 3: Calculate the magnitude of the electric force F12.
F12 =8.99 ×109N·m2/C2·5.0×10−6C·3.0×10−6C
(0.30 m)2
F12 = 29965.00 N
Step 4: Calculate the electric force on charge q2due to charge q3.
F23 =k· |q2|·|q3|
r2
23
Step 5: Calculate the direction of the electric force on charge q2due to
charge q3. The force will be attractive because the charges have opposite signs,
so the force vector points in the negative x-direction.
Step 6: Calculate the magnitude of the electric force F23.
F23 =8.99 ×109N·m2/C2·3.0×10−6C·2.0×10−6C
(0.15 m)2
F23 = 35984.17 N
Step 7: Calculate the net electric force on charge q2.
Fnet =F12 −F23
Fnet = 29965.00 N−35984.17 N
Fnet =−6029.17 N
Therefore, the net electric force on charge q2is −6029.17 N in the negative
x-direction.
Question 25
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are separated by a distance
of 10.0m. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−4.0µC=−4.0×10−6C
q2= 8.0µC= 8.0×10−6C
21
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: F= magnitude of the electric force k= Coulomb’s constant (8.99 ×
109N·m2/C2)q1,q2= magnitudes of the charges r= distance between the
charges
Plugging in the values:
F= (8.99 ×109)·| − 4.0×10−6·8.0×10−6|
10.02
Step 3: Calculate the magnitude of the electric force.
F= (8.99 ×109)·32 ×10−12
100
F= (8.99 ×109)·3.2×10−13
F= 2.8776 ×10−3N
Therefore, the magnitude of the electric force between the charges is 2.8776×
10−3N.
22
F= 14.98 ×10−3N
F= 0.01498 N
Therefore, the electric force that charge q1exerts on charge q2is 0.01498 N.
Question 2
Question
Three point charges are arranged at the vertices of an equilateral triangle as
shown below:
Q
−Q Q
If each charge magnitude is Q= 2 mC, what is the magnitude of the electric
force experienced by the charge at the top vertex due to the other two charges?
Solution
Step 1: Begin by labeling the charges and distances in the triangle. Let the
distance between adjacent charges be a.
Q
a
−Q Q
a
Step 2: Calculate the electric force experienced by the charge at the top
vertex due to the charge on the bottom left.
F1=kQ2
a2
Step 3: Calculate the electric force experienced by the charge at the top
vertex due to the charge on the bottom right.
F2=kQ2
(2a)2
Step 4: Determine the net force experienced by the charge at the top vertex.
Since the two forces are acting in opposite directions along the same line, the
net force is the difference between them.
2
Fnet =F1−F2
=kQ2
a2−kQ2
(2a)2
=kQ21
a2−1
4a2
=kQ23
4a2
Step 5: Substitute the known values k= 8.99×109N·m2/C2and Q= 2 mC,
and afor an equilateral triangle (a=a√3/3).
Fnet = (8.99 ×109)·(2 ×10−3)2·
3
4a√3
32
= (8.99 ×109)·4×10−6·
3
a√3
32
= 3√3×104N
Therefore, the magnitude of the electric force experienced by the charge at
the top vertex due to the other two charges is 3√3×104N.
Question 3
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are placed 20.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges from microcoulombs to coulombs.
q1=−4.0µC=−4.0×10−6C
q2= 8.0µC= 8.0×10−6C
Step 2: Determine the distance between the charges in meters.
r= 20.0cm = 20.0×10−2m= 0.20 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k·|q1·q2|
r2
3
where k= 8.99 ×109N·m2/C2is the Coulomb’s constant.
Step 4: Substitute the values into the equation to find the electric force.
F= (8.99 ×109)·|(−4.0×10−6)·(8.0×10−6)|
(0.20)2
= 8.99 ×109·32 ×10−12
0.04
= 8.99 ×109·8×10−11
= 71.92 N
Therefore, the magnitude of the electric force between the charges is 71.92
N.
Question 5
Question
Three point charges are arranged as shown in the diagram below: a +2.0 µC
charge at the origin, a -3.0 µC charge at (4.0 m, 0), and a +4.0 µC charge at
(0, 3.0 m). Calculate the magnitude of the net electric force acting on the +4.0
µC charge due to the other two charges.
Solution
1. Let’s denote the charges at the origin, (4.0 m, 0), and (0, 3.0 m) respectively
as q1= +2.0µC, q2=−3.0µC, and q3= +4.0µC.
2. We will calculate the net electric force on the +4.0µC charge, q3, due to
charges q1and q2using the principle of superposition. The electric force exerted
by each point charge can be calculated using the formula:
Felec =k|q1q2|
r2
3. First, we will calculate the force on q3due to q1:
F1=k|q1q3|
r2
13
4. Given that k= 8.9875×109Nm2
C2and r13 = 3.0m, we can substitute these
values into the formula.
5. Solving for F1:
F1= 8.9875 ×109Nm2
C2·(2.0µC)(4.0µC)
(3.0m)2
6. Next, we will calculate the force on q3due to q2:
F2=k|q2q3|
r2
23
4
7. Given that r23 = 4.0m, we can substitute the known values into the
formula.
8. Solving for F2:
F2= 8.9875 ×109Nm2
C2·(3.0µC)(4.0µC)
(4.0m)2
9. The net electric force on q3would be the vector sum of F1and F2.
10. Calculate the magnitudes and directions of F1and F2, then sum them
to find the net force acting on q3.
Question 7
Question
Three point charges are arranged along the x-axis. Charge q1=−2.0µC is
located at the origin, charge q2= 3.0µC is located at x= 4.0m, and charge
q3= 1.0µC is located at x= 8.0m. Calculate the net force on charge q3due
to charges q1and q2.
Solution
Step 1: Calculate the force F31 on charge q3due to charge q1.
Since the charges are on the x-axis, we can use Coulomb’s Law to calculate
the force:
F31 =k|q1||q3|
r2
31
where k= 8.99 ×109N·m2/C2is the Coulomb’s constant, q1=−2.0µC =
−2.0×10−6Cis the charge at the origin, and q3= 1.0µC = 1.0×10−6Cis the
charge at x= 8.0m.
The distance r31 between charges q3(at x= 8.0m) and q1(at the origin) is:
r31 = 8.0m
Therefore, the force F31 is:
F31 = (8.99 ×109)| − 2.0×10−6×1.0×10−6|
(8.0)2
Calculating the force, we get:
F31 = 4.49 ×10−4N
Step 2: Calculate the force F32 on charge q3due to charge q2.
Using Coulomb’s Law again, the force F32 is:
F32 =k|q2||q3|
r2
32
5
where q2= 3.0µC = 3.0×10−6Cis the charge at x= 4.0m.
The distance r32 between charges q3(at x= 8.0m) and q2(at x= 4.0m) is:
r32 = 8.0−4.0 = 4.0m
Plugging in the values, we get:
F32 = (8.99 ×109)3.0×10−6×1.0×10−6
(4.0)2
Solving for the force F32, we find:
F32 = 6.74 ×10−4N
Step 3: Calculate the net force Fnet on charge q3.
The net force Fnet is the vector sum of forces F31 and F32. Since F31 is to
the left (negative x direction) and F32 is to the right (positive x direction), the
net force is:
Fnet =F32 −F31
Fnet = 6.74 ×10−4−4.49 ×10−4
Fnet = 2.25 ×10−4N
Therefore, the net force on charge q3due to charges q1and q2is 2.25×
Question 8
Question
Two point charges, q1=−3µC and q2= 5 µC, are placed 10 cm apart in air.
Calculate the magnitude and direction of the electric force that each charge
exerts on the other.
Solution
Step 1: Calculate the magnitude of the electric force each charge exerts on the
other 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 k= 8.99 ×109Nm2/C2is Coulomb’s constant, q1and q2are the magni-
tudes of the charges, and ris the distance between the charges.
Plugging in the values:
F=(8.99 ×109)·3×10−6·5×10−6
0.12
6
F=4.495 ×104
0.01
F= 4495 N
Therefore, the magnitude of the electric force that each charge exerts on the
other is 4495 N.
Step 2: Determine the direction of the electric force on each charge. Since
the charges are opposite in sign, there will be an attractive force between them.
The force on q1due to q2will be directed toward q2, and the force on q2due to
q1will be directed toward q1.
Question 9
Question
Two point charges, q1=−4µC and q2= 8 µC, are located 10 cm apart on
the x-axis. Calculate the magnitude and direction of the electric force on each
charge.
Solution
Step 1: Calculate the distance between the two charges. Given that the charges
are located 10 cm (or 0.1 m) apart on the x-axis, the distance (r) between them
is 0.1 m.
Step 2: Calculate the magnitude of the electric force on q1. The magnitude
of the electric force between two point charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the electrostatic constant (8.99 ×109N·m2/C2).
Substitute the given values into the formula:
F=(8.99 ×109N·m2/C2)·(4 ×10−6C)·(8 ×10−6C)
(0.1m)2
F=287.68 ×10−6
0.01
F= 28.768 N
So, the magnitude of the electric force acting on q1is 28.768 N.
Step 3: Determine the direction of the force on q1. Since q1is negative, the
force on it will be attractive and directed towards q2.
Step 4: Calculate the magnitude of the electric force on q2. The magnitude
of the electric force on q2will be the same as that on q1, i.e., 28.768 N.
Step 5: Determine the direction of the force on q2. Since q2is positive, the
force on it will be repulsive and directed away from q1.
7
Question 10
Question
Three point charges are arranged in an equilateral triangle with sides of length
a. The charges are +q,−2q, and +3q. Calculate the net electric force on the
+qcharge due to the other two charges.
Solution
To find the net electric force on the +qcharge, we need to calculate the in-
dividual forces on it due to the other two charges and then sum these forces
vectorially.
Step 1: Calculate the force on the +qcharge due to the −2qcharge. The
magnitude of the electric force between two point charges q1and q2separated
by a distance ris given by Coulomb’s Law:
F=k·|q1q2|
r2,
where kis the Coulomb constant (8.99 ×109N m2/C2).
The distance between the +qand −2qcharges is a. Therefore, the magnitude
of the force on the +qcharge due to the −2qcharge is:
F+qdue to −2q=k·|+q(−2q)|
a2.
Step 2: Calculate the force on the +qcharge due to the +3qcharge. The
distance between the +qand +3qcharges is also a. The magnitude of the force
on the +qcharge due to the +3qcharge is:
F+qdue to +3q=k·|+q(3q)|
a2.
Step 3: Find the net force on the +qcharge. The net force on the +q
charge is the vector sum of the forces from the −2qand +3qcharges. Since the
forces are acting along the sides of an equilateral triangle, they will have the
same magnitude but will be in opposite directions.
Therefore, the net force is:
Fnet =F+qdue to +3q−F+qdue to −2q.
Substitute the expressions for the forces into the equation above and simplify
to find the net force on the +qcharge.
Question 11
Question
Two point charges are fixed in place along the x-axis. Charge q1=−4.0µC is
located at x= 0 m and charge q2= 7.0µC is located at x= 2.0m. Find the
8
electric force on q2due to q1.
Solution
Step 1: Calculate the distance between the two charges. Given that q2is located
at x= 2.0m and q1is located at x= 0 m, the distance rbetween the two charges
is:
r=x2−x1= 2.0m−0m= 2.0m
Step 2: Calculate the electric force. The magnitude of the electric force F
between two point charges can be calculated using Coulomb’s Law:
F=k· |q1|·|q2|
r2
where
•k≈8.99 ×109N m2/C2is the Coulomb constant,
•|q1|= 4.0×10−6C,
•|q2|= 7.0×10−6C, and
•r= 2.0m.
Plugging in the values, we get:
F=(8.99 ×109N m2/C2)·(4.0×10−6C)·(7.0×10−6C)
(2.0m)2
F=(8.99 ×103)·(4.0) ·(7.0)
4.0
F= 8.99 ×4×7 = 251.32 N
Therefore, the electric force on q2due to q1is 251.32 N directed towards q1.
Question 12
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between these two
charges.
9
Solution
Step 1: Convert the charges to coulombs: The charges in microcoulombs can be
converted to coulombs by multiplying by 10−6.q1=−4.0µC = −4.0×10−6C
=−4.0µCq2= 6.0µC = 6.0×10−6C = 6.0µC
Step 2: Find the distance between the charges in meters: Given that the
charges are placed 10.0 cm apart, convert the distance to meters by dividing by
100. r= 10.0cm = 10.0×10−2m = 0.10 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
The electric force between two point charges q1and q2separated by a distance
ris given by Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
Substitute the values into the formula:
F=(8.99 ×109N m2/C2)·|−4.0×10−6C·6.0×10−6C|
(0.10 m)2
Step 4: Calculate the magnitude of the electric force:
F=(8.99 ×109)·(4.0×6.0) ×10−6
0.01 N
F=8.99 ×109·24
0.01 N
F= 2.16 ×10−2N
Therefore, the magnitude of the electric force between the charges q1=−4.0
µC and q2= 6.0µC placed 10.0 cm apart is 2.16 ×10−2N.
Question 13
Question
Three charges are arranged along the x-axis. Charge q1= 4 nC is at the origin,
charge q2=−6nC is located at x= 5 m, and charge q3= 8 nC is placed at
x= 10 m. Calculate the electric force acting on charge q3due to the other two
charges.
Solution
Step 1: Calculate the electric force on charge q3due to charge q1using Coulomb’s
law:
F13 =k|q1q3|
r2
13
10
Step 2: Calculate the distance r13 between charge q1and charge q3:
r13 = 10 m
Step 3: Substitute the given values into Coulomb’s law:
F13 =(8.99 ×109N m2/C2)(4 ×10−9C)(8 ×10−9C)
(10 m)2
Step 4: Calculate the electric force on charge q3due to charge q1:
F13 =(35.96 ×101)(8 ×10−9)
100
F13 =287.68 ×10−8
100
F13 = 2.8768 ×10−6N
Step 5: Calculate the electric force on charge q3due to charge q2using the
same process:
F23 =k|q2q3|
r2
23
Step 6: Calculate the distance r23 between charge q2and charge q3:
r23 = 5 m
Step 7: Substitute the given values into Coulomb’s law:
F23 =(8.99 ×109)(−6×10−9)(8 ×10−9)
(5)2
Step 8: Calculate the electric force on charge q3due to charge q2:
F23 =−43.92 ×101
25
F23 =−1.7568 ×10−6N
Step 9: Finally, calculate the total electric force on charge q3by summing
the forces F13 and F23:
Ftotal =F13 +F23
Ftotal = 2.8768 ×10−6−1.7568 ×10−6
Ftotal = 1.12 ×10−6N
Therefore, the total electric force on charge q3is 1.12 ×10−6N, directed
towards charge q1.
11
Question 14
Question
Two point charges, q1=−8nC and q2= 4 nC, are placed 6cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs using the conversion factor 1nC =
10−9C.
q1=−8×10−9C
q2= 4 ×10−9C
Step 2: Calculate the distance between the charges in meters.
6cm = 6 ×10−2m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law:
F=k·|q1·q2|
r2where k= 8.99 ×109N m2/C2is the electrostatic constant.
F=(8.99 ×109)·|−8×10−9·4×10−9|
(6 ×10−2)2
=8.99 ×109·32 ×10−18
36 ×10−4
=287.68 ×10−9
36 ×10−4
= 7.99 ×10−5N
Therefore, the magnitude of the electric force between the charges is 7.99 ×
10−5N.
Question 15
Question
Three point charges are arranged on the vertices of a right-angled triangle as
shown:
Q1(+3 C)
Q2(−2C)
Q3(+4 C)
The side lengths are 3 m, 4 m, and 5 m. Calculate the electric force on Q1
due to Q2and Q3.
12
Solution
Step 1: Calculate the electric force on Q1due to Q2. Given: Q1= +3 C,
Q2=−2C, r= 4 m.
The electric force between two point charges is given by Coulomb’s Law:
F=k·|Q1|·|Q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the formula to find the electric force on Q1
due to Q2:
FQ1Q2= 8.99 ×109·3·2
42= 1.3485 ×109N
Step 2: Calculate the electric force on Q1due to Q3. Given: Q1= +3 C,
Q3= +4 C, r= 5 m.
Using Coulomb’s Law again, the electric force on Q1due to Q3is:
FQ1Q3= 8.99 ×109·3·4
52= 2.8792 ×109N
Step 3: Calculate the net electric force on Q1. The electric force on Q1due
to Q2is attractive (since Q2is negative) and the force due to Q3is repulsive
(since Q3is positive).
Therefore, the net force on Q1is:
Fnet =FQ1Q3−FQ1Q2= 2.8792 ×109−1.3485 ×109= 1.5307 ×109N
So, the net electric force on Q1is 1.5307 ×109N.
Question 16
Question
Two point charges, q1=−2µC and q2= 4 µC, are placed 10 cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
To calculate the magnitude of the electric force between the charges, we can use
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 Coulomb’s constant
(8.99 ×109N·m2/C2), - q1and q2are the magnitudes of the point charges
involved, - ris the distance between the charges.
13
Given: - q1=−2µC, - q2= 4 µC, - r= 0.10 m.
Step 1: Calculate the electric force using Coulomb’s Law.
Plugging in the values:
F= (8.99 ×109)·|(−2×10−6)·(4 ×10−6)|
(0.10)2
F= 8.99 ×109·8×10−12
0.01
F= 8.99 ×109·8×10−10
F= 71.92 ×10−1
F= 7.192 N
Therefore, the magnitude of the electric force between the charges is 7.192
N.
Question 17
Question
Two point charges, q1=−3µC and q2= 6µC, are placed 10 cm apart. Calculate
the magnitude and direction of the electric force that q1exerts on q2.
Solution
Step 1: Determine the distance between the charges. Given that the charges
are 10 cm apart, we convert this distance to meters:
d= 10 cm = 0.10 m
Step 2: Calculate the magnitude of the electric force between the charges
using Coulomb’s law:
F=k·
q1·q2
d2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the values, we have:
F= 8.99 ×109N m2/C2·| − 3×10−6C·6×10−6C|
(0.10 m)2
F= 8.99 ×109N m2/C2·1.8×10−5C2/(0.01 m2)
F= 8.99 ×109N·1.8×10−3m/0.01 m2
14
F= 8.99 ×109·1.8×10−2N= 1.6182 ×108N
Step 3: Determine the direction of the force. The force is repulsive since the
charges have opposite signs, therefore the force points away from q1and towards
q2.
So, the magnitude of the electric force that q1exerts on q2is 1.6182 ×108
N, directed away from q1and towards q2.
Question 18
Question
Two point charges, +4.0µC and −2.0µC, are placed 10.0 cm apart in a vacuum.
Calculate the magnitude of the electric force between these two charges.
Solution
To calculate the magnitude of the electric force between the two charges, we
will use Coulomb’s Law, which states that the magnitude of the electric force
between two point charges is given by:
F=k·|q1·q2|
r2
where: - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the two charges, and -
ris the distance between the charges.
Given: - q1= +4.0µC= 4.0×10−6C, - q2=−2.0µC=−2.0×10−6C, and
-r= 10.0cm = 0.10 m.
Step 1: Substitute the given values into Coulomb’s Law formula:
F= (8.99 ×109)·|4.0×10−6· −2.0×10−6|
(0.10)2
F= (8.99 ×109)·8.0×10−12
0.01
Step 2: Calculate the magnitude of the electric force:
F= (8.99 ×109)·8.0×10−10
F= 7.192 ×10−1N
Therefore, the magnitude of the electric force between the two charges is
0.7192 N.
15
Question 19
Question
Three point charges are placed at the vertices of an equilateral triangle as shown
below:
−q
q
−q
If the charges qare positive and the charges −qare negative, calculate the
magnitude and direction of the electric force experienced by the charge at point
B.
Given: Each charge q= 2 µC, the side length of the equilateral triangle is 2
m, and the Coulomb’s constant k= 9 ×109N m2/C2.
Solution
Step 1: Calculate the distance between the charges at point B and point A or C.
Given that the side length of the equilateral triangle is 2 m, the distance between
the charges at points A and B (or C) can be calculated using the Pythagorean
theorem:
d=q22+ (2√3)2=√52 = 2√13 m
Step 2: Calculate the electric force between the charges qand −qat points B
and A (or C). The electric force between two point charges is given by Coulomb’s
Law:
F=k|q1q2|
r2
where kis the Coulomb’s constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
Substitute the given values into the equation to calculate the force:
F= 9 ×109×(2 ×10−6)(2 ×10−6)
(2√13)2
F= 36 ×10−12 ×4×10−12
52
F=144
52 ×10−24
16
F≈2.769 ×10−23 N
The force between the charges at point B and point A (or C) is approximately
2.769 ×10−23 N.
Step 3: Determine the direction of the electric force. Since the charges at
points B and A (or C) have opposite signs (attraction between positive and
negative charges), the electric force experienced by the charge at point B is
directed towards the charge at point A (or C).
Question 20
Question
Two point charges, q1= 4 µC and q2=−6µC, are placed 1 meter apart.
Determine the magnitude and direction of the electric force experienced by
each charge due to the other charge.
Solution
Step 1: Calculate the magnitude of the electric force on q1due to q2. The
magnitude of the electric force between two point charges is given by Coulomb’s
law:
F=k|q1q2|
r2
where kis Coulomb’s constant (8.99 ×109N·m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Plugging in the values, we get:
F1,2= (8.99 ×109)|4×10−6× −6×10−6|
(1)2
F1,2= 4314 N
The force on q1due to q2is attractive and points towards q2.
Step 2: Calculate the magnitude of the electric force on q2due to q1. Since
the forces between charges are equal in magnitude and opposite in direction by
Newton’s third law, the magnitude of the force on q2due to q1is also 4314 N,
but in the opposite direction (towards q1).
Therefore, the magnitude of the electric force experienced by each charge
due to the other charge is 4314 N, with the direction of each force being towards
the other charge.
Question 21
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. The charges are +2 µC,−3µC, and +4 µC. Calculate the mag-
17
nitude and direction of the net electric force on the charge located at the top
vertex.
+2 µC
−3µC
+4 µC
Solution
Step 1: Calculate the electric force between all pairs of charges using Coulomb’s
law: The electric force between two charges q1and q2separated by a distance
ris given by
F=k·|q1q2|
r2,
where k= 8.99 ×109N m2/C2is Coulomb’s constant.
We denote the charge at the top vertex as q1= +2 µC, the charge at the
right vertex as q2=−3µC, and the charge at the left vertex as q3= +4 µC.
The distance between any two charges in an equilateral triangle is the length
of a side, denoted as d. For an equilateral triangle, d= 2r√3where ris the
length from a vertex to the centroid. Given that the side of the triangle is
d= 4 cm, we have r=d
2√3=4
2√3=2√3
3cm. So, r=2√3
3cm.
Now, we calculate the distances between charges:
r12 =r13 =r23 =2√3
3cm.
The electric forces between each pair of charges are:
F12 =k·|+ 2 ×(−3)|
(2√3/3)2=18k√3
4,
F13 =k·|+ 2 ×4|
(2√3/3)2=8k√3
4,
F23 =k·| − 3×4|
(2√3/3)2=12k√3
4.
18
Question 22
Question
Two point charges, q1=−3.00 µC and q2= 1.50 µC, are placed 20.0cm apart.
Calculate the magnitude of the electric force between them.
Solution
To calculate the magnitude of the electric force between the two charges, we
can use Coulomb’s law. Coulomb’s law states that the magnitude of the electric
force between two point charges is given by the equation:
F=k·|q1|·|q2|
r2,
where: - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the two charges, and -
ris the distance between the charges.
Let’s substitute the given values into the equation and solve for F.
Step 1: Calculate the distance between the charges in meters Given:
r= 20.0cm = 0.20 m
Step 2: Calculate the electric force using Coulomb’s law Given:
q1=−3.00 µC=−3.00 ×10−6C Given: q2= 1.50 µC= 1.50 ×10−6C
Plugging the values into Coulomb’s law:
F=k·|q1|·|q2|
r2
F= (8.99 ×109N m2/C2)·3.00 ×10−6C·1.50 ×10−6C
(0.20 m)2
Step 3: Calculate the electric force
F= 8.99 ×109×3.00 ×1.50
0.202N
F= 8.99 ×109×4.50
0.04 N
F= 8.99 ×109×112.5N
F= 1.013 ×1012 N
Therefore, the magnitude of the electric force between the two charges is
1.013 ×1012 N.
Question 23
Question
Two point charges, Q1= 3µC and Q2=−4µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
19
Solution
Step 1: Convert the given charges to standard units.
Given: Q1= 3µC and Q2=−4µC.
To convert from µC to C, we use the conversion factor 1µC = 10−6C.
So, Q1= 3µC = 3 ×10−6Cand Q2=−4µC =−4×10−6C.
Step 2: Determine the distance between the charges.
Given that the charges are placed 10 cm apart, we convert 10 cm to meters
using the conversion factor 1cm = 10−2m.
So, the distance between the charges is 10 cm = 10 ×10−2m= 0.1m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s law.
The magnitude of the electric force between point charges is given by Coulomb’s
law:
F=k·|Q1·Q2|
r2,
where kis the electrostatic constant (k≈8.9875 ×109N·m2/C2), Q1and Q2
are the magnitudes of the charges, and ris the distance between the charges.
Substitute the given values:
F= 8.9875 ×109N·m2/C2×|3×10−6· −4×10−6|
(0.1)2
F= 8.9875 ×109×12 ×10−12
0.01
F= 8.9875 ×109×1.2×10−9
F= 10.785 ×100N
F= 10.785 N
Therefore, the magnitude of the electric force between the charges is 10.785
N.
Question 24
Question
Three point charges are arranged in a line. Charge q1= +5.0µC is at the
origin, charge q2=−3.0µC is located at x= 0.30 m, and charge q3= +2.0µC
is located at x= 0.45 m. Calculate the net electric force on charge q2.
Solution
Step 1: Calculate the electric force on charge q2due to charge q1.
F12 =k· |q1|·|q2|
r2
12
20
Step 2: Calculate the direction of the electric force on charge q2due to
charge q1. The force will be repulsive because the charges have opposite signs,
so the force vector points in the positive x-direction.
Step 3: Calculate the magnitude of the electric force F12.
F12 =8.99 ×109N·m2/C2·5.0×10−6C·3.0×10−6C
(0.30 m)2
F12 = 29965.00 N
Step 4: Calculate the electric force on charge q2due to charge q3.
F23 =k· |q2|·|q3|
r2
23
Step 5: Calculate the direction of the electric force on charge q2due to
charge q3. The force will be attractive because the charges have opposite signs,
so the force vector points in the negative x-direction.
Step 6: Calculate the magnitude of the electric force F23.
F23 =8.99 ×109N·m2/C2·3.0×10−6C·2.0×10−6C
(0.15 m)2
F23 = 35984.17 N
Step 7: Calculate the net electric force on charge q2.
Fnet =F12 −F23
Fnet = 29965.00 N−35984.17 N
Fnet =−6029.17 N
Therefore, the net electric force on charge q2is −6029.17 N in the negative
x-direction.
Question 25
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are separated by a distance
of 10.0m. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−4.0µC=−4.0×10−6C
q2= 8.0µC= 8.0×10−6C
21
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: F= magnitude of the electric force k= Coulomb’s constant (8.99 ×
109N·m2/C2)q1,q2= magnitudes of the charges r= distance between the
charges
Plugging in the values:
F= (8.99 ×109)·| − 4.0×10−6·8.0×10−6|
10.02
Step 3: Calculate the magnitude of the electric force.
F= (8.99 ×109)·32 ×10−12
100
F= (8.99 ×109)·3.2×10−13
F= 2.8776 ×10−3N
Therefore, the magnitude of the electric force between the charges is 2.8776×
10−3N.
22
F= 14.98 ×10−3N
F= 0.01498 N
Therefore, the electric force that charge q1exerts on charge q2is 0.01498 N.
Question 2
Question
Three point charges are arranged at the vertices of an equilateral triangle as
shown below:
Q
−Q Q
If each charge magnitude is Q= 2 mC, what is the magnitude of the electric
force experienced by the charge at the top vertex due to the other two charges?
Solution
Step 1: Begin by labeling the charges and distances in the triangle. Let the
distance between adjacent charges be a.
Q
a
−Q Q
a
Step 2: Calculate the electric force experienced by the charge at the top
vertex due to the charge on the bottom left.
F1=kQ2
a2
Step 3: Calculate the electric force experienced by the charge at the top
vertex due to the charge on the bottom right.
F2=kQ2
(2a)2
Step 4: Determine the net force experienced by the charge at the top vertex.
Since the two forces are acting in opposite directions along the same line, the
net force is the difference between them.
2
Fnet =F1−F2
=kQ2
a2−kQ2
(2a)2
=kQ21
a2−1
4a2
=kQ23
4a2
Step 5: Substitute the known values k= 8.99×109N·m2/C2and Q= 2 mC,
and afor an equilateral triangle (a=a√3/3).
Fnet = (8.99 ×109)·(2 ×10−3)2·
3
4a√3
32
= (8.99 ×109)·4×10−6·
3
a√3
32
= 3√3×104N
Therefore, the magnitude of the electric force experienced by the charge at
the top vertex due to the other two charges is 3√3×104N.
Question 3
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are placed 20.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges from microcoulombs to coulombs.
q1=−4.0µC=−4.0×10−6C
q2= 8.0µC= 8.0×10−6C
Step 2: Determine the distance between the charges in meters.
r= 20.0cm = 20.0×10−2m= 0.20 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k·|q1·q2|
r2
3
where k= 8.99 ×109N·m2/C2is the Coulomb’s constant.
Step 4: Substitute the values into the equation to find the electric force.
F= (8.99 ×109)·|(−4.0×10−6)·(8.0×10−6)|
(0.20)2
= 8.99 ×109·32 ×10−12
0.04
= 8.99 ×109·8×10−11
= 71.92 N
Therefore, the magnitude of the electric force between the charges is 71.92
N.
Question 5
Question
Three point charges are arranged as shown in the diagram below: a +2.0 µC
charge at the origin, a -3.0 µC charge at (4.0 m, 0), and a +4.0 µC charge at
(0, 3.0 m). Calculate the magnitude of the net electric force acting on the +4.0
µC charge due to the other two charges.
Solution
1. Let’s denote the charges at the origin, (4.0 m, 0), and (0, 3.0 m) respectively
as q1= +2.0µC, q2=−3.0µC, and q3= +4.0µC.
2. We will calculate the net electric force on the +4.0µC charge, q3, due to
charges q1and q2using the principle of superposition. The electric force exerted
by each point charge can be calculated using the formula:
Felec =k|q1q2|
r2
3. First, we will calculate the force on q3due to q1:
F1=k|q1q3|
r2
13
4. Given that k= 8.9875×109Nm2
C2and r13 = 3.0m, we can substitute these
values into the formula.
5. Solving for F1:
F1= 8.9875 ×109Nm2
C2·(2.0µC)(4.0µC)
(3.0m)2
6. Next, we will calculate the force on q3due to q2:
F2=k|q2q3|
r2
23
4
7. Given that r23 = 4.0m, we can substitute the known values into the
formula.
8. Solving for F2:
F2= 8.9875 ×109Nm2
C2·(3.0µC)(4.0µC)
(4.0m)2
9. The net electric force on q3would be the vector sum of F1and F2.
10. Calculate the magnitudes and directions of F1and F2, then sum them
to find the net force acting on q3.
Question 7
Question
Three point charges are arranged along the x-axis. Charge q1=−2.0µC is
located at the origin, charge q2= 3.0µC is located at x= 4.0m, and charge
q3= 1.0µC is located at x= 8.0m. Calculate the net force on charge q3due
to charges q1and q2.
Solution
Step 1: Calculate the force F31 on charge q3due to charge q1.
Since the charges are on the x-axis, we can use Coulomb’s Law to calculate
the force:
F31 =k|q1||q3|
r2
31
where k= 8.99 ×109N·m2/C2is the Coulomb’s constant, q1=−2.0µC =
−2.0×10−6Cis the charge at the origin, and q3= 1.0µC = 1.0×10−6Cis the
charge at x= 8.0m.
The distance r31 between charges q3(at x= 8.0m) and q1(at the origin) is:
r31 = 8.0m
Therefore, the force F31 is:
F31 = (8.99 ×109)| − 2.0×10−6×1.0×10−6|
(8.0)2
Calculating the force, we get:
F31 = 4.49 ×10−4N
Step 2: Calculate the force F32 on charge q3due to charge q2.
Using Coulomb’s Law again, the force F32 is:
F32 =k|q2||q3|
r2
32
5
where q2= 3.0µC = 3.0×10−6Cis the charge at x= 4.0m.
The distance r32 between charges q3(at x= 8.0m) and q2(at x= 4.0m) is:
r32 = 8.0−4.0 = 4.0m
Plugging in the values, we get:
F32 = (8.99 ×109)3.0×10−6×1.0×10−6
(4.0)2
Solving for the force F32, we find:
F32 = 6.74 ×10−4N
Step 3: Calculate the net force Fnet on charge q3.
The net force Fnet is the vector sum of forces F31 and F32. Since F31 is to
the left (negative x direction) and F32 is to the right (positive x direction), the
net force is:
Fnet =F32 −F31
Fnet = 6.74 ×10−4−4.49 ×10−4
Fnet = 2.25 ×10−4N
Therefore, the net force on charge q3due to charges q1and q2is 2.25×
Question 8
Question
Two point charges, q1=−3µC and q2= 5 µC, are placed 10 cm apart in air.
Calculate the magnitude and direction of the electric force that each charge
exerts on the other.
Solution
Step 1: Calculate the magnitude of the electric force each charge exerts on the
other 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 k= 8.99 ×109Nm2/C2is Coulomb’s constant, q1and q2are the magni-
tudes of the charges, and ris the distance between the charges.
Plugging in the values:
F=(8.99 ×109)·3×10−6·5×10−6
0.12
6
F=4.495 ×104
0.01
F= 4495 N
Therefore, the magnitude of the electric force that each charge exerts on the
other is 4495 N.
Step 2: Determine the direction of the electric force on each charge. Since
the charges are opposite in sign, there will be an attractive force between them.
The force on q1due to q2will be directed toward q2, and the force on q2due to
q1will be directed toward q1.
Question 9
Question
Two point charges, q1=−4µC and q2= 8 µC, are located 10 cm apart on
the x-axis. Calculate the magnitude and direction of the electric force on each
charge.
Solution
Step 1: Calculate the distance between the two charges. Given that the charges
are located 10 cm (or 0.1 m) apart on the x-axis, the distance (r) between them
is 0.1 m.
Step 2: Calculate the magnitude of the electric force on q1. The magnitude
of the electric force between two point charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the electrostatic constant (8.99 ×109N·m2/C2).
Substitute the given values into the formula:
F=(8.99 ×109N·m2/C2)·(4 ×10−6C)·(8 ×10−6C)
(0.1m)2
F=287.68 ×10−6
0.01
F= 28.768 N
So, the magnitude of the electric force acting on q1is 28.768 N.
Step 3: Determine the direction of the force on q1. Since q1is negative, the
force on it will be attractive and directed towards q2.
Step 4: Calculate the magnitude of the electric force on q2. The magnitude
of the electric force on q2will be the same as that on q1, i.e., 28.768 N.
Step 5: Determine the direction of the force on q2. Since q2is positive, the
force on it will be repulsive and directed away from q1.
7
Question 10
Question
Three point charges are arranged in an equilateral triangle with sides of length
a. The charges are +q,−2q, and +3q. Calculate the net electric force on the
+qcharge due to the other two charges.
Solution
To find the net electric force on the +qcharge, we need to calculate the in-
dividual forces on it due to the other two charges and then sum these forces
vectorially.
Step 1: Calculate the force on the +qcharge due to the −2qcharge. The
magnitude of the electric force between two point charges q1and q2separated
by a distance ris given by Coulomb’s Law:
F=k·|q1q2|
r2,
where kis the Coulomb constant (8.99 ×109N m2/C2).
The distance between the +qand −2qcharges is a. Therefore, the magnitude
of the force on the +qcharge due to the −2qcharge is:
F+qdue to −2q=k·|+q(−2q)|
a2.
Step 2: Calculate the force on the +qcharge due to the +3qcharge. The
distance between the +qand +3qcharges is also a. The magnitude of the force
on the +qcharge due to the +3qcharge is:
F+qdue to +3q=k·|+q(3q)|
a2.
Step 3: Find the net force on the +qcharge. The net force on the +q
charge is the vector sum of the forces from the −2qand +3qcharges. Since the
forces are acting along the sides of an equilateral triangle, they will have the
same magnitude but will be in opposite directions.
Therefore, the net force is:
Fnet =F+qdue to +3q−F+qdue to −2q.
Substitute the expressions for the forces into the equation above and simplify
to find the net force on the +qcharge.
Question 11
Question
Two point charges are fixed in place along the x-axis. Charge q1=−4.0µC is
located at x= 0 m and charge q2= 7.0µC is located at x= 2.0m. Find the
8
electric force on q2due to q1.
Solution
Step 1: Calculate the distance between the two charges. Given that q2is located
at x= 2.0m and q1is located at x= 0 m, the distance rbetween the two charges
is:
r=x2−x1= 2.0m−0m= 2.0m
Step 2: Calculate the electric force. The magnitude of the electric force F
between two point charges can be calculated using Coulomb’s Law:
F=k· |q1|·|q2|
r2
where
•k≈8.99 ×109N m2/C2is the Coulomb constant,
•|q1|= 4.0×10−6C,
•|q2|= 7.0×10−6C, and
•r= 2.0m.
Plugging in the values, we get:
F=(8.99 ×109N m2/C2)·(4.0×10−6C)·(7.0×10−6C)
(2.0m)2
F=(8.99 ×103)·(4.0) ·(7.0)
4.0
F= 8.99 ×4×7 = 251.32 N
Therefore, the electric force on q2due to q1is 251.32 N directed towards q1.
Question 12
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between these two
charges.
9
Solution
Step 1: Convert the charges to coulombs: The charges in microcoulombs can be
converted to coulombs by multiplying by 10−6.q1=−4.0µC = −4.0×10−6C
=−4.0µCq2= 6.0µC = 6.0×10−6C = 6.0µC
Step 2: Find the distance between the charges in meters: Given that the
charges are placed 10.0 cm apart, convert the distance to meters by dividing by
100. r= 10.0cm = 10.0×10−2m = 0.10 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
The electric force between two point charges q1and q2separated by a distance
ris given by Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
Substitute the values into the formula:
F=(8.99 ×109N m2/C2)·|−4.0×10−6C·6.0×10−6C|
(0.10 m)2
Step 4: Calculate the magnitude of the electric force:
F=(8.99 ×109)·(4.0×6.0) ×10−6
0.01 N
F=8.99 ×109·24
0.01 N
F= 2.16 ×10−2N
Therefore, the magnitude of the electric force between the charges q1=−4.0
µC and q2= 6.0µC placed 10.0 cm apart is 2.16 ×10−2N.
Question 13
Question
Three charges are arranged along the x-axis. Charge q1= 4 nC is at the origin,
charge q2=−6nC is located at x= 5 m, and charge q3= 8 nC is placed at
x= 10 m. Calculate the electric force acting on charge q3due to the other two
charges.
Solution
Step 1: Calculate the electric force on charge q3due to charge q1using Coulomb’s
law:
F13 =k|q1q3|
r2
13
10
Step 2: Calculate the distance r13 between charge q1and charge q3:
r13 = 10 m
Step 3: Substitute the given values into Coulomb’s law:
F13 =(8.99 ×109N m2/C2)(4 ×10−9C)(8 ×10−9C)
(10 m)2
Step 4: Calculate the electric force on charge q3due to charge q1:
F13 =(35.96 ×101)(8 ×10−9)
100
F13 =287.68 ×10−8
100
F13 = 2.8768 ×10−6N
Step 5: Calculate the electric force on charge q3due to charge q2using the
same process:
F23 =k|q2q3|
r2
23
Step 6: Calculate the distance r23 between charge q2and charge q3:
r23 = 5 m
Step 7: Substitute the given values into Coulomb’s law:
F23 =(8.99 ×109)(−6×10−9)(8 ×10−9)
(5)2
Step 8: Calculate the electric force on charge q3due to charge q2:
F23 =−43.92 ×101
25
F23 =−1.7568 ×10−6N
Step 9: Finally, calculate the total electric force on charge q3by summing
the forces F13 and F23:
Ftotal =F13 +F23
Ftotal = 2.8768 ×10−6−1.7568 ×10−6
Ftotal = 1.12 ×10−6N
Therefore, the total electric force on charge q3is 1.12 ×10−6N, directed
towards charge q1.
11
Question 14
Question
Two point charges, q1=−8nC and q2= 4 nC, are placed 6cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs using the conversion factor 1nC =
10−9C.
q1=−8×10−9C
q2= 4 ×10−9C
Step 2: Calculate the distance between the charges in meters.
6cm = 6 ×10−2m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law:
F=k·|q1·q2|
r2where k= 8.99 ×109N m2/C2is the electrostatic constant.
F=(8.99 ×109)·|−8×10−9·4×10−9|
(6 ×10−2)2
=8.99 ×109·32 ×10−18
36 ×10−4
=287.68 ×10−9
36 ×10−4
= 7.99 ×10−5N
Therefore, the magnitude of the electric force between the charges is 7.99 ×
10−5N.
Question 15
Question
Three point charges are arranged on the vertices of a right-angled triangle as
shown:
Q1(+3 C)
Q2(−2C)
Q3(+4 C)
The side lengths are 3 m, 4 m, and 5 m. Calculate the electric force on Q1
due to Q2and Q3.
12
Solution
Step 1: Calculate the electric force on Q1due to Q2. Given: Q1= +3 C,
Q2=−2C, r= 4 m.
The electric force between two point charges is given by Coulomb’s Law:
F=k·|Q1|·|Q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the formula to find the electric force on Q1
due to Q2:
FQ1Q2= 8.99 ×109·3·2
42= 1.3485 ×109N
Step 2: Calculate the electric force on Q1due to Q3. Given: Q1= +3 C,
Q3= +4 C, r= 5 m.
Using Coulomb’s Law again, the electric force on Q1due to Q3is:
FQ1Q3= 8.99 ×109·3·4
52= 2.8792 ×109N
Step 3: Calculate the net electric force on Q1. The electric force on Q1due
to Q2is attractive (since Q2is negative) and the force due to Q3is repulsive
(since Q3is positive).
Therefore, the net force on Q1is:
Fnet =FQ1Q3−FQ1Q2= 2.8792 ×109−1.3485 ×109= 1.5307 ×109N
So, the net electric force on Q1is 1.5307 ×109N.
Question 16
Question
Two point charges, q1=−2µC and q2= 4 µC, are placed 10 cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
To calculate the magnitude of the electric force between the charges, we can use
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 Coulomb’s constant
(8.99 ×109N·m2/C2), - q1and q2are the magnitudes of the point charges
involved, - ris the distance between the charges.
13
Given: - q1=−2µC, - q2= 4 µC, - r= 0.10 m.
Step 1: Calculate the electric force using Coulomb’s Law.
Plugging in the values:
F= (8.99 ×109)·|(−2×10−6)·(4 ×10−6)|
(0.10)2
F= 8.99 ×109·8×10−12
0.01
F= 8.99 ×109·8×10−10
F= 71.92 ×10−1
F= 7.192 N
Therefore, the magnitude of the electric force between the charges is 7.192
N.
Question 17
Question
Two point charges, q1=−3µC and q2= 6µC, are placed 10 cm apart. Calculate
the magnitude and direction of the electric force that q1exerts on q2.
Solution
Step 1: Determine the distance between the charges. Given that the charges
are 10 cm apart, we convert this distance to meters:
d= 10 cm = 0.10 m
Step 2: Calculate the magnitude of the electric force between the charges
using Coulomb’s law:
F=k·
q1·q2
d2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the values, we have:
F= 8.99 ×109N m2/C2·| − 3×10−6C·6×10−6C|
(0.10 m)2
F= 8.99 ×109N m2/C2·1.8×10−5C2/(0.01 m2)
F= 8.99 ×109N·1.8×10−3m/0.01 m2
14
F= 8.99 ×109·1.8×10−2N= 1.6182 ×108N
Step 3: Determine the direction of the force. The force is repulsive since the
charges have opposite signs, therefore the force points away from q1and towards
q2.
So, the magnitude of the electric force that q1exerts on q2is 1.6182 ×108
N, directed away from q1and towards q2.
Question 18
Question
Two point charges, +4.0µC and −2.0µC, are placed 10.0 cm apart in a vacuum.
Calculate the magnitude of the electric force between these two charges.
Solution
To calculate the magnitude of the electric force between the two charges, we
will use Coulomb’s Law, which states that the magnitude of the electric force
between two point charges is given by:
F=k·|q1·q2|
r2
where: - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the two charges, and -
ris the distance between the charges.
Given: - q1= +4.0µC= 4.0×10−6C, - q2=−2.0µC=−2.0×10−6C, and
-r= 10.0cm = 0.10 m.
Step 1: Substitute the given values into Coulomb’s Law formula:
F= (8.99 ×109)·|4.0×10−6· −2.0×10−6|
(0.10)2
F= (8.99 ×109)·8.0×10−12
0.01
Step 2: Calculate the magnitude of the electric force:
F= (8.99 ×109)·8.0×10−10
F= 7.192 ×10−1N
Therefore, the magnitude of the electric force between the two charges is
0.7192 N.
15
Question 19
Question
Three point charges are placed at the vertices of an equilateral triangle as shown
below:
−q
q
−q
If the charges qare positive and the charges −qare negative, calculate the
magnitude and direction of the electric force experienced by the charge at point
B.
Given: Each charge q= 2 µC, the side length of the equilateral triangle is 2
m, and the Coulomb’s constant k= 9 ×109N m2/C2.
Solution
Step 1: Calculate the distance between the charges at point B and point A or C.
Given that the side length of the equilateral triangle is 2 m, the distance between
the charges at points A and B (or C) can be calculated using the Pythagorean
theorem:
d=q22+ (2√3)2=√52 = 2√13 m
Step 2: Calculate the electric force between the charges qand −qat points B
and A (or C). The electric force between two point charges is given by Coulomb’s
Law:
F=k|q1q2|
r2
where kis the Coulomb’s constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
Substitute the given values into the equation to calculate the force:
F= 9 ×109×(2 ×10−6)(2 ×10−6)
(2√13)2
F= 36 ×10−12 ×4×10−12
52
F=144
52 ×10−24
16
F≈2.769 ×10−23 N
The force between the charges at point B and point A (or C) is approximately
2.769 ×10−23 N.
Step 3: Determine the direction of the electric force. Since the charges at
points B and A (or C) have opposite signs (attraction between positive and
negative charges), the electric force experienced by the charge at point B is
directed towards the charge at point A (or C).
Question 20
Question
Two point charges, q1= 4 µC and q2=−6µC, are placed 1 meter apart.
Determine the magnitude and direction of the electric force experienced by
each charge due to the other charge.
Solution
Step 1: Calculate the magnitude of the electric force on q1due to q2. The
magnitude of the electric force between two point charges is given by Coulomb’s
law:
F=k|q1q2|
r2
where kis Coulomb’s constant (8.99 ×109N·m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Plugging in the values, we get:
F1,2= (8.99 ×109)|4×10−6× −6×10−6|
(1)2
F1,2= 4314 N
The force on q1due to q2is attractive and points towards q2.
Step 2: Calculate the magnitude of the electric force on q2due to q1. Since
the forces between charges are equal in magnitude and opposite in direction by
Newton’s third law, the magnitude of the force on q2due to q1is also 4314 N,
but in the opposite direction (towards q1).
Therefore, the magnitude of the electric force experienced by each charge
due to the other charge is 4314 N, with the direction of each force being towards
the other charge.
Question 21
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. The charges are +2 µC,−3µC, and +4 µC. Calculate the mag-
17
nitude and direction of the net electric force on the charge located at the top
vertex.
+2 µC
−3µC
+4 µC
Solution
Step 1: Calculate the electric force between all pairs of charges using Coulomb’s
law: The electric force between two charges q1and q2separated by a distance
ris given by
F=k·|q1q2|
r2,
where k= 8.99 ×109N m2/C2is Coulomb’s constant.
We denote the charge at the top vertex as q1= +2 µC, the charge at the
right vertex as q2=−3µC, and the charge at the left vertex as q3= +4 µC.
The distance between any two charges in an equilateral triangle is the length
of a side, denoted as d. For an equilateral triangle, d= 2r√3where ris the
length from a vertex to the centroid. Given that the side of the triangle is
d= 4 cm, we have r=d
2√3=4
2√3=2√3
3cm. So, r=2√3
3cm.
Now, we calculate the distances between charges:
r12 =r13 =r23 =2√3
3cm.
The electric forces between each pair of charges are:
F12 =k·|+ 2 ×(−3)|
(2√3/3)2=18k√3
4,
F13 =k·|+ 2 ×4|
(2√3/3)2=8k√3
4,
F23 =k·| − 3×4|
(2√3/3)2=12k√3
4.
18
Question 22
Question
Two point charges, q1=−3.00 µC and q2= 1.50 µC, are placed 20.0cm apart.
Calculate the magnitude of the electric force between them.
Solution
To calculate the magnitude of the electric force between the two charges, we
can use Coulomb’s law. Coulomb’s law states that the magnitude of the electric
force between two point charges is given by the equation:
F=k·|q1|·|q2|
r2,
where: - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the two charges, and -
ris the distance between the charges.
Let’s substitute the given values into the equation and solve for F.
Step 1: Calculate the distance between the charges in meters Given:
r= 20.0cm = 0.20 m
Step 2: Calculate the electric force using Coulomb’s law Given:
q1=−3.00 µC=−3.00 ×10−6C Given: q2= 1.50 µC= 1.50 ×10−6C
Plugging the values into Coulomb’s law:
F=k·|q1|·|q2|
r2
F= (8.99 ×109N m2/C2)·3.00 ×10−6C·1.50 ×10−6C
(0.20 m)2
Step 3: Calculate the electric force
F= 8.99 ×109×3.00 ×1.50
0.202N
F= 8.99 ×109×4.50
0.04 N
F= 8.99 ×109×112.5N
F= 1.013 ×1012 N
Therefore, the magnitude of the electric force between the two charges is
1.013 ×1012 N.
Question 23
Question
Two point charges, Q1= 3µC and Q2=−4µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
19
Solution
Step 1: Convert the given charges to standard units.
Given: Q1= 3µC and Q2=−4µC.
To convert from µC to C, we use the conversion factor 1µC = 10−6C.
So, Q1= 3µC = 3 ×10−6Cand Q2=−4µC =−4×10−6C.
Step 2: Determine the distance between the charges.
Given that the charges are placed 10 cm apart, we convert 10 cm to meters
using the conversion factor 1cm = 10−2m.
So, the distance between the charges is 10 cm = 10 ×10−2m= 0.1m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s law.
The magnitude of the electric force between point charges is given by Coulomb’s
law:
F=k·|Q1·Q2|
r2,
where kis the electrostatic constant (k≈8.9875 ×109N·m2/C2), Q1and Q2
are the magnitudes of the charges, and ris the distance between the charges.
Substitute the given values:
F= 8.9875 ×109N·m2/C2×|3×10−6· −4×10−6|
(0.1)2
F= 8.9875 ×109×12 ×10−12
0.01
F= 8.9875 ×109×1.2×10−9
F= 10.785 ×100N
F= 10.785 N
Therefore, the magnitude of the electric force between the charges is 10.785
N.
Question 24
Question
Three point charges are arranged in a line. Charge q1= +5.0µC is at the
origin, charge q2=−3.0µC is located at x= 0.30 m, and charge q3= +2.0µC
is located at x= 0.45 m. Calculate the net electric force on charge q2.
Solution
Step 1: Calculate the electric force on charge q2due to charge q1.
F12 =k· |q1|·|q2|
r2
12
20
Step 2: Calculate the direction of the electric force on charge q2due to
charge q1. The force will be repulsive because the charges have opposite signs,
so the force vector points in the positive x-direction.
Step 3: Calculate the magnitude of the electric force F12.
F12 =8.99 ×109N·m2/C2·5.0×10−6C·3.0×10−6C
(0.30 m)2
F12 = 29965.00 N
Step 4: Calculate the electric force on charge q2due to charge q3.
F23 =k· |q2|·|q3|
r2
23
Step 5: Calculate the direction of the electric force on charge q2due to
charge q3. The force will be attractive because the charges have opposite signs,
so the force vector points in the negative x-direction.
Step 6: Calculate the magnitude of the electric force F23.
F23 =8.99 ×109N·m2/C2·3.0×10−6C·2.0×10−6C
(0.15 m)2
F23 = 35984.17 N
Step 7: Calculate the net electric force on charge q2.
Fnet =F12 −F23
Fnet = 29965.00 N−35984.17 N
Fnet =−6029.17 N
Therefore, the net electric force on charge q2is −6029.17 N in the negative
x-direction.
Question 25
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are separated by a distance
of 10.0m. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−4.0µC=−4.0×10−6C
q2= 8.0µC= 8.0×10−6C
21
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: F= magnitude of the electric force k= Coulomb’s constant (8.99 ×
109N·m2/C2)q1,q2= magnitudes of the charges r= distance between the
charges
Plugging in the values:
F= (8.99 ×109)·| − 4.0×10−6·8.0×10−6|
10.02
Step 3: Calculate the magnitude of the electric force.
F= (8.99 ×109)·32 ×10−12
100
F= (8.99 ×109)·3.2×10−13
F= 2.8776 ×10−3N
Therefore, the magnitude of the electric force between the charges is 2.8776×
10−3N.
22
F= 14.98 ×10−3N
F= 0.01498 N
Therefore, the electric force that charge q1exerts on charge q2is 0.01498 N.
Question 2
Question
Three point charges are arranged at the vertices of an equilateral triangle as
shown below:
Q
−Q Q
If each charge magnitude is Q= 2 mC, what is the magnitude of the electric
force experienced by the charge at the top vertex due to the other two charges?
Solution
Step 1: Begin by labeling the charges and distances in the triangle. Let the
distance between adjacent charges be a.
Q
a
−Q Q
a
Step 2: Calculate the electric force experienced by the charge at the top
vertex due to the charge on the bottom left.
F1=kQ2
a2
Step 3: Calculate the electric force experienced by the charge at the top
vertex due to the charge on the bottom right.
F2=kQ2
(2a)2
Step 4: Determine the net force experienced by the charge at the top vertex.
Since the two forces are acting in opposite directions along the same line, the
net force is the difference between them.
2
Fnet =F1−F2
=kQ2
a2−kQ2
(2a)2
=kQ21
a2−1
4a2
=kQ23
4a2
Step 5: Substitute the known values k= 8.99×109N·m2/C2and Q= 2 mC,
and afor an equilateral triangle (a=a√3/3).
Fnet = (8.99 ×109)·(2 ×10−3)2·
3
4a√3
32
= (8.99 ×109)·4×10−6·
3
a√3
32
= 3√3×104N
Therefore, the magnitude of the electric force experienced by the charge at
the top vertex due to the other two charges is 3√3×104N.
Question 3
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are placed 20.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges from microcoulombs to coulombs.
q1=−4.0µC=−4.0×10−6C
q2= 8.0µC= 8.0×10−6C
Step 2: Determine the distance between the charges in meters.
r= 20.0cm = 20.0×10−2m= 0.20 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k·|q1·q2|
r2
3
where k= 8.99 ×109N·m2/C2is the Coulomb’s constant.
Step 4: Substitute the values into the equation to find the electric force.
F= (8.99 ×109)·|(−4.0×10−6)·(8.0×10−6)|
(0.20)2
= 8.99 ×109·32 ×10−12
0.04
= 8.99 ×109·8×10−11
= 71.92 N
Therefore, the magnitude of the electric force between the charges is 71.92
N.
Question 5
Question
Three point charges are arranged as shown in the diagram below: a +2.0 µC
charge at the origin, a -3.0 µC charge at (4.0 m, 0), and a +4.0 µC charge at
(0, 3.0 m). Calculate the magnitude of the net electric force acting on the +4.0
µC charge due to the other two charges.
Solution
1. Let’s denote the charges at the origin, (4.0 m, 0), and (0, 3.0 m) respectively
as q1= +2.0µC, q2=−3.0µC, and q3= +4.0µC.
2. We will calculate the net electric force on the +4.0µC charge, q3, due to
charges q1and q2using the principle of superposition. The electric force exerted
by each point charge can be calculated using the formula:
Felec =k|q1q2|
r2
3. First, we will calculate the force on q3due to q1:
F1=k|q1q3|
r2
13
4. Given that k= 8.9875×109Nm2
C2and r13 = 3.0m, we can substitute these
values into the formula.
5. Solving for F1:
F1= 8.9875 ×109Nm2
C2·(2.0µC)(4.0µC)
(3.0m)2
6. Next, we will calculate the force on q3due to q2:
F2=k|q2q3|
r2
23
4
7. Given that r23 = 4.0m, we can substitute the known values into the
formula.
8. Solving for F2:
F2= 8.9875 ×109Nm2
C2·(3.0µC)(4.0µC)
(4.0m)2
9. The net electric force on q3would be the vector sum of F1and F2.
10. Calculate the magnitudes and directions of F1and F2, then sum them
to find the net force acting on q3.
Question 7
Question
Three point charges are arranged along the x-axis. Charge q1=−2.0µC is
located at the origin, charge q2= 3.0µC is located at x= 4.0m, and charge
q3= 1.0µC is located at x= 8.0m. Calculate the net force on charge q3due
to charges q1and q2.
Solution
Step 1: Calculate the force F31 on charge q3due to charge q1.
Since the charges are on the x-axis, we can use Coulomb’s Law to calculate
the force:
F31 =k|q1||q3|
r2
31
where k= 8.99 ×109N·m2/C2is the Coulomb’s constant, q1=−2.0µC =
−2.0×10−6Cis the charge at the origin, and q3= 1.0µC = 1.0×10−6Cis the
charge at x= 8.0m.
The distance r31 between charges q3(at x= 8.0m) and q1(at the origin) is:
r31 = 8.0m
Therefore, the force F31 is:
F31 = (8.99 ×109)| − 2.0×10−6×1.0×10−6|
(8.0)2
Calculating the force, we get:
F31 = 4.49 ×10−4N
Step 2: Calculate the force F32 on charge q3due to charge q2.
Using Coulomb’s Law again, the force F32 is:
F32 =k|q2||q3|
r2
32
5
where q2= 3.0µC = 3.0×10−6Cis the charge at x= 4.0m.
The distance r32 between charges q3(at x= 8.0m) and q2(at x= 4.0m) is:
r32 = 8.0−4.0 = 4.0m
Plugging in the values, we get:
F32 = (8.99 ×109)3.0×10−6×1.0×10−6
(4.0)2
Solving for the force F32, we find:
F32 = 6.74 ×10−4N
Step 3: Calculate the net force Fnet on charge q3.
The net force Fnet is the vector sum of forces F31 and F32. Since F31 is to
the left (negative x direction) and F32 is to the right (positive x direction), the
net force is:
Fnet =F32 −F31
Fnet = 6.74 ×10−4−4.49 ×10−4
Fnet = 2.25 ×10−4N
Therefore, the net force on charge q3due to charges q1and q2is 2.25×
Question 8
Question
Two point charges, q1=−3µC and q2= 5 µC, are placed 10 cm apart in air.
Calculate the magnitude and direction of the electric force that each charge
exerts on the other.
Solution
Step 1: Calculate the magnitude of the electric force each charge exerts on the
other 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 k= 8.99 ×109Nm2/C2is Coulomb’s constant, q1and q2are the magni-
tudes of the charges, and ris the distance between the charges.
Plugging in the values:
F=(8.99 ×109)·3×10−6·5×10−6
0.12
6
F=4.495 ×104
0.01
F= 4495 N
Therefore, the magnitude of the electric force that each charge exerts on the
other is 4495 N.
Step 2: Determine the direction of the electric force on each charge. Since
the charges are opposite in sign, there will be an attractive force between them.
The force on q1due to q2will be directed toward q2, and the force on q2due to
q1will be directed toward q1.
Question 9
Question
Two point charges, q1=−4µC and q2= 8 µC, are located 10 cm apart on
the x-axis. Calculate the magnitude and direction of the electric force on each
charge.
Solution
Step 1: Calculate the distance between the two charges. Given that the charges
are located 10 cm (or 0.1 m) apart on the x-axis, the distance (r) between them
is 0.1 m.
Step 2: Calculate the magnitude of the electric force on q1. The magnitude
of the electric force between two point charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the electrostatic constant (8.99 ×109N·m2/C2).
Substitute the given values into the formula:
F=(8.99 ×109N·m2/C2)·(4 ×10−6C)·(8 ×10−6C)
(0.1m)2
F=287.68 ×10−6
0.01
F= 28.768 N
So, the magnitude of the electric force acting on q1is 28.768 N.
Step 3: Determine the direction of the force on q1. Since q1is negative, the
force on it will be attractive and directed towards q2.
Step 4: Calculate the magnitude of the electric force on q2. The magnitude
of the electric force on q2will be the same as that on q1, i.e., 28.768 N.
Step 5: Determine the direction of the force on q2. Since q2is positive, the
force on it will be repulsive and directed away from q1.
7
Question 10
Question
Three point charges are arranged in an equilateral triangle with sides of length
a. The charges are +q,−2q, and +3q. Calculate the net electric force on the
+qcharge due to the other two charges.
Solution
To find the net electric force on the +qcharge, we need to calculate the in-
dividual forces on it due to the other two charges and then sum these forces
vectorially.
Step 1: Calculate the force on the +qcharge due to the −2qcharge. The
magnitude of the electric force between two point charges q1and q2separated
by a distance ris given by Coulomb’s Law:
F=k·|q1q2|
r2,
where kis the Coulomb constant (8.99 ×109N m2/C2).
The distance between the +qand −2qcharges is a. Therefore, the magnitude
of the force on the +qcharge due to the −2qcharge is:
F+qdue to −2q=k·|+q(−2q)|
a2.
Step 2: Calculate the force on the +qcharge due to the +3qcharge. The
distance between the +qand +3qcharges is also a. The magnitude of the force
on the +qcharge due to the +3qcharge is:
F+qdue to +3q=k·|+q(3q)|
a2.
Step 3: Find the net force on the +qcharge. The net force on the +q
charge is the vector sum of the forces from the −2qand +3qcharges. Since the
forces are acting along the sides of an equilateral triangle, they will have the
same magnitude but will be in opposite directions.
Therefore, the net force is:
Fnet =F+qdue to +3q−F+qdue to −2q.
Substitute the expressions for the forces into the equation above and simplify
to find the net force on the +qcharge.
Question 11
Question
Two point charges are fixed in place along the x-axis. Charge q1=−4.0µC is
located at x= 0 m and charge q2= 7.0µC is located at x= 2.0m. Find the
8
electric force on q2due to q1.
Solution
Step 1: Calculate the distance between the two charges. Given that q2is located
at x= 2.0m and q1is located at x= 0 m, the distance rbetween the two charges
is:
r=x2−x1= 2.0m−0m= 2.0m
Step 2: Calculate the electric force. The magnitude of the electric force F
between two point charges can be calculated using Coulomb’s Law:
F=k· |q1|·|q2|
r2
where
•k≈8.99 ×109N m2/C2is the Coulomb constant,
•|q1|= 4.0×10−6C,
•|q2|= 7.0×10−6C, and
•r= 2.0m.
Plugging in the values, we get:
F=(8.99 ×109N m2/C2)·(4.0×10−6C)·(7.0×10−6C)
(2.0m)2
F=(8.99 ×103)·(4.0) ·(7.0)
4.0
F= 8.99 ×4×7 = 251.32 N
Therefore, the electric force on q2due to q1is 251.32 N directed towards q1.
Question 12
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between these two
charges.
9
Solution
Step 1: Convert the charges to coulombs: The charges in microcoulombs can be
converted to coulombs by multiplying by 10−6.q1=−4.0µC = −4.0×10−6C
=−4.0µCq2= 6.0µC = 6.0×10−6C = 6.0µC
Step 2: Find the distance between the charges in meters: Given that the
charges are placed 10.0 cm apart, convert the distance to meters by dividing by
100. r= 10.0cm = 10.0×10−2m = 0.10 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
The electric force between two point charges q1and q2separated by a distance
ris given by Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
Substitute the values into the formula:
F=(8.99 ×109N m2/C2)·|−4.0×10−6C·6.0×10−6C|
(0.10 m)2
Step 4: Calculate the magnitude of the electric force:
F=(8.99 ×109)·(4.0×6.0) ×10−6
0.01 N
F=8.99 ×109·24
0.01 N
F= 2.16 ×10−2N
Therefore, the magnitude of the electric force between the charges q1=−4.0
µC and q2= 6.0µC placed 10.0 cm apart is 2.16 ×10−2N.
Question 13
Question
Three charges are arranged along the x-axis. Charge q1= 4 nC is at the origin,
charge q2=−6nC is located at x= 5 m, and charge q3= 8 nC is placed at
x= 10 m. Calculate the electric force acting on charge q3due to the other two
charges.
Solution
Step 1: Calculate the electric force on charge q3due to charge q1using Coulomb’s
law:
F13 =k|q1q3|
r2
13
10
Step 2: Calculate the distance r13 between charge q1and charge q3:
r13 = 10 m
Step 3: Substitute the given values into Coulomb’s law:
F13 =(8.99 ×109N m2/C2)(4 ×10−9C)(8 ×10−9C)
(10 m)2
Step 4: Calculate the electric force on charge q3due to charge q1:
F13 =(35.96 ×101)(8 ×10−9)
100
F13 =287.68 ×10−8
100
F13 = 2.8768 ×10−6N
Step 5: Calculate the electric force on charge q3due to charge q2using the
same process:
F23 =k|q2q3|
r2
23
Step 6: Calculate the distance r23 between charge q2and charge q3:
r23 = 5 m
Step 7: Substitute the given values into Coulomb’s law:
F23 =(8.99 ×109)(−6×10−9)(8 ×10−9)
(5)2
Step 8: Calculate the electric force on charge q3due to charge q2:
F23 =−43.92 ×101
25
F23 =−1.7568 ×10−6N
Step 9: Finally, calculate the total electric force on charge q3by summing
the forces F13 and F23:
Ftotal =F13 +F23
Ftotal = 2.8768 ×10−6−1.7568 ×10−6
Ftotal = 1.12 ×10−6N
Therefore, the total electric force on charge q3is 1.12 ×10−6N, directed
towards charge q1.
11
Question 14
Question
Two point charges, q1=−8nC and q2= 4 nC, are placed 6cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs using the conversion factor 1nC =
10−9C.
q1=−8×10−9C
q2= 4 ×10−9C
Step 2: Calculate the distance between the charges in meters.
6cm = 6 ×10−2m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law:
F=k·|q1·q2|
r2where k= 8.99 ×109N m2/C2is the electrostatic constant.
F=(8.99 ×109)·|−8×10−9·4×10−9|
(6 ×10−2)2
=8.99 ×109·32 ×10−18
36 ×10−4
=287.68 ×10−9
36 ×10−4
= 7.99 ×10−5N
Therefore, the magnitude of the electric force between the charges is 7.99 ×
10−5N.
Question 15
Question
Three point charges are arranged on the vertices of a right-angled triangle as
shown:
Q1(+3 C)
Q2(−2C)
Q3(+4 C)
The side lengths are 3 m, 4 m, and 5 m. Calculate the electric force on Q1
due to Q2and Q3.
12
Solution
Step 1: Calculate the electric force on Q1due to Q2. Given: Q1= +3 C,
Q2=−2C, r= 4 m.
The electric force between two point charges is given by Coulomb’s Law:
F=k·|Q1|·|Q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the formula to find the electric force on Q1
due to Q2:
FQ1Q2= 8.99 ×109·3·2
42= 1.3485 ×109N
Step 2: Calculate the electric force on Q1due to Q3. Given: Q1= +3 C,
Q3= +4 C, r= 5 m.
Using Coulomb’s Law again, the electric force on Q1due to Q3is:
FQ1Q3= 8.99 ×109·3·4
52= 2.8792 ×109N
Step 3: Calculate the net electric force on Q1. The electric force on Q1due
to Q2is attractive (since Q2is negative) and the force due to Q3is repulsive
(since Q3is positive).
Therefore, the net force on Q1is:
Fnet =FQ1Q3−FQ1Q2= 2.8792 ×109−1.3485 ×109= 1.5307 ×109N
So, the net electric force on Q1is 1.5307 ×109N.
Question 16
Question
Two point charges, q1=−2µC and q2= 4 µC, are placed 10 cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
To calculate the magnitude of the electric force between the charges, we can use
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 Coulomb’s constant
(8.99 ×109N·m2/C2), - q1and q2are the magnitudes of the point charges
involved, - ris the distance between the charges.
13
Given: - q1=−2µC, - q2= 4 µC, - r= 0.10 m.
Step 1: Calculate the electric force using Coulomb’s Law.
Plugging in the values:
F= (8.99 ×109)·|(−2×10−6)·(4 ×10−6)|
(0.10)2
F= 8.99 ×109·8×10−12
0.01
F= 8.99 ×109·8×10−10
F= 71.92 ×10−1
F= 7.192 N
Therefore, the magnitude of the electric force between the charges is 7.192
N.
Question 17
Question
Two point charges, q1=−3µC and q2= 6µC, are placed 10 cm apart. Calculate
the magnitude and direction of the electric force that q1exerts on q2.
Solution
Step 1: Determine the distance between the charges. Given that the charges
are 10 cm apart, we convert this distance to meters:
d= 10 cm = 0.10 m
Step 2: Calculate the magnitude of the electric force between the charges
using Coulomb’s law:
F=k·
q1·q2
d2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the values, we have:
F= 8.99 ×109N m2/C2·| − 3×10−6C·6×10−6C|
(0.10 m)2
F= 8.99 ×109N m2/C2·1.8×10−5C2/(0.01 m2)
F= 8.99 ×109N·1.8×10−3m/0.01 m2
14
F= 8.99 ×109·1.8×10−2N= 1.6182 ×108N
Step 3: Determine the direction of the force. The force is repulsive since the
charges have opposite signs, therefore the force points away from q1and towards
q2.
So, the magnitude of the electric force that q1exerts on q2is 1.6182 ×108
N, directed away from q1and towards q2.
Question 18
Question
Two point charges, +4.0µC and −2.0µC, are placed 10.0 cm apart in a vacuum.
Calculate the magnitude of the electric force between these two charges.
Solution
To calculate the magnitude of the electric force between the two charges, we
will use Coulomb’s Law, which states that the magnitude of the electric force
between two point charges is given by:
F=k·|q1·q2|
r2
where: - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the two charges, and -
ris the distance between the charges.
Given: - q1= +4.0µC= 4.0×10−6C, - q2=−2.0µC=−2.0×10−6C, and
-r= 10.0cm = 0.10 m.
Step 1: Substitute the given values into Coulomb’s Law formula:
F= (8.99 ×109)·|4.0×10−6· −2.0×10−6|
(0.10)2
F= (8.99 ×109)·8.0×10−12
0.01
Step 2: Calculate the magnitude of the electric force:
F= (8.99 ×109)·8.0×10−10
F= 7.192 ×10−1N
Therefore, the magnitude of the electric force between the two charges is
0.7192 N.
15
Question 19
Question
Three point charges are placed at the vertices of an equilateral triangle as shown
below:
−q
q
−q
If the charges qare positive and the charges −qare negative, calculate the
magnitude and direction of the electric force experienced by the charge at point
B.
Given: Each charge q= 2 µC, the side length of the equilateral triangle is 2
m, and the Coulomb’s constant k= 9 ×109N m2/C2.
Solution
Step 1: Calculate the distance between the charges at point B and point A or C.
Given that the side length of the equilateral triangle is 2 m, the distance between
the charges at points A and B (or C) can be calculated using the Pythagorean
theorem:
d=q22+ (2√3)2=√52 = 2√13 m
Step 2: Calculate the electric force between the charges qand −qat points B
and A (or C). The electric force between two point charges is given by Coulomb’s
Law:
F=k|q1q2|
r2
where kis the Coulomb’s constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
Substitute the given values into the equation to calculate the force:
F= 9 ×109×(2 ×10−6)(2 ×10−6)
(2√13)2
F= 36 ×10−12 ×4×10−12
52
F=144
52 ×10−24
16
F≈2.769 ×10−23 N
The force between the charges at point B and point A (or C) is approximately
2.769 ×10−23 N.
Step 3: Determine the direction of the electric force. Since the charges at
points B and A (or C) have opposite signs (attraction between positive and
negative charges), the electric force experienced by the charge at point B is
directed towards the charge at point A (or C).
Question 20
Question
Two point charges, q1= 4 µC and q2=−6µC, are placed 1 meter apart.
Determine the magnitude and direction of the electric force experienced by
each charge due to the other charge.
Solution
Step 1: Calculate the magnitude of the electric force on q1due to q2. The
magnitude of the electric force between two point charges is given by Coulomb’s
law:
F=k|q1q2|
r2
where kis Coulomb’s constant (8.99 ×109N·m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Plugging in the values, we get:
F1,2= (8.99 ×109)|4×10−6× −6×10−6|
(1)2
F1,2= 4314 N
The force on q1due to q2is attractive and points towards q2.
Step 2: Calculate the magnitude of the electric force on q2due to q1. Since
the forces between charges are equal in magnitude and opposite in direction by
Newton’s third law, the magnitude of the force on q2due to q1is also 4314 N,
but in the opposite direction (towards q1).
Therefore, the magnitude of the electric force experienced by each charge
due to the other charge is 4314 N, with the direction of each force being towards
the other charge.
Question 21
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. The charges are +2 µC,−3µC, and +4 µC. Calculate the mag-
17
nitude and direction of the net electric force on the charge located at the top
vertex.
+2 µC
−3µC
+4 µC
Solution
Step 1: Calculate the electric force between all pairs of charges using Coulomb’s
law: The electric force between two charges q1and q2separated by a distance
ris given by
F=k·|q1q2|
r2,
where k= 8.99 ×109N m2/C2is Coulomb’s constant.
We denote the charge at the top vertex as q1= +2 µC, the charge at the
right vertex as q2=−3µC, and the charge at the left vertex as q3= +4 µC.
The distance between any two charges in an equilateral triangle is the length
of a side, denoted as d. For an equilateral triangle, d= 2r√3where ris the
length from a vertex to the centroid. Given that the side of the triangle is
d= 4 cm, we have r=d
2√3=4
2√3=2√3
3cm. So, r=2√3
3cm.
Now, we calculate the distances between charges:
r12 =r13 =r23 =2√3
3cm.
The electric forces between each pair of charges are:
F12 =k·|+ 2 ×(−3)|
(2√3/3)2=18k√3
4,
F13 =k·|+ 2 ×4|
(2√3/3)2=8k√3
4,
F23 =k·| − 3×4|
(2√3/3)2=12k√3
4.
18
Question 22
Question
Two point charges, q1=−3.00 µC and q2= 1.50 µC, are placed 20.0cm apart.
Calculate the magnitude of the electric force between them.
Solution
To calculate the magnitude of the electric force between the two charges, we
can use Coulomb’s law. Coulomb’s law states that the magnitude of the electric
force between two point charges is given by the equation:
F=k·|q1|·|q2|
r2,
where: - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the two charges, and -
ris the distance between the charges.
Let’s substitute the given values into the equation and solve for F.
Step 1: Calculate the distance between the charges in meters Given:
r= 20.0cm = 0.20 m
Step 2: Calculate the electric force using Coulomb’s law Given:
q1=−3.00 µC=−3.00 ×10−6C Given: q2= 1.50 µC= 1.50 ×10−6C
Plugging the values into Coulomb’s law:
F=k·|q1|·|q2|
r2
F= (8.99 ×109N m2/C2)·3.00 ×10−6C·1.50 ×10−6C
(0.20 m)2
Step 3: Calculate the electric force
F= 8.99 ×109×3.00 ×1.50
0.202N
F= 8.99 ×109×4.50
0.04 N
F= 8.99 ×109×112.5N
F= 1.013 ×1012 N
Therefore, the magnitude of the electric force between the two charges is
1.013 ×1012 N.
Question 23
Question
Two point charges, Q1= 3µC and Q2=−4µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
19
Solution
Step 1: Convert the given charges to standard units.
Given: Q1= 3µC and Q2=−4µC.
To convert from µC to C, we use the conversion factor 1µC = 10−6C.
So, Q1= 3µC = 3 ×10−6Cand Q2=−4µC =−4×10−6C.
Step 2: Determine the distance between the charges.
Given that the charges are placed 10 cm apart, we convert 10 cm to meters
using the conversion factor 1cm = 10−2m.
So, the distance between the charges is 10 cm = 10 ×10−2m= 0.1m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s law.
The magnitude of the electric force between point charges is given by Coulomb’s
law:
F=k·|Q1·Q2|
r2,
where kis the electrostatic constant (k≈8.9875 ×109N·m2/C2), Q1and Q2
are the magnitudes of the charges, and ris the distance between the charges.
Substitute the given values:
F= 8.9875 ×109N·m2/C2×|3×10−6· −4×10−6|
(0.1)2
F= 8.9875 ×109×12 ×10−12
0.01
F= 8.9875 ×109×1.2×10−9
F= 10.785 ×100N
F= 10.785 N
Therefore, the magnitude of the electric force between the charges is 10.785
N.
Question 24
Question
Three point charges are arranged in a line. Charge q1= +5.0µC is at the
origin, charge q2=−3.0µC is located at x= 0.30 m, and charge q3= +2.0µC
is located at x= 0.45 m. Calculate the net electric force on charge q2.
Solution
Step 1: Calculate the electric force on charge q2due to charge q1.
F12 =k· |q1|·|q2|
r2
12
20
Step 2: Calculate the direction of the electric force on charge q2due to
charge q1. The force will be repulsive because the charges have opposite signs,
so the force vector points in the positive x-direction.
Step 3: Calculate the magnitude of the electric force F12.
F12 =8.99 ×109N·m2/C2·5.0×10−6C·3.0×10−6C
(0.30 m)2
F12 = 29965.00 N
Step 4: Calculate the electric force on charge q2due to charge q3.
F23 =k· |q2|·|q3|
r2
23
Step 5: Calculate the direction of the electric force on charge q2due to
charge q3. The force will be attractive because the charges have opposite signs,
so the force vector points in the negative x-direction.
Step 6: Calculate the magnitude of the electric force F23.
F23 =8.99 ×109N·m2/C2·3.0×10−6C·2.0×10−6C
(0.15 m)2
F23 = 35984.17 N
Step 7: Calculate the net electric force on charge q2.
Fnet =F12 −F23
Fnet = 29965.00 N−35984.17 N
Fnet =−6029.17 N
Therefore, the net electric force on charge q2is −6029.17 N in the negative
x-direction.
Question 25
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are separated by a distance
of 10.0m. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−4.0µC=−4.0×10−6C
q2= 8.0µC= 8.0×10−6C
21
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: F= magnitude of the electric force k= Coulomb’s constant (8.99 ×
109N·m2/C2)q1,q2= magnitudes of the charges r= distance between the
charges
Plugging in the values:
F= (8.99 ×109)·| − 4.0×10−6·8.0×10−6|
10.02
Step 3: Calculate the magnitude of the electric force.
F= (8.99 ×109)·32 ×10−12
100
F= (8.99 ×109)·3.2×10−13
F= 2.8776 ×10−3N
Therefore, the magnitude of the electric force between the charges is 2.8776×
10−3N.
22
F= 14.98 ×10−3N
F= 0.01498 N
Therefore, the electric force that charge q1exerts on charge q2is 0.01498 N.
Question 2
Question
Three point charges are arranged at the vertices of an equilateral triangle as
shown below:
Q
−Q Q
If each charge magnitude is Q= 2 mC, what is the magnitude of the electric
force experienced by the charge at the top vertex due to the other two charges?
Solution
Step 1: Begin by labeling the charges and distances in the triangle. Let the
distance between adjacent charges be a.
Q
a
−Q Q
a
Step 2: Calculate the electric force experienced by the charge at the top
vertex due to the charge on the bottom left.
F1=kQ2
a2
Step 3: Calculate the electric force experienced by the charge at the top
vertex due to the charge on the bottom right.
F2=kQ2
(2a)2
Step 4: Determine the net force experienced by the charge at the top vertex.
Since the two forces are acting in opposite directions along the same line, the
net force is the difference between them.
2
Fnet =F1−F2
=kQ2
a2−kQ2
(2a)2
=kQ21
a2−1
4a2
=kQ23
4a2
Step 5: Substitute the known values k= 8.99×109N·m2/C2and Q= 2 mC,
and afor an equilateral triangle (a=a√3/3).
Fnet = (8.99 ×109)·(2 ×10−3)2·
3
4a√3
32
= (8.99 ×109)·4×10−6·
3
a√3
32
= 3√3×104N
Therefore, the magnitude of the electric force experienced by the charge at
the top vertex due to the other two charges is 3√3×104N.
Question 3
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are placed 20.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges from microcoulombs to coulombs.
q1=−4.0µC=−4.0×10−6C
q2= 8.0µC= 8.0×10−6C
Step 2: Determine the distance between the charges in meters.
r= 20.0cm = 20.0×10−2m= 0.20 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k·|q1·q2|
r2
3
where k= 8.99 ×109N·m2/C2is the Coulomb’s constant.
Step 4: Substitute the values into the equation to find the electric force.
F= (8.99 ×109)·|(−4.0×10−6)·(8.0×10−6)|
(0.20)2
= 8.99 ×109·32 ×10−12
0.04
= 8.99 ×109·8×10−11
= 71.92 N
Therefore, the magnitude of the electric force between the charges is 71.92
N.
Question 5
Question
Three point charges are arranged as shown in the diagram below: a +2.0 µC
charge at the origin, a -3.0 µC charge at (4.0 m, 0), and a +4.0 µC charge at
(0, 3.0 m). Calculate the magnitude of the net electric force acting on the +4.0
µC charge due to the other two charges.
Solution
1. Let’s denote the charges at the origin, (4.0 m, 0), and (0, 3.0 m) respectively
as q1= +2.0µC, q2=−3.0µC, and q3= +4.0µC.
2. We will calculate the net electric force on the +4.0µC charge, q3, due to
charges q1and q2using the principle of superposition. The electric force exerted
by each point charge can be calculated using the formula:
Felec =k|q1q2|
r2
3. First, we will calculate the force on q3due to q1:
F1=k|q1q3|
r2
13
4. Given that k= 8.9875×109Nm2
C2and r13 = 3.0m, we can substitute these
values into the formula.
5. Solving for F1:
F1= 8.9875 ×109Nm2
C2·(2.0µC)(4.0µC)
(3.0m)2
6. Next, we will calculate the force on q3due to q2:
F2=k|q2q3|
r2
23
4
7. Given that r23 = 4.0m, we can substitute the known values into the
formula.
8. Solving for F2:
F2= 8.9875 ×109Nm2
C2·(3.0µC)(4.0µC)
(4.0m)2
9. The net electric force on q3would be the vector sum of F1and F2.
10. Calculate the magnitudes and directions of F1and F2, then sum them
to find the net force acting on q3.
Question 7
Question
Three point charges are arranged along the x-axis. Charge q1=−2.0µC is
located at the origin, charge q2= 3.0µC is located at x= 4.0m, and charge
q3= 1.0µC is located at x= 8.0m. Calculate the net force on charge q3due
to charges q1and q2.
Solution
Step 1: Calculate the force F31 on charge q3due to charge q1.
Since the charges are on the x-axis, we can use Coulomb’s Law to calculate
the force:
F31 =k|q1||q3|
r2
31
where k= 8.99 ×109N·m2/C2is the Coulomb’s constant, q1=−2.0µC =
−2.0×10−6Cis the charge at the origin, and q3= 1.0µC = 1.0×10−6Cis the
charge at x= 8.0m.
The distance r31 between charges q3(at x= 8.0m) and q1(at the origin) is:
r31 = 8.0m
Therefore, the force F31 is:
F31 = (8.99 ×109)| − 2.0×10−6×1.0×10−6|
(8.0)2
Calculating the force, we get:
F31 = 4.49 ×10−4N
Step 2: Calculate the force F32 on charge q3due to charge q2.
Using Coulomb’s Law again, the force F32 is:
F32 =k|q2||q3|
r2
32
5
where q2= 3.0µC = 3.0×10−6Cis the charge at x= 4.0m.
The distance r32 between charges q3(at x= 8.0m) and q2(at x= 4.0m) is:
r32 = 8.0−4.0 = 4.0m
Plugging in the values, we get:
F32 = (8.99 ×109)3.0×10−6×1.0×10−6
(4.0)2
Solving for the force F32, we find:
F32 = 6.74 ×10−4N
Step 3: Calculate the net force Fnet on charge q3.
The net force Fnet is the vector sum of forces F31 and F32. Since F31 is to
the left (negative x direction) and F32 is to the right (positive x direction), the
net force is:
Fnet =F32 −F31
Fnet = 6.74 ×10−4−4.49 ×10−4
Fnet = 2.25 ×10−4N
Therefore, the net force on charge q3due to charges q1and q2is 2.25×
Question 8
Question
Two point charges, q1=−3µC and q2= 5 µC, are placed 10 cm apart in air.
Calculate the magnitude and direction of the electric force that each charge
exerts on the other.
Solution
Step 1: Calculate the magnitude of the electric force each charge exerts on the
other 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 k= 8.99 ×109Nm2/C2is Coulomb’s constant, q1and q2are the magni-
tudes of the charges, and ris the distance between the charges.
Plugging in the values:
F=(8.99 ×109)·3×10−6·5×10−6
0.12
6
F=4.495 ×104
0.01
F= 4495 N
Therefore, the magnitude of the electric force that each charge exerts on the
other is 4495 N.
Step 2: Determine the direction of the electric force on each charge. Since
the charges are opposite in sign, there will be an attractive force between them.
The force on q1due to q2will be directed toward q2, and the force on q2due to
q1will be directed toward q1.
Question 9
Question
Two point charges, q1=−4µC and q2= 8 µC, are located 10 cm apart on
the x-axis. Calculate the magnitude and direction of the electric force on each
charge.
Solution
Step 1: Calculate the distance between the two charges. Given that the charges
are located 10 cm (or 0.1 m) apart on the x-axis, the distance (r) between them
is 0.1 m.
Step 2: Calculate the magnitude of the electric force on q1. The magnitude
of the electric force between two point charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the electrostatic constant (8.99 ×109N·m2/C2).
Substitute the given values into the formula:
F=(8.99 ×109N·m2/C2)·(4 ×10−6C)·(8 ×10−6C)
(0.1m)2
F=287.68 ×10−6
0.01
F= 28.768 N
So, the magnitude of the electric force acting on q1is 28.768 N.
Step 3: Determine the direction of the force on q1. Since q1is negative, the
force on it will be attractive and directed towards q2.
Step 4: Calculate the magnitude of the electric force on q2. The magnitude
of the electric force on q2will be the same as that on q1, i.e., 28.768 N.
Step 5: Determine the direction of the force on q2. Since q2is positive, the
force on it will be repulsive and directed away from q1.
7
Question 10
Question
Three point charges are arranged in an equilateral triangle with sides of length
a. The charges are +q,−2q, and +3q. Calculate the net electric force on the
+qcharge due to the other two charges.
Solution
To find the net electric force on the +qcharge, we need to calculate the in-
dividual forces on it due to the other two charges and then sum these forces
vectorially.
Step 1: Calculate the force on the +qcharge due to the −2qcharge. The
magnitude of the electric force between two point charges q1and q2separated
by a distance ris given by Coulomb’s Law:
F=k·|q1q2|
r2,
where kis the Coulomb constant (8.99 ×109N m2/C2).
The distance between the +qand −2qcharges is a. Therefore, the magnitude
of the force on the +qcharge due to the −2qcharge is:
F+qdue to −2q=k·|+q(−2q)|
a2.
Step 2: Calculate the force on the +qcharge due to the +3qcharge. The
distance between the +qand +3qcharges is also a. The magnitude of the force
on the +qcharge due to the +3qcharge is:
F+qdue to +3q=k·|+q(3q)|
a2.
Step 3: Find the net force on the +qcharge. The net force on the +q
charge is the vector sum of the forces from the −2qand +3qcharges. Since the
forces are acting along the sides of an equilateral triangle, they will have the
same magnitude but will be in opposite directions.
Therefore, the net force is:
Fnet =F+qdue to +3q−F+qdue to −2q.
Substitute the expressions for the forces into the equation above and simplify
to find the net force on the +qcharge.
Question 11
Question
Two point charges are fixed in place along the x-axis. Charge q1=−4.0µC is
located at x= 0 m and charge q2= 7.0µC is located at x= 2.0m. Find the
8
electric force on q2due to q1.
Solution
Step 1: Calculate the distance between the two charges. Given that q2is located
at x= 2.0m and q1is located at x= 0 m, the distance rbetween the two charges
is:
r=x2−x1= 2.0m−0m= 2.0m
Step 2: Calculate the electric force. The magnitude of the electric force F
between two point charges can be calculated using Coulomb’s Law:
F=k· |q1|·|q2|
r2
where
•k≈8.99 ×109N m2/C2is the Coulomb constant,
•|q1|= 4.0×10−6C,
•|q2|= 7.0×10−6C, and
•r= 2.0m.
Plugging in the values, we get:
F=(8.99 ×109N m2/C2)·(4.0×10−6C)·(7.0×10−6C)
(2.0m)2
F=(8.99 ×103)·(4.0) ·(7.0)
4.0
F= 8.99 ×4×7 = 251.32 N
Therefore, the electric force on q2due to q1is 251.32 N directed towards q1.
Question 12
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between these two
charges.
9
Solution
Step 1: Convert the charges to coulombs: The charges in microcoulombs can be
converted to coulombs by multiplying by 10−6.q1=−4.0µC = −4.0×10−6C
=−4.0µCq2= 6.0µC = 6.0×10−6C = 6.0µC
Step 2: Find the distance between the charges in meters: Given that the
charges are placed 10.0 cm apart, convert the distance to meters by dividing by
100. r= 10.0cm = 10.0×10−2m = 0.10 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
The electric force between two point charges q1and q2separated by a distance
ris given by Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
Substitute the values into the formula:
F=(8.99 ×109N m2/C2)·|−4.0×10−6C·6.0×10−6C|
(0.10 m)2
Step 4: Calculate the magnitude of the electric force:
F=(8.99 ×109)·(4.0×6.0) ×10−6
0.01 N
F=8.99 ×109·24
0.01 N
F= 2.16 ×10−2N
Therefore, the magnitude of the electric force between the charges q1=−4.0
µC and q2= 6.0µC placed 10.0 cm apart is 2.16 ×10−2N.
Question 13
Question
Three charges are arranged along the x-axis. Charge q1= 4 nC is at the origin,
charge q2=−6nC is located at x= 5 m, and charge q3= 8 nC is placed at
x= 10 m. Calculate the electric force acting on charge q3due to the other two
charges.
Solution
Step 1: Calculate the electric force on charge q3due to charge q1using Coulomb’s
law:
F13 =k|q1q3|
r2
13
10
Step 2: Calculate the distance r13 between charge q1and charge q3:
r13 = 10 m
Step 3: Substitute the given values into Coulomb’s law:
F13 =(8.99 ×109N m2/C2)(4 ×10−9C)(8 ×10−9C)
(10 m)2
Step 4: Calculate the electric force on charge q3due to charge q1:
F13 =(35.96 ×101)(8 ×10−9)
100
F13 =287.68 ×10−8
100
F13 = 2.8768 ×10−6N
Step 5: Calculate the electric force on charge q3due to charge q2using the
same process:
F23 =k|q2q3|
r2
23
Step 6: Calculate the distance r23 between charge q2and charge q3:
r23 = 5 m
Step 7: Substitute the given values into Coulomb’s law:
F23 =(8.99 ×109)(−6×10−9)(8 ×10−9)
(5)2
Step 8: Calculate the electric force on charge q3due to charge q2:
F23 =−43.92 ×101
25
F23 =−1.7568 ×10−6N
Step 9: Finally, calculate the total electric force on charge q3by summing
the forces F13 and F23:
Ftotal =F13 +F23
Ftotal = 2.8768 ×10−6−1.7568 ×10−6
Ftotal = 1.12 ×10−6N
Therefore, the total electric force on charge q3is 1.12 ×10−6N, directed
towards charge q1.
11
Question 14
Question
Two point charges, q1=−8nC and q2= 4 nC, are placed 6cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs using the conversion factor 1nC =
10−9C.
q1=−8×10−9C
q2= 4 ×10−9C
Step 2: Calculate the distance between the charges in meters.
6cm = 6 ×10−2m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law:
F=k·|q1·q2|
r2where k= 8.99 ×109N m2/C2is the electrostatic constant.
F=(8.99 ×109)·|−8×10−9·4×10−9|
(6 ×10−2)2
=8.99 ×109·32 ×10−18
36 ×10−4
=287.68 ×10−9
36 ×10−4
= 7.99 ×10−5N
Therefore, the magnitude of the electric force between the charges is 7.99 ×
10−5N.
Question 15
Question
Three point charges are arranged on the vertices of a right-angled triangle as
shown:
Q1(+3 C)
Q2(−2C)
Q3(+4 C)
The side lengths are 3 m, 4 m, and 5 m. Calculate the electric force on Q1
due to Q2and Q3.
12
Solution
Step 1: Calculate the electric force on Q1due to Q2. Given: Q1= +3 C,
Q2=−2C, r= 4 m.
The electric force between two point charges is given by Coulomb’s Law:
F=k·|Q1|·|Q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the formula to find the electric force on Q1
due to Q2:
FQ1Q2= 8.99 ×109·3·2
42= 1.3485 ×109N
Step 2: Calculate the electric force on Q1due to Q3. Given: Q1= +3 C,
Q3= +4 C, r= 5 m.
Using Coulomb’s Law again, the electric force on Q1due to Q3is:
FQ1Q3= 8.99 ×109·3·4
52= 2.8792 ×109N
Step 3: Calculate the net electric force on Q1. The electric force on Q1due
to Q2is attractive (since Q2is negative) and the force due to Q3is repulsive
(since Q3is positive).
Therefore, the net force on Q1is:
Fnet =FQ1Q3−FQ1Q2= 2.8792 ×109−1.3485 ×109= 1.5307 ×109N
So, the net electric force on Q1is 1.5307 ×109N.
Question 16
Question
Two point charges, q1=−2µC and q2= 4 µC, are placed 10 cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
To calculate the magnitude of the electric force between the charges, we can use
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 Coulomb’s constant
(8.99 ×109N·m2/C2), - q1and q2are the magnitudes of the point charges
involved, - ris the distance between the charges.
13
Given: - q1=−2µC, - q2= 4 µC, - r= 0.10 m.
Step 1: Calculate the electric force using Coulomb’s Law.
Plugging in the values:
F= (8.99 ×109)·|(−2×10−6)·(4 ×10−6)|
(0.10)2
F= 8.99 ×109·8×10−12
0.01
F= 8.99 ×109·8×10−10
F= 71.92 ×10−1
F= 7.192 N
Therefore, the magnitude of the electric force between the charges is 7.192
N.
Question 17
Question
Two point charges, q1=−3µC and q2= 6µC, are placed 10 cm apart. Calculate
the magnitude and direction of the electric force that q1exerts on q2.
Solution
Step 1: Determine the distance between the charges. Given that the charges
are 10 cm apart, we convert this distance to meters:
d= 10 cm = 0.10 m
Step 2: Calculate the magnitude of the electric force between the charges
using Coulomb’s law:
F=k·
q1·q2
d2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the values, we have:
F= 8.99 ×109N m2/C2·| − 3×10−6C·6×10−6C|
(0.10 m)2
F= 8.99 ×109N m2/C2·1.8×10−5C2/(0.01 m2)
F= 8.99 ×109N·1.8×10−3m/0.01 m2
14
F= 8.99 ×109·1.8×10−2N= 1.6182 ×108N
Step 3: Determine the direction of the force. The force is repulsive since the
charges have opposite signs, therefore the force points away from q1and towards
q2.
So, the magnitude of the electric force that q1exerts on q2is 1.6182 ×108
N, directed away from q1and towards q2.
Question 18
Question
Two point charges, +4.0µC and −2.0µC, are placed 10.0 cm apart in a vacuum.
Calculate the magnitude of the electric force between these two charges.
Solution
To calculate the magnitude of the electric force between the two charges, we
will use Coulomb’s Law, which states that the magnitude of the electric force
between two point charges is given by:
F=k·|q1·q2|
r2
where: - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the two charges, and -
ris the distance between the charges.
Given: - q1= +4.0µC= 4.0×10−6C, - q2=−2.0µC=−2.0×10−6C, and
-r= 10.0cm = 0.10 m.
Step 1: Substitute the given values into Coulomb’s Law formula:
F= (8.99 ×109)·|4.0×10−6· −2.0×10−6|
(0.10)2
F= (8.99 ×109)·8.0×10−12
0.01
Step 2: Calculate the magnitude of the electric force:
F= (8.99 ×109)·8.0×10−10
F= 7.192 ×10−1N
Therefore, the magnitude of the electric force between the two charges is
0.7192 N.
15
Question 19
Question
Three point charges are placed at the vertices of an equilateral triangle as shown
below:
−q
q
−q
If the charges qare positive and the charges −qare negative, calculate the
magnitude and direction of the electric force experienced by the charge at point
B.
Given: Each charge q= 2 µC, the side length of the equilateral triangle is 2
m, and the Coulomb’s constant k= 9 ×109N m2/C2.
Solution
Step 1: Calculate the distance between the charges at point B and point A or C.
Given that the side length of the equilateral triangle is 2 m, the distance between
the charges at points A and B (or C) can be calculated using the Pythagorean
theorem:
d=q22+ (2√3)2=√52 = 2√13 m
Step 2: Calculate the electric force between the charges qand −qat points B
and A (or C). The electric force between two point charges is given by Coulomb’s
Law:
F=k|q1q2|
r2
where kis the Coulomb’s constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
Substitute the given values into the equation to calculate the force:
F= 9 ×109×(2 ×10−6)(2 ×10−6)
(2√13)2
F= 36 ×10−12 ×4×10−12
52
F=144
52 ×10−24
16
F≈2.769 ×10−23 N
The force between the charges at point B and point A (or C) is approximately
2.769 ×10−23 N.
Step 3: Determine the direction of the electric force. Since the charges at
points B and A (or C) have opposite signs (attraction between positive and
negative charges), the electric force experienced by the charge at point B is
directed towards the charge at point A (or C).
Question 20
Question
Two point charges, q1= 4 µC and q2=−6µC, are placed 1 meter apart.
Determine the magnitude and direction of the electric force experienced by
each charge due to the other charge.
Solution
Step 1: Calculate the magnitude of the electric force on q1due to q2. The
magnitude of the electric force between two point charges is given by Coulomb’s
law:
F=k|q1q2|
r2
where kis Coulomb’s constant (8.99 ×109N·m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Plugging in the values, we get:
F1,2= (8.99 ×109)|4×10−6× −6×10−6|
(1)2
F1,2= 4314 N
The force on q1due to q2is attractive and points towards q2.
Step 2: Calculate the magnitude of the electric force on q2due to q1. Since
the forces between charges are equal in magnitude and opposite in direction by
Newton’s third law, the magnitude of the force on q2due to q1is also 4314 N,
but in the opposite direction (towards q1).
Therefore, the magnitude of the electric force experienced by each charge
due to the other charge is 4314 N, with the direction of each force being towards
the other charge.
Question 21
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. The charges are +2 µC,−3µC, and +4 µC. Calculate the mag-
17
nitude and direction of the net electric force on the charge located at the top
vertex.
+2 µC
−3µC
+4 µC
Solution
Step 1: Calculate the electric force between all pairs of charges using Coulomb’s
law: The electric force between two charges q1and q2separated by a distance
ris given by
F=k·|q1q2|
r2,
where k= 8.99 ×109N m2/C2is Coulomb’s constant.
We denote the charge at the top vertex as q1= +2 µC, the charge at the
right vertex as q2=−3µC, and the charge at the left vertex as q3= +4 µC.
The distance between any two charges in an equilateral triangle is the length
of a side, denoted as d. For an equilateral triangle, d= 2r√3where ris the
length from a vertex to the centroid. Given that the side of the triangle is
d= 4 cm, we have r=d
2√3=4
2√3=2√3
3cm. So, r=2√3
3cm.
Now, we calculate the distances between charges:
r12 =r13 =r23 =2√3
3cm.
The electric forces between each pair of charges are:
F12 =k·|+ 2 ×(−3)|
(2√3/3)2=18k√3
4,
F13 =k·|+ 2 ×4|
(2√3/3)2=8k√3
4,
F23 =k·| − 3×4|
(2√3/3)2=12k√3
4.
18
Question 22
Question
Two point charges, q1=−3.00 µC and q2= 1.50 µC, are placed 20.0cm apart.
Calculate the magnitude of the electric force between them.
Solution
To calculate the magnitude of the electric force between the two charges, we
can use Coulomb’s law. Coulomb’s law states that the magnitude of the electric
force between two point charges is given by the equation:
F=k·|q1|·|q2|
r2,
where: - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the two charges, and -
ris the distance between the charges.
Let’s substitute the given values into the equation and solve for F.
Step 1: Calculate the distance between the charges in meters Given:
r= 20.0cm = 0.20 m
Step 2: Calculate the electric force using Coulomb’s law Given:
q1=−3.00 µC=−3.00 ×10−6C Given: q2= 1.50 µC= 1.50 ×10−6C
Plugging the values into Coulomb’s law:
F=k·|q1|·|q2|
r2
F= (8.99 ×109N m2/C2)·3.00 ×10−6C·1.50 ×10−6C
(0.20 m)2
Step 3: Calculate the electric force
F= 8.99 ×109×3.00 ×1.50
0.202N
F= 8.99 ×109×4.50
0.04 N
F= 8.99 ×109×112.5N
F= 1.013 ×1012 N
Therefore, the magnitude of the electric force between the two charges is
1.013 ×1012 N.
Question 23
Question
Two point charges, Q1= 3µC and Q2=−4µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
19
Solution
Step 1: Convert the given charges to standard units.
Given: Q1= 3µC and Q2=−4µC.
To convert from µC to C, we use the conversion factor 1µC = 10−6C.
So, Q1= 3µC = 3 ×10−6Cand Q2=−4µC =−4×10−6C.
Step 2: Determine the distance between the charges.
Given that the charges are placed 10 cm apart, we convert 10 cm to meters
using the conversion factor 1cm = 10−2m.
So, the distance between the charges is 10 cm = 10 ×10−2m= 0.1m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s law.
The magnitude of the electric force between point charges is given by Coulomb’s
law:
F=k·|Q1·Q2|
r2,
where kis the electrostatic constant (k≈8.9875 ×109N·m2/C2), Q1and Q2
are the magnitudes of the charges, and ris the distance between the charges.
Substitute the given values:
F= 8.9875 ×109N·m2/C2×|3×10−6· −4×10−6|
(0.1)2
F= 8.9875 ×109×12 ×10−12
0.01
F= 8.9875 ×109×1.2×10−9
F= 10.785 ×100N
F= 10.785 N
Therefore, the magnitude of the electric force between the charges is 10.785
N.
Question 24
Question
Three point charges are arranged in a line. Charge q1= +5.0µC is at the
origin, charge q2=−3.0µC is located at x= 0.30 m, and charge q3= +2.0µC
is located at x= 0.45 m. Calculate the net electric force on charge q2.
Solution
Step 1: Calculate the electric force on charge q2due to charge q1.
F12 =k· |q1|·|q2|
r2
12
20
Step 2: Calculate the direction of the electric force on charge q2due to
charge q1. The force will be repulsive because the charges have opposite signs,
so the force vector points in the positive x-direction.
Step 3: Calculate the magnitude of the electric force F12.
F12 =8.99 ×109N·m2/C2·5.0×10−6C·3.0×10−6C
(0.30 m)2
F12 = 29965.00 N
Step 4: Calculate the electric force on charge q2due to charge q3.
F23 =k· |q2|·|q3|
r2
23
Step 5: Calculate the direction of the electric force on charge q2due to
charge q3. The force will be attractive because the charges have opposite signs,
so the force vector points in the negative x-direction.
Step 6: Calculate the magnitude of the electric force F23.
F23 =8.99 ×109N·m2/C2·3.0×10−6C·2.0×10−6C
(0.15 m)2
F23 = 35984.17 N
Step 7: Calculate the net electric force on charge q2.
Fnet =F12 −F23
Fnet = 29965.00 N−35984.17 N
Fnet =−6029.17 N
Therefore, the net electric force on charge q2is −6029.17 N in the negative
x-direction.
Question 25
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are separated by a distance
of 10.0m. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−4.0µC=−4.0×10−6C
q2= 8.0µC= 8.0×10−6C
21
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: F= magnitude of the electric force k= Coulomb’s constant (8.99 ×
109N·m2/C2)q1,q2= magnitudes of the charges r= distance between the
charges
Plugging in the values:
F= (8.99 ×109)·| − 4.0×10−6·8.0×10−6|
10.02
Step 3: Calculate the magnitude of the electric force.
F= (8.99 ×109)·32 ×10−12
100
F= (8.99 ×109)·3.2×10−13
F= 2.8776 ×10−3N
Therefore, the magnitude of the electric force between the charges is 2.8776×
10−3N.
22
F= 14.98 ×10−3N
F= 0.01498 N
Therefore, the electric force that charge q1exerts on charge q2is 0.01498 N.
Question 2
Question
Three point charges are arranged at the vertices of an equilateral triangle as
shown below:
Q
−Q Q
If each charge magnitude is Q= 2 mC, what is the magnitude of the electric
force experienced by the charge at the top vertex due to the other two charges?
Solution
Step 1: Begin by labeling the charges and distances in the triangle. Let the
distance between adjacent charges be a.
Q
a
−Q Q
a
Step 2: Calculate the electric force experienced by the charge at the top
vertex due to the charge on the bottom left.
F1=kQ2
a2
Step 3: Calculate the electric force experienced by the charge at the top
vertex due to the charge on the bottom right.
F2=kQ2
(2a)2
Step 4: Determine the net force experienced by the charge at the top vertex.
Since the two forces are acting in opposite directions along the same line, the
net force is the difference between them.
2
Fnet =F1−F2
=kQ2
a2−kQ2
(2a)2
=kQ21
a2−1
4a2
=kQ23
4a2
Step 5: Substitute the known values k= 8.99×109N·m2/C2and Q= 2 mC,
and afor an equilateral triangle (a=a√3/3).
Fnet = (8.99 ×109)·(2 ×10−3)2·
3
4a√3
32
= (8.99 ×109)·4×10−6·
3
a√3
32
= 3√3×104N
Therefore, the magnitude of the electric force experienced by the charge at
the top vertex due to the other two charges is 3√3×104N.
Question 3
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are placed 20.0 cm apart in
a vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges from microcoulombs to coulombs.
q1=−4.0µC=−4.0×10−6C
q2= 8.0µC= 8.0×10−6C
Step 2: Determine the distance between the charges in meters.
r= 20.0cm = 20.0×10−2m= 0.20 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k·|q1·q2|
r2
3
where k= 8.99 ×109N·m2/C2is the Coulomb’s constant.
Step 4: Substitute the values into the equation to find the electric force.
F= (8.99 ×109)·|(−4.0×10−6)·(8.0×10−6)|
(0.20)2
= 8.99 ×109·32 ×10−12
0.04
= 8.99 ×109·8×10−11
= 71.92 N
Therefore, the magnitude of the electric force between the charges is 71.92
N.
Question 5
Question
Three point charges are arranged as shown in the diagram below: a +2.0 µC
charge at the origin, a -3.0 µC charge at (4.0 m, 0), and a +4.0 µC charge at
(0, 3.0 m). Calculate the magnitude of the net electric force acting on the +4.0
µC charge due to the other two charges.
Solution
1. Let’s denote the charges at the origin, (4.0 m, 0), and (0, 3.0 m) respectively
as q1= +2.0µC, q2=−3.0µC, and q3= +4.0µC.
2. We will calculate the net electric force on the +4.0µC charge, q3, due to
charges q1and q2using the principle of superposition. The electric force exerted
by each point charge can be calculated using the formula:
Felec =k|q1q2|
r2
3. First, we will calculate the force on q3due to q1:
F1=k|q1q3|
r2
13
4. Given that k= 8.9875×109Nm2
C2and r13 = 3.0m, we can substitute these
values into the formula.
5. Solving for F1:
F1= 8.9875 ×109Nm2
C2·(2.0µC)(4.0µC)
(3.0m)2
6. Next, we will calculate the force on q3due to q2:
F2=k|q2q3|
r2
23
4
7. Given that r23 = 4.0m, we can substitute the known values into the
formula.
8. Solving for F2:
F2= 8.9875 ×109Nm2
C2·(3.0µC)(4.0µC)
(4.0m)2
9. The net electric force on q3would be the vector sum of F1and F2.
10. Calculate the magnitudes and directions of F1and F2, then sum them
to find the net force acting on q3.
Question 7
Question
Three point charges are arranged along the x-axis. Charge q1=−2.0µC is
located at the origin, charge q2= 3.0µC is located at x= 4.0m, and charge
q3= 1.0µC is located at x= 8.0m. Calculate the net force on charge q3due
to charges q1and q2.
Solution
Step 1: Calculate the force F31 on charge q3due to charge q1.
Since the charges are on the x-axis, we can use Coulomb’s Law to calculate
the force:
F31 =k|q1||q3|
r2
31
where k= 8.99 ×109N·m2/C2is the Coulomb’s constant, q1=−2.0µC =
−2.0×10−6Cis the charge at the origin, and q3= 1.0µC = 1.0×10−6Cis the
charge at x= 8.0m.
The distance r31 between charges q3(at x= 8.0m) and q1(at the origin) is:
r31 = 8.0m
Therefore, the force F31 is:
F31 = (8.99 ×109)| − 2.0×10−6×1.0×10−6|
(8.0)2
Calculating the force, we get:
F31 = 4.49 ×10−4N
Step 2: Calculate the force F32 on charge q3due to charge q2.
Using Coulomb’s Law again, the force F32 is:
F32 =k|q2||q3|
r2
32
5
where q2= 3.0µC = 3.0×10−6Cis the charge at x= 4.0m.
The distance r32 between charges q3(at x= 8.0m) and q2(at x= 4.0m) is:
r32 = 8.0−4.0 = 4.0m
Plugging in the values, we get:
F32 = (8.99 ×109)3.0×10−6×1.0×10−6
(4.0)2
Solving for the force F32, we find:
F32 = 6.74 ×10−4N
Step 3: Calculate the net force Fnet on charge q3.
The net force Fnet is the vector sum of forces F31 and F32. Since F31 is to
the left (negative x direction) and F32 is to the right (positive x direction), the
net force is:
Fnet =F32 −F31
Fnet = 6.74 ×10−4−4.49 ×10−4
Fnet = 2.25 ×10−4N
Therefore, the net force on charge q3due to charges q1and q2is 2.25×
Question 8
Question
Two point charges, q1=−3µC and q2= 5 µC, are placed 10 cm apart in air.
Calculate the magnitude and direction of the electric force that each charge
exerts on the other.
Solution
Step 1: Calculate the magnitude of the electric force each charge exerts on the
other 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 k= 8.99 ×109Nm2/C2is Coulomb’s constant, q1and q2are the magni-
tudes of the charges, and ris the distance between the charges.
Plugging in the values:
F=(8.99 ×109)·3×10−6·5×10−6
0.12
6
F=4.495 ×104
0.01
F= 4495 N
Therefore, the magnitude of the electric force that each charge exerts on the
other is 4495 N.
Step 2: Determine the direction of the electric force on each charge. Since
the charges are opposite in sign, there will be an attractive force between them.
The force on q1due to q2will be directed toward q2, and the force on q2due to
q1will be directed toward q1.
Question 9
Question
Two point charges, q1=−4µC and q2= 8 µC, are located 10 cm apart on
the x-axis. Calculate the magnitude and direction of the electric force on each
charge.
Solution
Step 1: Calculate the distance between the two charges. Given that the charges
are located 10 cm (or 0.1 m) apart on the x-axis, the distance (r) between them
is 0.1 m.
Step 2: Calculate the magnitude of the electric force on q1. The magnitude
of the electric force between two point charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2
where kis the electrostatic constant (8.99 ×109N·m2/C2).
Substitute the given values into the formula:
F=(8.99 ×109N·m2/C2)·(4 ×10−6C)·(8 ×10−6C)
(0.1m)2
F=287.68 ×10−6
0.01
F= 28.768 N
So, the magnitude of the electric force acting on q1is 28.768 N.
Step 3: Determine the direction of the force on q1. Since q1is negative, the
force on it will be attractive and directed towards q2.
Step 4: Calculate the magnitude of the electric force on q2. The magnitude
of the electric force on q2will be the same as that on q1, i.e., 28.768 N.
Step 5: Determine the direction of the force on q2. Since q2is positive, the
force on it will be repulsive and directed away from q1.
7
Question 10
Question
Three point charges are arranged in an equilateral triangle with sides of length
a. The charges are +q,−2q, and +3q. Calculate the net electric force on the
+qcharge due to the other two charges.
Solution
To find the net electric force on the +qcharge, we need to calculate the in-
dividual forces on it due to the other two charges and then sum these forces
vectorially.
Step 1: Calculate the force on the +qcharge due to the −2qcharge. The
magnitude of the electric force between two point charges q1and q2separated
by a distance ris given by Coulomb’s Law:
F=k·|q1q2|
r2,
where kis the Coulomb constant (8.99 ×109N m2/C2).
The distance between the +qand −2qcharges is a. Therefore, the magnitude
of the force on the +qcharge due to the −2qcharge is:
F+qdue to −2q=k·|+q(−2q)|
a2.
Step 2: Calculate the force on the +qcharge due to the +3qcharge. The
distance between the +qand +3qcharges is also a. The magnitude of the force
on the +qcharge due to the +3qcharge is:
F+qdue to +3q=k·|+q(3q)|
a2.
Step 3: Find the net force on the +qcharge. The net force on the +q
charge is the vector sum of the forces from the −2qand +3qcharges. Since the
forces are acting along the sides of an equilateral triangle, they will have the
same magnitude but will be in opposite directions.
Therefore, the net force is:
Fnet =F+qdue to +3q−F+qdue to −2q.
Substitute the expressions for the forces into the equation above and simplify
to find the net force on the +qcharge.
Question 11
Question
Two point charges are fixed in place along the x-axis. Charge q1=−4.0µC is
located at x= 0 m and charge q2= 7.0µC is located at x= 2.0m. Find the
8
electric force on q2due to q1.
Solution
Step 1: Calculate the distance between the two charges. Given that q2is located
at x= 2.0m and q1is located at x= 0 m, the distance rbetween the two charges
is:
r=x2−x1= 2.0m−0m= 2.0m
Step 2: Calculate the electric force. The magnitude of the electric force F
between two point charges can be calculated using Coulomb’s Law:
F=k· |q1|·|q2|
r2
where
•k≈8.99 ×109N m2/C2is the Coulomb constant,
•|q1|= 4.0×10−6C,
•|q2|= 7.0×10−6C, and
•r= 2.0m.
Plugging in the values, we get:
F=(8.99 ×109N m2/C2)·(4.0×10−6C)·(7.0×10−6C)
(2.0m)2
F=(8.99 ×103)·(4.0) ·(7.0)
4.0
F= 8.99 ×4×7 = 251.32 N
Therefore, the electric force on q2due to q1is 251.32 N directed towards q1.
Question 12
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between these two
charges.
9
Solution
Step 1: Convert the charges to coulombs: The charges in microcoulombs can be
converted to coulombs by multiplying by 10−6.q1=−4.0µC = −4.0×10−6C
=−4.0µCq2= 6.0µC = 6.0×10−6C = 6.0µC
Step 2: Find the distance between the charges in meters: Given that the
charges are placed 10.0 cm apart, convert the distance to meters by dividing by
100. r= 10.0cm = 10.0×10−2m = 0.10 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
The electric force between two point charges q1and q2separated by a distance
ris given by Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant with a value of 8.99 ×109N m2/C2.
Substitute the values into the formula:
F=(8.99 ×109N m2/C2)·|−4.0×10−6C·6.0×10−6C|
(0.10 m)2
Step 4: Calculate the magnitude of the electric force:
F=(8.99 ×109)·(4.0×6.0) ×10−6
0.01 N
F=8.99 ×109·24
0.01 N
F= 2.16 ×10−2N
Therefore, the magnitude of the electric force between the charges q1=−4.0
µC and q2= 6.0µC placed 10.0 cm apart is 2.16 ×10−2N.
Question 13
Question
Three charges are arranged along the x-axis. Charge q1= 4 nC is at the origin,
charge q2=−6nC is located at x= 5 m, and charge q3= 8 nC is placed at
x= 10 m. Calculate the electric force acting on charge q3due to the other two
charges.
Solution
Step 1: Calculate the electric force on charge q3due to charge q1using Coulomb’s
law:
F13 =k|q1q3|
r2
13
10
Step 2: Calculate the distance r13 between charge q1and charge q3:
r13 = 10 m
Step 3: Substitute the given values into Coulomb’s law:
F13 =(8.99 ×109N m2/C2)(4 ×10−9C)(8 ×10−9C)
(10 m)2
Step 4: Calculate the electric force on charge q3due to charge q1:
F13 =(35.96 ×101)(8 ×10−9)
100
F13 =287.68 ×10−8
100
F13 = 2.8768 ×10−6N
Step 5: Calculate the electric force on charge q3due to charge q2using the
same process:
F23 =k|q2q3|
r2
23
Step 6: Calculate the distance r23 between charge q2and charge q3:
r23 = 5 m
Step 7: Substitute the given values into Coulomb’s law:
F23 =(8.99 ×109)(−6×10−9)(8 ×10−9)
(5)2
Step 8: Calculate the electric force on charge q3due to charge q2:
F23 =−43.92 ×101
25
F23 =−1.7568 ×10−6N
Step 9: Finally, calculate the total electric force on charge q3by summing
the forces F13 and F23:
Ftotal =F13 +F23
Ftotal = 2.8768 ×10−6−1.7568 ×10−6
Ftotal = 1.12 ×10−6N
Therefore, the total electric force on charge q3is 1.12 ×10−6N, directed
towards charge q1.
11
Question 14
Question
Two point charges, q1=−8nC and q2= 4 nC, are placed 6cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs using the conversion factor 1nC =
10−9C.
q1=−8×10−9C
q2= 4 ×10−9C
Step 2: Calculate the distance between the charges in meters.
6cm = 6 ×10−2m
Step 3: Calculate the magnitude of the electric force using Coulomb’s law:
F=k·|q1·q2|
r2where k= 8.99 ×109N m2/C2is the electrostatic constant.
F=(8.99 ×109)·|−8×10−9·4×10−9|
(6 ×10−2)2
=8.99 ×109·32 ×10−18
36 ×10−4
=287.68 ×10−9
36 ×10−4
= 7.99 ×10−5N
Therefore, the magnitude of the electric force between the charges is 7.99 ×
10−5N.
Question 15
Question
Three point charges are arranged on the vertices of a right-angled triangle as
shown:
Q1(+3 C)
Q2(−2C)
Q3(+4 C)
The side lengths are 3 m, 4 m, and 5 m. Calculate the electric force on Q1
due to Q2and Q3.
12
Solution
Step 1: Calculate the electric force on Q1due to Q2. Given: Q1= +3 C,
Q2=−2C, r= 4 m.
The electric force between two point charges is given by Coulomb’s Law:
F=k·|Q1|·|Q2|
r2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the given values into the formula to find the electric force on Q1
due to Q2:
FQ1Q2= 8.99 ×109·3·2
42= 1.3485 ×109N
Step 2: Calculate the electric force on Q1due to Q3. Given: Q1= +3 C,
Q3= +4 C, r= 5 m.
Using Coulomb’s Law again, the electric force on Q1due to Q3is:
FQ1Q3= 8.99 ×109·3·4
52= 2.8792 ×109N
Step 3: Calculate the net electric force on Q1. The electric force on Q1due
to Q2is attractive (since Q2is negative) and the force due to Q3is repulsive
(since Q3is positive).
Therefore, the net force on Q1is:
Fnet =FQ1Q3−FQ1Q2= 2.8792 ×109−1.3485 ×109= 1.5307 ×109N
So, the net electric force on Q1is 1.5307 ×109N.
Question 16
Question
Two point charges, q1=−2µC and q2= 4 µC, are placed 10 cm apart. Calculate
the magnitude of the electric force between the charges.
Solution
To calculate the magnitude of the electric force between the charges, we can use
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 Coulomb’s constant
(8.99 ×109N·m2/C2), - q1and q2are the magnitudes of the point charges
involved, - ris the distance between the charges.
13
Given: - q1=−2µC, - q2= 4 µC, - r= 0.10 m.
Step 1: Calculate the electric force using Coulomb’s Law.
Plugging in the values:
F= (8.99 ×109)·|(−2×10−6)·(4 ×10−6)|
(0.10)2
F= 8.99 ×109·8×10−12
0.01
F= 8.99 ×109·8×10−10
F= 71.92 ×10−1
F= 7.192 N
Therefore, the magnitude of the electric force between the charges is 7.192
N.
Question 17
Question
Two point charges, q1=−3µC and q2= 6µC, are placed 10 cm apart. Calculate
the magnitude and direction of the electric force that q1exerts on q2.
Solution
Step 1: Determine the distance between the charges. Given that the charges
are 10 cm apart, we convert this distance to meters:
d= 10 cm = 0.10 m
Step 2: Calculate the magnitude of the electric force between the charges
using Coulomb’s law:
F=k·
q1·q2
d2
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Plugging in the values, we have:
F= 8.99 ×109N m2/C2·| − 3×10−6C·6×10−6C|
(0.10 m)2
F= 8.99 ×109N m2/C2·1.8×10−5C2/(0.01 m2)
F= 8.99 ×109N·1.8×10−3m/0.01 m2
14
F= 8.99 ×109·1.8×10−2N= 1.6182 ×108N
Step 3: Determine the direction of the force. The force is repulsive since the
charges have opposite signs, therefore the force points away from q1and towards
q2.
So, the magnitude of the electric force that q1exerts on q2is 1.6182 ×108
N, directed away from q1and towards q2.
Question 18
Question
Two point charges, +4.0µC and −2.0µC, are placed 10.0 cm apart in a vacuum.
Calculate the magnitude of the electric force between these two charges.
Solution
To calculate the magnitude of the electric force between the two charges, we
will use Coulomb’s Law, which states that the magnitude of the electric force
between two point charges is given by:
F=k·|q1·q2|
r2
where: - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the two charges, and -
ris the distance between the charges.
Given: - q1= +4.0µC= 4.0×10−6C, - q2=−2.0µC=−2.0×10−6C, and
-r= 10.0cm = 0.10 m.
Step 1: Substitute the given values into Coulomb’s Law formula:
F= (8.99 ×109)·|4.0×10−6· −2.0×10−6|
(0.10)2
F= (8.99 ×109)·8.0×10−12
0.01
Step 2: Calculate the magnitude of the electric force:
F= (8.99 ×109)·8.0×10−10
F= 7.192 ×10−1N
Therefore, the magnitude of the electric force between the two charges is
0.7192 N.
15
Question 19
Question
Three point charges are placed at the vertices of an equilateral triangle as shown
below:
−q
q
−q
If the charges qare positive and the charges −qare negative, calculate the
magnitude and direction of the electric force experienced by the charge at point
B.
Given: Each charge q= 2 µC, the side length of the equilateral triangle is 2
m, and the Coulomb’s constant k= 9 ×109N m2/C2.
Solution
Step 1: Calculate the distance between the charges at point B and point A or C.
Given that the side length of the equilateral triangle is 2 m, the distance between
the charges at points A and B (or C) can be calculated using the Pythagorean
theorem:
d=q22+ (2√3)2=√52 = 2√13 m
Step 2: Calculate the electric force between the charges qand −qat points B
and A (or C). The electric force between two point charges is given by Coulomb’s
Law:
F=k|q1q2|
r2
where kis the Coulomb’s constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
Substitute the given values into the equation to calculate the force:
F= 9 ×109×(2 ×10−6)(2 ×10−6)
(2√13)2
F= 36 ×10−12 ×4×10−12
52
F=144
52 ×10−24
16
F≈2.769 ×10−23 N
The force between the charges at point B and point A (or C) is approximately
2.769 ×10−23 N.
Step 3: Determine the direction of the electric force. Since the charges at
points B and A (or C) have opposite signs (attraction between positive and
negative charges), the electric force experienced by the charge at point B is
directed towards the charge at point A (or C).
Question 20
Question
Two point charges, q1= 4 µC and q2=−6µC, are placed 1 meter apart.
Determine the magnitude and direction of the electric force experienced by
each charge due to the other charge.
Solution
Step 1: Calculate the magnitude of the electric force on q1due to q2. The
magnitude of the electric force between two point charges is given by Coulomb’s
law:
F=k|q1q2|
r2
where kis Coulomb’s constant (8.99 ×109N·m2/C2), q1and q2are the mag-
nitudes of the charges, and ris the distance between the charges.
Plugging in the values, we get:
F1,2= (8.99 ×109)|4×10−6× −6×10−6|
(1)2
F1,2= 4314 N
The force on q1due to q2is attractive and points towards q2.
Step 2: Calculate the magnitude of the electric force on q2due to q1. Since
the forces between charges are equal in magnitude and opposite in direction by
Newton’s third law, the magnitude of the force on q2due to q1is also 4314 N,
but in the opposite direction (towards q1).
Therefore, the magnitude of the electric force experienced by each charge
due to the other charge is 4314 N, with the direction of each force being towards
the other charge.
Question 21
Question
Three point charges are arranged on the vertices of an equilateral triangle as
shown below. The charges are +2 µC,−3µC, and +4 µC. Calculate the mag-
17
nitude and direction of the net electric force on the charge located at the top
vertex.
+2 µC
−3µC
+4 µC
Solution
Step 1: Calculate the electric force between all pairs of charges using Coulomb’s
law: The electric force between two charges q1and q2separated by a distance
ris given by
F=k·|q1q2|
r2,
where k= 8.99 ×109N m2/C2is Coulomb’s constant.
We denote the charge at the top vertex as q1= +2 µC, the charge at the
right vertex as q2=−3µC, and the charge at the left vertex as q3= +4 µC.
The distance between any two charges in an equilateral triangle is the length
of a side, denoted as d. For an equilateral triangle, d= 2r√3where ris the
length from a vertex to the centroid. Given that the side of the triangle is
d= 4 cm, we have r=d
2√3=4
2√3=2√3
3cm. So, r=2√3
3cm.
Now, we calculate the distances between charges:
r12 =r13 =r23 =2√3
3cm.
The electric forces between each pair of charges are:
F12 =k·|+ 2 ×(−3)|
(2√3/3)2=18k√3
4,
F13 =k·|+ 2 ×4|
(2√3/3)2=8k√3
4,
F23 =k·| − 3×4|
(2√3/3)2=12k√3
4.
18
Question 22
Question
Two point charges, q1=−3.00 µC and q2= 1.50 µC, are placed 20.0cm apart.
Calculate the magnitude of the electric force between them.
Solution
To calculate the magnitude of the electric force between the two charges, we
can use Coulomb’s law. Coulomb’s law states that the magnitude of the electric
force between two point charges is given by the equation:
F=k·|q1|·|q2|
r2,
where: - Fis the magnitude of the electric force, - kis Coulomb’s constant
(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the two charges, and -
ris the distance between the charges.
Let’s substitute the given values into the equation and solve for F.
Step 1: Calculate the distance between the charges in meters Given:
r= 20.0cm = 0.20 m
Step 2: Calculate the electric force using Coulomb’s law Given:
q1=−3.00 µC=−3.00 ×10−6C Given: q2= 1.50 µC= 1.50 ×10−6C
Plugging the values into Coulomb’s law:
F=k·|q1|·|q2|
r2
F= (8.99 ×109N m2/C2)·3.00 ×10−6C·1.50 ×10−6C
(0.20 m)2
Step 3: Calculate the electric force
F= 8.99 ×109×3.00 ×1.50
0.202N
F= 8.99 ×109×4.50
0.04 N
F= 8.99 ×109×112.5N
F= 1.013 ×1012 N
Therefore, the magnitude of the electric force between the two charges is
1.013 ×1012 N.
Question 23
Question
Two point charges, Q1= 3µC and Q2=−4µC, are placed 10 cm apart in air.
Calculate the magnitude of the electric force between the charges.
19
Solution
Step 1: Convert the given charges to standard units.
Given: Q1= 3µC and Q2=−4µC.
To convert from µC to C, we use the conversion factor 1µC = 10−6C.
So, Q1= 3µC = 3 ×10−6Cand Q2=−4µC =−4×10−6C.
Step 2: Determine the distance between the charges.
Given that the charges are placed 10 cm apart, we convert 10 cm to meters
using the conversion factor 1cm = 10−2m.
So, the distance between the charges is 10 cm = 10 ×10−2m= 0.1m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s law.
The magnitude of the electric force between point charges is given by Coulomb’s
law:
F=k·|Q1·Q2|
r2,
where kis the electrostatic constant (k≈8.9875 ×109N·m2/C2), Q1and Q2
are the magnitudes of the charges, and ris the distance between the charges.
Substitute the given values:
F= 8.9875 ×109N·m2/C2×|3×10−6· −4×10−6|
(0.1)2
F= 8.9875 ×109×12 ×10−12
0.01
F= 8.9875 ×109×1.2×10−9
F= 10.785 ×100N
F= 10.785 N
Therefore, the magnitude of the electric force between the charges is 10.785
N.
Question 24
Question
Three point charges are arranged in a line. Charge q1= +5.0µC is at the
origin, charge q2=−3.0µC is located at x= 0.30 m, and charge q3= +2.0µC
is located at x= 0.45 m. Calculate the net electric force on charge q2.
Solution
Step 1: Calculate the electric force on charge q2due to charge q1.
F12 =k· |q1|·|q2|
r2
12
20
Step 2: Calculate the direction of the electric force on charge q2due to
charge q1. The force will be repulsive because the charges have opposite signs,
so the force vector points in the positive x-direction.
Step 3: Calculate the magnitude of the electric force F12.
F12 =8.99 ×109N·m2/C2·5.0×10−6C·3.0×10−6C
(0.30 m)2
F12 = 29965.00 N
Step 4: Calculate the electric force on charge q2due to charge q3.
F23 =k· |q2|·|q3|
r2
23
Step 5: Calculate the direction of the electric force on charge q2due to
charge q3. The force will be attractive because the charges have opposite signs,
so the force vector points in the negative x-direction.
Step 6: Calculate the magnitude of the electric force F23.
F23 =8.99 ×109N·m2/C2·3.0×10−6C·2.0×10−6C
(0.15 m)2
F23 = 35984.17 N
Step 7: Calculate the net electric force on charge q2.
Fnet =F12 −F23
Fnet = 29965.00 N−35984.17 N
Fnet =−6029.17 N
Therefore, the net electric force on charge q2is −6029.17 N in the negative
x-direction.
Question 25
Question
Two point charges, q1=−4.0µC and q2= 8.0µC, are separated by a distance
of 10.0m. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to coulombs. Given: q1=−4.0µC=−4.0×10−6C
q2= 8.0µC= 8.0×10−6C
21
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: F= magnitude of the electric force k= Coulomb’s constant (8.99 ×
109N·m2/C2)q1,q2= magnitudes of the charges r= distance between the
charges
Plugging in the values:
F= (8.99 ×109)·| − 4.0×10−6·8.0×10−6|
10.02
Step 3: Calculate the magnitude of the electric force.
F= (8.99 ×109)·32 ×10−12
100
F= (8.99 ×109)·3.2×10−13
F= 2.8776 ×10−3N
Therefore, the magnitude of the electric force between the charges is 2.8776×
10−3N.
22
