PHYS 232 - UNIVERSITY PHYSICS
II - Calculation of electric forces
between point charges
Question Bank - Set 2
Liberty University
Question 1
Question
Two point charges, q1= 2 µC and q2=−3µC, are placed 5 cm apart in air.
Calculate the magnitude and direction of the electric force on each charge.
Solution
Step 1: Calculate the magnitude of the electric force on charge q1. Given:
q1= 2 µC, q2=−3µC, r= 0.05 m. The electric force between two point
charges is given by Coulomb’s Law:
F=k· |q1·q2|
r2,
where kis the Coulomb constant (8.99 ×109N m2/C2).
Plugging in the values, we get:
F=(8.99 ×109)· |2×10−6× −3×10−6|
(0.05)2
F=8.99 ×109·6×10−12
0.0025
F=53.94 ×10−3
0.0025 = 21.576 N
Therefore, the magnitude of the electric force on charge q1is 21.576 N.
Step 2: Determine the direction of the electric force on charge q1. Since
charge q2is negative, the force on q1will be attractive. Therefore, the direction
of the electric force on charge q1is towards charge q2.
Step 3: Calculate the magnitude of the electric force on charge q2. Since the
magnitudes of the charges are the same, the magnitude of the force on q2will
be the same as on q1, which is 21.576 N.
Step 4: Determine the direction of the electric force on charge q2. Since
charge q1is positive, the force on q2will be attractive also. Therefore, the
direction of the electric force on charge q2is towards charge q1.
Question 3
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length 2. The charges are +2µC at the top vertex, −3µC at the bottom
left vertex, and +5µC at the bottom right vertex. Calculate the magnitude and
direction of the net electrostatic force acting on the +2µC charge due to the
other charges.
Solution
To find the net electrostatic force acting on the +2µC charge, we need to cal-
culate the individual forces due to each of the other charges and then add them
as vectors.
Step 1: Calculate the force due to the −3µC charge at the bottom left ver-
tex. The magnitude of the force between two point charges q1and q2separated
by distance ris given by Coulomb’s Law:
F=k|q1q2|
r2
where kis the Coulomb’s constant (k≈8.99 ×109Nm2/C2).
For the +2µC charge and the −3µC charge, the distance between them is the
height of the equilateral triangle, h=√3. So, the force between these charges
is:
Fleft =(8.99 ×109)|2×10−6×3×10−6|
(√3)2
Step 2: Calculate the direction of the force due to the −3µC charge. Since
the −3µC charge is negative, the force on the +2µC charge will be attractive,
i.e., towards the −3µC charge. Therefore, the direction is towards the bottom
left vertex.
Step 3: Calculate the force due to the +5µC charge at the bottom right
vertex. Using Coulomb’s Law, the force between the +2µC charge and the
2
+5µC charge is:
Fright =(8.99 ×109)|2×10−6×5×10−6|
(√3)2
Step 4: Calculate the direction of the force due to the +5µC charge. Since
both charges are positive, the force on the +2µC charge will be repulsive, i.e.,
away from the +5µC charge. Therefore, the direction is away from the bottom
right vertex.
Step 5: Find the net force on the +2µC charge. To find the net force, we
need to add these forces as vectors. Let the force due to the −3µC charge be
Fleft and the force due to the +5µC charge be
Fright. The net force is:
Fnet =
Fleft +
Fright
Now, calculate the magnitude and direction of the net force.
Question 4
Question
Two point charges, q1= +2 µC and q2=−3µC, are placed 10 cm apart.
Calculate the magnitude and direction of the electric force each charge exerts
on the other. Assume the charges are located on the x-axis, with q1at the origin
and q2at x= 10 cm.
Solution
Step 1: Calculate the distance between the charges in meters. Given that the
charges are 10 cm apart, the distance rin meters is:
r= 10 cm = 0.10 m
Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force between two point charges is given by Coulomb’s law:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging in the given values:
F=(8.99 ×109)(2 ×10−6)(3 ×10−6)
(0.10)2
Calculating:
F=53.94
0.01 = 5394 N
3
Therefore, the magnitude of the electric force between the charges is 5394
N.
Step 3: Determine the direction of the force. Since q1is positive and q2is
negative, the force between them is attractive and directed from q2towards q1.
Thus, the electric force Fon q1points towards q1. The electric force Fon
q2points towards q2.
In conclusion, the magnitude of the electric force each charge exerts on the
other is 5394 N, and the direction of the force on q1is towards q1, while the
direction of the force on q2is towards q2.
Question 5
Question
Two point charges, q1 = +2.0 µC and q2 = -4.0 µC, are placed 10.0 cm apart
in air. Calculate the magnitude and direction of the electric force that q1 exerts
on q2.
Solution
Step 1: Convert µC to Coulombs. Given that 1µC = 10−6C, we can convert the
charges as follows: q1 = 2.0µC = 2.0×10−6C q2 = −4.0µC =−4.0×10−6C
Step 2: Calculate the distance between the two charges. Given that the
charges are placed 10.0 cm apart, we convert this distance to meters: d=
10.0cm = 10.0×10−2m= 0.10m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
The electric force between the charges is given by Coulomb’s Law: F=k·|q1·q2|
d2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the values: F= (8.99 ×109)·|2.0×10−6·−4.0×10−6|
0.102
F= 71.92 ×10−12/0.01
F= 7.192 ×10−10N
Step 4: Find the direction of the force. Since q1 is positive and q2 is negative,
the force between them will be attractive. Therefore, the force exerted by q1
on q2 is directed towards q1.
The magnitude of the electric force that q1 exerts on q2 is 7.192 ×10−10N
and the direction is towards q1.
Question 6
Question
Two point charges, +2 µC and −3µC, are placed 6 cm apart in air. Calculate
the magnitude and direction of the electric force that one charge exerts on the
other.
4
Solution
Step 1: Convert the charges to Coulombs. We have +2 µC= 2 ×10−6C and
−3µC=−3×10−6C.
Step 2: Calculate the electric force using Coulomb’s Law:
The magnitude of the electric force between two point charges is given by:
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 charges, - ris the distance between the
charges.
Plugging in the values, we get:
F=8.99 ×109· |2×10−6· −3×10−6|
(0.06)2
Step 3: Calculate the magnitude of the electric force:
F=8.99 ×109·6×10−12
0.0036
F=53.94 ×10−3
0.0036
F= 14950 N
Step 4: Determine the direction of the electric force: The force is attractive
since one charge is positive and the other is negative. It acts along the line
joining the charges, from the positive charge to the negative charge.
Therefore, the magnitude of the electric force that the positive charge exerts
on the negative charge is 14950 N, directed from the positive charge to the
negative charge.
Question 7
Question
Three point charges are arranged as shown: - Charge q1= +2.0µC is located at
the origin. - Charge q2=−3.0µC is located at coordinates (0,0.6m). - Charge
q3= +1.0µC is located at coordinates (0.8m,0).
Calculate the net electric force acting on q1due to q2and q3.
Solution
Step 1: Find the electric force between q1and q2: The electric force F12 between
two point charges is given by Coulomb’s law:
F12 =k|q1||q2|
r2
12
5
where - kis Coulomb’s constant (8.99 ×109N m2/C2), - q1and q2are the
magnitudes of the point charges, and - r12 is the distance between the two
charges.
Given that q1= 2.0µC,q2=−3.0µC, and r12 = 0.6m, we have:
F12 =(8.99 ×109)×(2.0×10−6)×(3.0×10−6)
(0.6)2
Step 2: Calculate F12:
F12 =8.99 ×2.0×3.0
(0.6)2×103N
Step 3: Simplify the expression:
F12 =53.94
0.36 ×103N
F12 = 149.83 ×103N
F12 = 1.50 ×105N
Step 4: Find the direction of F12: The direction of F12 is repulsive since the
charges q1and q2are of opposite sign.
Step 5: Repeat steps 1-4 for the electric force F13 between q1and q3and
then find the net electric force acting on q1.
Question 8
Question
Two point charges, q1= +2.0µC and q2=−4.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the two
charges.
Solution
Step 1: Convert the charges to the standard unit of Coulombs. Given: q1=
+2.0µC= 2.0×10−6C, and q2=−4.0µC=−4.0×10−6C.
Step 2: Recall the formula for the magnitude of electric force between two
point charges:
F=k·|q1·q2|
r2
where kis the Coulomb constant (k= 8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the separation distance between the charges.
Step 3: Substitute the given values into the formula and solve for the electric
force.
F= (8.99 ×109)·|2.0×10−6· −4.0×10−6|
(0.10)2
6
F= (8.99 ×109)·8.0×10−12
0.01
F= 8.99 ×109·8.0×10−10
F= 71.92 ×10−1
F= 7.192 N
Step 4: Therefore, the magnitude of the electric force between the two
charges is 7.192 N.
Question 9
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed on the x-axis at
positions x=−2.0m and x= 2.0m, respectively. Calculate the magnitude and
direction of the electric force on q1due to q2.
Solution
Step 1: Calculate the distance between the two charges. Given: q1=−4.0µC=
−4.0×10−6Cq2= 6.0µC= 6.0×10−6Cx1=−2.0mx2= 2.0m
The distance rbetween the two charges is:
r=|x2−x1|=|2.0−(−2.0)|= 4.0m
Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force Fbetween two charges is given by Coulomb’s law:
F=k·|q1·q2|
r2
where k= 8.99 ×109N m2C−2is the Coulomb’s constant.
Plugging in the values:
F= 8.99 ×109·| − 4.0×10−6·6.0×10−6|
(4.0)2
F= 8.99 ×109·24 ×10−12
16
F= 8.99 ×109·1.5×10−12
F≈13.485 N
Step 3: Determine the direction of the force. The force on q1due to q2
is attractive since the charges have opposite signs. Therefore, the force acts
towards the positive xdirection.
So, the magnitude of the electric force on q1due to q2is approximately
13.485 N, and it acts in the positive xdirection.
7
Question 10
Question
Two point charges, q1=−3.5µC and q2= 1.8µC, are placed 12 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to SI units.
q1=−3.5µC =−3.5×10−6C
q2= 1.8µC = 1.8×10−6C
Step 2: Determine the distance between the charges in meters.
d= 12 cm = 0.12 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
F=k|q1||q2|
d2
Where: k= 8.99 ×109Nm2/C2is the Coulomb constant
Step 4: Substitute the given values into the formula and calculate.
F=(8.99 ×109)(3.5×10−6)(1.8×10−6)
0.122
F=56.735 ×10−15
0.0144
F= 56.735 ×10−15 ×69.444
F= 3938 ×10−15 N
F= 3.938 ×10−12 N
Therefore, the magnitude of the electric force between the charges is 3.938 ×
10−12 N.
Question 11
Question
Two point charges, Q1=−2.0µC and Q2= 4.0µC, are placed 0.10 meters
apart in a vacuum. Calculate the magnitude and direction of the net electric
force on Q1.
8
Solution
Given: Q1=−2.0µC
Q2= 4.0µC
Distance between charges, r= 0.10 m
k= 8.99 ×109N·m2/C2(Coulomb’s constant)
Step 1: Calculate the electric force between the two charges using Coulomb’s
law:
F=k·|Q1·Q2|
r2
F= 8.99 ×109·2.0×10−6·4.0×10−6
0.12
Step 2: Calculate the magnitude of electric force:
F= 8.99 ×109·8.0×10−12
0.01
F= 8.99 ×109·8.0×10−10
F= 7.192 ×10−1N
Step 3: Determine the direction of the net electric force: Since Q1and Q2
have opposite charges, the electric force on Q1will be attractive towards Q2.
Therefore, the net electric force on Q1is directed towards Q2.
Thus, the magnitude of the net electric force on Q1is 7.192 ×10−1N and
the direction is towards Q2.
Question 12
Question
Three point charges are placed at the corners of an equilateral triangle with
side length d= 2 m. The charges are +4µC,−3µC and +5µC. Calculate the
magnitude and direction of the net force on the +4µC charge due to the other
two charges.
Solution
Step 1: Calculate the distance between the +4µC charge and the −3µC charge
using the Pythagorean theorem. The distance r1can be found by dividing the
equilateral triangle into two right triangles.
r1=v
u
u
td
22
+ √3d
2+d!2
=r(1)2+√3+12
9
Step 2: Calculate the distance between the +4µC charge and the +5µC
charge. The distance r2can be found in the same manner.
r2=v
u
u
td
22
+ √3d
2!2
=r(1)2+√32
Step 3: Calculate the magnitude of the force F1between the +4µC and
−3µC charges using Coulomb’s law.
F1=k|q1||q2|
r2
1
Step 4: Calculate the magnitude of the force F2between the +4µC and
+5µC charges in the same way.
F2=k|q1||q3|
r2
2
Step 5: Determine the direction of the net force on the +4µC charge. Since
the forces are vectors, we need to consider their directions. The force F1will be
directed away from the −3µC charge, and the force F2will be directed towards
the +5µC charge. Calculate the net force by considering the vector sum of F1
and F2.
Question 13
Question
Three point charges are arranged as shown in the diagram below. Each charge
has a magnitude of 3.00 µC. Calculate the magnitude and direction of the net
force on the charge q3.
q1q2
q3
Solution
Step 1: Calculate the distances between the charges using the Pythagorean
theorem. Let the length of each side of the equilateral triangle be L.
L=p22+ 1.7322
L=√4+3
L=√7
10
Step 2: Calculate the electric force between charges using Coulomb’s law:
F=k|q1q2|
r2where - kis the Coulomb constant (8.99 ×109N·m2/C2), - q1
and q2are the magnitudes of the charges, and - ris the distance between the
charges.
Step 3: Calculate the force between q1and q3:
F13 = 8.99 ×109(3 ×10−6)(3 ×10−6)
(√7)2
Step 4: Calculate the force between q2and q3:
F23 = 8.99 ×109(3 ×10−6)(3 ×10−6)
(√7)2
Step 5: The net force on q3is the vector sum of the forces F13 and F23.
Since the forces are acting at 120° angles to each other, the net force magnitude
can be calculated using the law of cosines:
Net force =qF2
13 +F2
23 + 2F13F23 cos(120◦)
Step 6: Now, plug in the values to find the magnitude and direction of the
net force on charge q3.
Question 14
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length a. The magnitudes of the charges are q,2q, and 3q. Calculate
the net electric force on the charge qdue to the other two charges.
Solution
Let’s denote the charges as follows: q1=q,q2= 2q, and q3= 3q. The distances
from q1to q2and q3are both a.
By Coulomb’s law, the magnitude of the electric force between two charges
qiand qjis Fij =k|qi|·|qj|
r2, where kis Coulomb’s constant (8.99 ×109N m2/C2)
and ris the distance between the charges.
Step 1: We start by finding the electric force on charge q1due to charge q2.
Given that |q1|=q,|q2|= 2q, and r=a, we have:
F12 =k|q1|·|q2|
r2= 8.99 ×109q·2q
a2
Step 2: Next, we find the electric force on charge q1due to charge q3. Given
that |q1|=q,|q3|= 3q, and r=a, we have:
F13 =k|q1|·|q3|
r2= 8.99 ×109q·3q
a2
11
Step 3: Now, we calculate the net electric force on charge q1by summing
the forces due to q2and q3.
Fnet =F12 +F13 = 8.99 ×1092q2
a2+ 8.99 ×1093q2
a2
Therefore, the net electric force on the charge qdue to the other two charges
is:
Fnet = 8.99 ×1095q2
a2N
Question 15
Question
Two point charges, q1= +4.0µC and q2=−6.0µC, are placed 0.10 m apart in
a vacuum. Calculate the magnitude of the electric force that each charge exerts
on the other.
Solution
Let’s denote the magnitude of the electric force between two point charges as F
and their separation distance as r. We can use Coulomb’s law to calculate the
electric force between these two charges. Coulomb’s law is given by:
F=k|q1q2|
r2,
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Step 1: Calculate the electric force that charge q1exerts on charge q2. This
force will be attractive since the charges are of opposite signs.
F1→2=k|q1q2|
r2.
Plugging in the given values:
F1→2= (8.99 ×109N m2/C2)|(4.0×10−6C)(−6.0×10−6C)|
(0.10 m)2.
Step 2: Calculate the magnitude of the electric force between q1and q2.
F1→2= 8.99 ×109×(4.0×10−6)(6.0×10−6)
0.01 .
F1→2= 8.99 ×109×2.4×10−11.
F1→2= 21.576 ×10−2N.
Therefore, the magnitude of the electric force that q1exerts on q2is 2.16 ×
10−1N.
12
Step 3: The force that charge q2exerts on charge q1will be equal in mag-
nitude but opposite in direction to the force calculated in Step 2.
Hence, the magnitude of the electric force that each charge exerts on the
other is 2.16 ×10−1N.
Question 16
Question
Three point charges are placed at the vertices of an equilateral triangle with
side length aas shown below. The charges are +q,−2q, and +4q. Calculate the
magnitude of the electric force on the +qcharge due to the other two charges.
+4q
−2q+q
Solution
Step 1: Calculate the electric force on the +qcharge due to the +4qcharge.
The electric force Fbetween two charges q1and q2separated by a distance ris
given by Coulomb’s law:
F=kq1q2
r2
where kis the Coulomb constant (8.9875 ×109N m2/C2).
In this case, the charge q1= +q,q2= +4q, and the distance between them is
the side length of the equilateral triangle a. Since the charges are located at the
vertices of an equilateral triangle, the distance between them is a
2. Therefore,
the electric force on the +qcharge due to the +4qcharge is:
F=kq(+4q)
(a
2)2=k4q2
a2
4
=k16q2
a2=72kq2
a2
Step 2: Calculate the electric force on the +qcharge due to the −2qcharge.
Similarly, the electric force on the +qcharge due to the −2qcharge can be
calculated:
F=kq(−2q)
a2=−2kq2
a2=−18kq2
a2
Step 3: Combine the two forces. The total force on the +qcharge is the
vector sum of the forces due to the +4qand −2qcharges. Since the forces are
in opposite directions along the same line:
Ftotal =72kq2
a2−18kq2
a2=54kq2
a2
Therefore, the magnitude of the electric force on the +qcharge due to the
other two charges is 54kq2
a2.
13
Question 17
Question
Three point charges are arranged as follows: a charge of +2.0µC at the origin,
a charge of −4.0µC at (0,3m), and a charge of +6.0µC at (4 m,0). Calculate
the net electric force on the charge at the origin.
Solution
To find the net electric force on the charge at the origin, we need to calculate
the individual electric forces due to each of the other charges and then sum
them up vectorially.
Step 1: Calculate the electric force due to the charge of +2.0µC at the
origin.
The electric force F1due to this charge can be calculated using Coulomb’s
law:
F1=kq1q
r2,
where kis the Coulomb constant, q1and qare the magnitudes of the two charges,
and ris the distance between the charges.
The distance between the charges is 0since the charge creating the force is
at the origin. Hence, r= 0.
Therefore, the force due to the charge of +2.0µC is 0.
Step 2: Calculate the electric force due to the charge of −4.0µC at (0,3m).
The electric force F2due to this charge can be calculated using Coulomb’s
law:
F2=kq2q
r2,
where q2and qare the magnitudes of the two charges, and ris the distance
between the charges.
The distance between the charges is √02+ 32= 3 m.
Plugging in the values and accounting for the direction of the force, which
would be along the negative y-axis, we get:
F2=−6.75 Nˆ
j.
Step 3: Calculate the electric force due to the charge of +6.0µC at (4 m,0).
The electric force F3due to this charge can be calculated using Coulomb’s
law:
F3=kq3q
r2,
where q3and qare the magnitudes of the two charges, and ris the distance
between the charges.
The distance between the charges is √42+ 02= 4 m.
Plugging in the values and accounting for the direction of the force, which
would be along the positive x-axis, we get:
F3= 6.75 Nˆ
i.
14
Step 4: Find the net electric force on the charge at the origin.
The net force on the charge at the origin is the vector sum of the individual
forces:
Fnet =F1+F2+F3.
Substitute the calculated values and summing the forces vectorially, we get:
Fnet =−6.75 Nˆ
j+ 6.75 Nˆ
i.
Therefore, the net electric force on the charge at the origin is −6.75 Nˆ
j+
6.75 Nˆ
i.
Question 18
Question
Two point charges, Q1=−8.0µC and Q2= 4.0µC, are placed at points A
and B, respectively, with a distance of 40 cm between them. Calculate the
magnitude and direction of the electric force that charge Q1exerts on charge
Q2. Given that the Coulomb constant is k= 8.99 ×109N m2/C2.
Solution
Step 1: Find the magnitude of the electric force using the equation:
F=k· |Q1|·|Q2|
r2
where Fis the electric force, kis the Coulomb constant, |Q1|and |Q2|are the
magnitudes of the charges, and ris the distance between the charges.
Plugging in the given values:
F=(8.99 ×109N m2/C2)·(8.0×10−6C)·(4.0×10−6C)
(0.40 m)2
F=8.99 ×8.0×4.0
0.16 ×103N
F= 8.99 ×8.0×4.0×103N
F= 2.8768 ×105N
Therefore, the magnitude of the electric force that charge Q1exerts on charge
Q2is 2.88 ×105N.
Step 2: Find the direction of the electric force. Since charge Q1is negative,
the force on charge Q2will be attractive, pulling Q2towards Q1. Thus, the
direction of the electric force is from charge Q2towards charge Q1.
15
Question 19
Question
Two point charges, q1= 8.0�C and q2=−10.0�C, are placed 40.0cm apart.
Determine the magnitude and direction of the electric force that q2exerts on
q1.
Solution
Step 1: First, we need to calculate the electric force between the two charges
using Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Plug in the given values:
F=(8.99 ×109)· |8.0×10−6·(−10.0×10−6)|
(0.40)2
Step 3: Calculate the magnitude of the electric force:
F=(8.99 ×109)·(8.0×10−6·10.0×10−6)
0.16
F=8.99 ×109·80 ×10−12
0.16
F=7.192 ×10−3
0.16 = 0.04495 N
Step 4: Now, to determine the direction of the force, we consider the fact
that q2exerts an attractive force on q1(since the charges have opposite signs).
Therefore, the force on q1points towards q2.
Step 5: So, the electric force that q2exerts on q1has a magnitude of 0.04495 N
and direction towards q2.
Question 20
Question
Two point charges, q1=−3.0µC and q2= 6.0µC, are placed 10.0cm apart.
Calculate the magnitude of the electric force that each charge exerts on the
other.
16
Solution
Step 1: Convert the given charges to coulombs.
q1=−3.0×10−6C
q2= 6.0×10−6C
Step 2: Calculate the distance between the charges in meters. Given: r=
10.0cm = 0.10 m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k|q1||q2|
r2
where k= 8.99 ×109N·m2/C2.
Step 4: Substitute the known values and calculate the magnitude of the
electric force.
F=8.99 ×109N·m2/C2×3.0×10−6C×6.0×10−6C
(0.10 m)2
F=161.82
0.01 = 16182 N
Therefore, the magnitude of the electric force that each charge exerts on the
other is 16182 N.
Question 21
Question
Two point charges, q1= 3.0×10−6C and q2=−4.0×10−6C, are placed 10
cm apart in a vacuum. Calculate the magnitude of the electric force between
the charges.
Solution
Step 1: Convert all given values to SI units. Given: q1= 3.0×10−6C, q2=
−4.0×10−6C, r= 10 cm = 0.10 m.
Step 2: Calculate the magnitude of the electric force between the charges
using Coulomb’s Law: The magnitude of the electric force between two point
charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2,
where k= 8.99 ×109N m2/C2is the electrostatic constant.
17
Substitute the given values into the formula:
F=8.99 ×109×3.0×10−6×4.0×10−6
(0.10)2.
Step 3: Calculate the electric force.
F=8.99 ×109×3.0×10−6×4.0×10−6
(0.10)2=107.88
0.01 = 10788 N.
Therefore, the magnitude of the electric force between the charges is 10788
N.
Question 22
Question
Three point charges are located at the corners of an equilateral triangle with
side length a. Each charge has a magnitude of q. Calculate the magnitude of
the net electric force on one of the charges due to the other two charges.
Solution
Let’s label the charges q1,q2, and q3at the vertices of the equilateral triangle.
We will find the magnitude of the net electric force on q1due to q2and q3.
Step 1: Calculate the electric force from q2on q1. The magnitude of the
electric force between two point charges can be calculated using Coulomb’s Law:
F=k|q1q2|
r2
where kis the Coulomb constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
In this case, the distance between q1and q2is a. The direction of the force
will be along the line connecting q1and q2.
Step 2: Calculate the electric force from q3on q1. Similarly, we can calculate
the electric force from q3on q1using Coulomb’s Law. The distance between q1
and q3is also ain this equilateral triangle configuration.
Step 3: Find the angle between the forces. The forces from q2and q3on q1
are equal in magnitude but opposite in direction. They form a 120◦angle with
respect to each other.
Step 4: Calculate the net electric force on q1. To find the net electric force
on q1, we can treat the forces from q2and q3as vectors. Since they are at 120◦
to each other, we can treat them as components of a single force vector.
The net force magnitude will be the vector sum of the forces from q2and q3:
Fnet =qF2
2+F2
3+ 2F2F3cos(120◦)
Solving this expression will give us the magnitude of the net electric force
on q1due to q2and q3.
18
Question 23
Question
Two point charges, q1= +3.00 nC and q2=−5.00 nC are located on the x-axis
at positions x= 2.00 m and x=−4.00 m respectively. Calculate the magnitude
and direction of the electric force on q1at x= 2.00 m due to the presence of
charge q2.
Solution
Step 1: Calculate the distance between the charges. Given that x1= 2.00 m and
x2=−4.00 m, we can find the distance between the charges using the formula:
r=|x2−x1|
r=| − 4.00 m−2.00 m|
r=| − 6.00 m|
r= 6.00 m
Therefore, the distance between the charges is 6.00 m.
Step 2: Calculate the electric force. The magnitude of the electric force
between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
where - kis the electrostatic constant 8.99 ×109N m2/C2-q1= +3.00 nC =
3.00 ×10−9C - q2=−5.00 nC =−5.00 ×10−9C - r= 6.00 m
Plugging in the values:
F= (8.99 ×109)|3.00 ×10−9× −5.00 ×10−9|
(6.00)2
F= (8.99 ×109)15.00 ×10−18
36
F= (8.99 ×109)×0.4167
F≈3.75 ×109N
Step 3: Determine the direction of the force. The force on q1is towards q2
because the charges are opposite in sign. Thus, the force is attractive.
Therefore, the magnitude of the electric force on q1at x= 2.00 m due to
the presence of charge q2at x=−4.00 m is approximately 3.75 ×109N and is
attractive.
19
Question 24
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 9.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between these two
charges.
Solution
Step 1: Convert the charges to Coulombs. Given that 1µC= 10−6C, we have:
q1=−4.0µC=−4.0×10−6C and q2= 6.0µC= 6.0×10−6C.
Step 2: Calculate the electric force using Coulomb’s Law. 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 Coulomb’s constant with a value of 8.9875 ×109N m2/C2,q1and q2
are the magnitudes of the charges, and ris the separation distance.
Step 3: Substitute the given values into Coulomb’s Law and solve for the
electric force. Substitute k= 8.9875 ×109N m2/C2,q1=−4.0×10−6C,
q2= 6.0×10−6C, and r= 9.0cm = 0.09 m into Coulomb’s Law:
F= 8.9875 ×109×|(−4.0×10−6)(6.0×10−6)|
(0.09)2
Step 4: Calculate the magnitude of the electric force. Performing the calcu-
lations:
F= 8.9875 ×109×24 ×10−12
0.0081
F= 8.9875 ×109×2.96 ×10−9
F= 26.6N
Therefore, the magnitude of the electric force between the two charges is
26.6N.
Question 25
Question
Two point charges q1=−3.0µC and q2= 5.0µC are placed 25 cm apart.
Calculate the magnitude of the electric force between the charges.
20
Solution
Step 1: Convert the charges to Coulombs.
Given: q1=−3.0µC=−3.0×10−6C
q2= 5.0µC= 5.0×10−6C
Step 2: Calculate the distance between the charges.
Given: r= 25 cm = 0.25 m
Step 3: Calculate the magnitude of the electric force.
The electric force between two point charges is given by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the electrostatic constant with a value of 8.99 ×109Nm2/C2.
Substitute the given values into the formula:
F= (8.99 ×109)·|(−3.0×10−6)·(5.0×10−6)|
(0.25)2
Step 4: Calculate the magnitude of the electric force.
F= 8.99 ×109·|(−3.0×10−6)·(5.0×10−6)|
(0.25)2
F= 8.99 ×109·15 ×10−12
0.0625
F= 8.99 ×109·2.4×10−10
F= 2.16 ×10−1N
Therefore, the magnitude of the electric force between the charges is 2.16 ×
10−1N.
21
Step 2: Determine the direction of the electric force on charge q1. Since
charge q2is negative, the force on q1will be attractive. Therefore, the direction
of the electric force on charge q1is towards charge q2.
Step 3: Calculate the magnitude of the electric force on charge q2. Since the
magnitudes of the charges are the same, the magnitude of the force on q2will
be the same as on q1, which is 21.576 N.
Step 4: Determine the direction of the electric force on charge q2. Since
charge q1is positive, the force on q2will be attractive also. Therefore, the
direction of the electric force on charge q2is towards charge q1.
Question 3
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length 2. The charges are +2µC at the top vertex, −3µC at the bottom
left vertex, and +5µC at the bottom right vertex. Calculate the magnitude and
direction of the net electrostatic force acting on the +2µC charge due to the
other charges.
Solution
To find the net electrostatic force acting on the +2µC charge, we need to cal-
culate the individual forces due to each of the other charges and then add them
as vectors.
Step 1: Calculate the force due to the −3µC charge at the bottom left ver-
tex. The magnitude of the force between two point charges q1and q2separated
by distance ris given by Coulomb’s Law:
F=k|q1q2|
r2
where kis the Coulomb’s constant (k≈8.99 ×109Nm2/C2).
For the +2µC charge and the −3µC charge, the distance between them is the
height of the equilateral triangle, h=√3. So, the force between these charges
is:
Fleft =(8.99 ×109)|2×10−6×3×10−6|
(√3)2
Step 2: Calculate the direction of the force due to the −3µC charge. Since
the −3µC charge is negative, the force on the +2µC charge will be attractive,
i.e., towards the −3µC charge. Therefore, the direction is towards the bottom
left vertex.
Step 3: Calculate the force due to the +5µC charge at the bottom right
vertex. Using Coulomb’s Law, the force between the +2µC charge and the
2
+5µC charge is:
Fright =(8.99 ×109)|2×10−6×5×10−6|
(√3)2
Step 4: Calculate the direction of the force due to the +5µC charge. Since
both charges are positive, the force on the +2µC charge will be repulsive, i.e.,
away from the +5µC charge. Therefore, the direction is away from the bottom
right vertex.
Step 5: Find the net force on the +2µC charge. To find the net force, we
need to add these forces as vectors. Let the force due to the −3µC charge be
Fleft and the force due to the +5µC charge be
Fright. The net force is:
Fnet =
Fleft +
Fright
Now, calculate the magnitude and direction of the net force.
Question 4
Question
Two point charges, q1= +2 µC and q2=−3µC, are placed 10 cm apart.
Calculate the magnitude and direction of the electric force each charge exerts
on the other. Assume the charges are located on the x-axis, with q1at the origin
and q2at x= 10 cm.
Solution
Step 1: Calculate the distance between the charges in meters. Given that the
charges are 10 cm apart, the distance rin meters is:
r= 10 cm = 0.10 m
Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force between two point charges is given by Coulomb’s law:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging in the given values:
F=(8.99 ×109)(2 ×10−6)(3 ×10−6)
(0.10)2
Calculating:
F=53.94
0.01 = 5394 N
3
Therefore, the magnitude of the electric force between the charges is 5394
N.
Step 3: Determine the direction of the force. Since q1is positive and q2is
negative, the force between them is attractive and directed from q2towards q1.
Thus, the electric force Fon q1points towards q1. The electric force Fon
q2points towards q2.
In conclusion, the magnitude of the electric force each charge exerts on the
other is 5394 N, and the direction of the force on q1is towards q1, while the
direction of the force on q2is towards q2.
Question 5
Question
Two point charges, q1 = +2.0 µC and q2 = -4.0 µC, are placed 10.0 cm apart
in air. Calculate the magnitude and direction of the electric force that q1 exerts
on q2.
Solution
Step 1: Convert µC to Coulombs. Given that 1µC = 10−6C, we can convert the
charges as follows: q1 = 2.0µC = 2.0×10−6C q2 = −4.0µC =−4.0×10−6C
Step 2: Calculate the distance between the two charges. Given that the
charges are placed 10.0 cm apart, we convert this distance to meters: d=
10.0cm = 10.0×10−2m= 0.10m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
The electric force between the charges is given by Coulomb’s Law: F=k·|q1·q2|
d2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the values: F= (8.99 ×109)·|2.0×10−6·−4.0×10−6|
0.102
F= 71.92 ×10−12/0.01
F= 7.192 ×10−10N
Step 4: Find the direction of the force. Since q1 is positive and q2 is negative,
the force between them will be attractive. Therefore, the force exerted by q1
on q2 is directed towards q1.
The magnitude of the electric force that q1 exerts on q2 is 7.192 ×10−10N
and the direction is towards q1.
Question 6
Question
Two point charges, +2 µC and −3µC, are placed 6 cm apart in air. Calculate
the magnitude and direction of the electric force that one charge exerts on the
other.
4
Solution
Step 1: Convert the charges to Coulombs. We have +2 µC= 2 ×10−6C and
−3µC=−3×10−6C.
Step 2: Calculate the electric force using Coulomb’s Law:
The magnitude of the electric force between two point charges is given by:
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 charges, - ris the distance between the
charges.
Plugging in the values, we get:
F=8.99 ×109· |2×10−6· −3×10−6|
(0.06)2
Step 3: Calculate the magnitude of the electric force:
F=8.99 ×109·6×10−12
0.0036
F=53.94 ×10−3
0.0036
F= 14950 N
Step 4: Determine the direction of the electric force: The force is attractive
since one charge is positive and the other is negative. It acts along the line
joining the charges, from the positive charge to the negative charge.
Therefore, the magnitude of the electric force that the positive charge exerts
on the negative charge is 14950 N, directed from the positive charge to the
negative charge.
Question 7
Question
Three point charges are arranged as shown: - Charge q1= +2.0µC is located at
the origin. - Charge q2=−3.0µC is located at coordinates (0,0.6m). - Charge
q3= +1.0µC is located at coordinates (0.8m,0).
Calculate the net electric force acting on q1due to q2and q3.
Solution
Step 1: Find the electric force between q1and q2: The electric force F12 between
two point charges is given by Coulomb’s law:
F12 =k|q1||q2|
r2
12
5
where - kis Coulomb’s constant (8.99 ×109N m2/C2), - q1and q2are the
magnitudes of the point charges, and - r12 is the distance between the two
charges.
Given that q1= 2.0µC,q2=−3.0µC, and r12 = 0.6m, we have:
F12 =(8.99 ×109)×(2.0×10−6)×(3.0×10−6)
(0.6)2
Step 2: Calculate F12:
F12 =8.99 ×2.0×3.0
(0.6)2×103N
Step 3: Simplify the expression:
F12 =53.94
0.36 ×103N
F12 = 149.83 ×103N
F12 = 1.50 ×105N
Step 4: Find the direction of F12: The direction of F12 is repulsive since the
charges q1and q2are of opposite sign.
Step 5: Repeat steps 1-4 for the electric force F13 between q1and q3and
then find the net electric force acting on q1.
Question 8
Question
Two point charges, q1= +2.0µC and q2=−4.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the two
charges.
Solution
Step 1: Convert the charges to the standard unit of Coulombs. Given: q1=
+2.0µC= 2.0×10−6C, and q2=−4.0µC=−4.0×10−6C.
Step 2: Recall the formula for the magnitude of electric force between two
point charges:
F=k·|q1·q2|
r2
where kis the Coulomb constant (k= 8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the separation distance between the charges.
Step 3: Substitute the given values into the formula and solve for the electric
force.
F= (8.99 ×109)·|2.0×10−6· −4.0×10−6|
(0.10)2
6
F= (8.99 ×109)·8.0×10−12
0.01
F= 8.99 ×109·8.0×10−10
F= 71.92 ×10−1
F= 7.192 N
Step 4: Therefore, the magnitude of the electric force between the two
charges is 7.192 N.
Question 9
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed on the x-axis at
positions x=−2.0m and x= 2.0m, respectively. Calculate the magnitude and
direction of the electric force on q1due to q2.
Solution
Step 1: Calculate the distance between the two charges. Given: q1=−4.0µC=
−4.0×10−6Cq2= 6.0µC= 6.0×10−6Cx1=−2.0mx2= 2.0m
The distance rbetween the two charges is:
r=|x2−x1|=|2.0−(−2.0)|= 4.0m
Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force Fbetween two charges is given by Coulomb’s law:
F=k·|q1·q2|
r2
where k= 8.99 ×109N m2C−2is the Coulomb’s constant.
Plugging in the values:
F= 8.99 ×109·| − 4.0×10−6·6.0×10−6|
(4.0)2
F= 8.99 ×109·24 ×10−12
16
F= 8.99 ×109·1.5×10−12
F≈13.485 N
Step 3: Determine the direction of the force. The force on q1due to q2
is attractive since the charges have opposite signs. Therefore, the force acts
towards the positive xdirection.
So, the magnitude of the electric force on q1due to q2is approximately
13.485 N, and it acts in the positive xdirection.
7
Question 10
Question
Two point charges, q1=−3.5µC and q2= 1.8µC, are placed 12 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to SI units.
q1=−3.5µC =−3.5×10−6C
q2= 1.8µC = 1.8×10−6C
Step 2: Determine the distance between the charges in meters.
d= 12 cm = 0.12 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
F=k|q1||q2|
d2
Where: k= 8.99 ×109Nm2/C2is the Coulomb constant
Step 4: Substitute the given values into the formula and calculate.
F=(8.99 ×109)(3.5×10−6)(1.8×10−6)
0.122
F=56.735 ×10−15
0.0144
F= 56.735 ×10−15 ×69.444
F= 3938 ×10−15 N
F= 3.938 ×10−12 N
Therefore, the magnitude of the electric force between the charges is 3.938 ×
10−12 N.
Question 11
Question
Two point charges, Q1=−2.0µC and Q2= 4.0µC, are placed 0.10 meters
apart in a vacuum. Calculate the magnitude and direction of the net electric
force on Q1.
8
Solution
Given: Q1=−2.0µC
Q2= 4.0µC
Distance between charges, r= 0.10 m
k= 8.99 ×109N·m2/C2(Coulomb’s constant)
Step 1: Calculate the electric force between the two charges using Coulomb’s
law:
F=k·|Q1·Q2|
r2
F= 8.99 ×109·2.0×10−6·4.0×10−6
0.12
Step 2: Calculate the magnitude of electric force:
F= 8.99 ×109·8.0×10−12
0.01
F= 8.99 ×109·8.0×10−10
F= 7.192 ×10−1N
Step 3: Determine the direction of the net electric force: Since Q1and Q2
have opposite charges, the electric force on Q1will be attractive towards Q2.
Therefore, the net electric force on Q1is directed towards Q2.
Thus, the magnitude of the net electric force on Q1is 7.192 ×10−1N and
the direction is towards Q2.
Question 12
Question
Three point charges are placed at the corners of an equilateral triangle with
side length d= 2 m. The charges are +4µC,−3µC and +5µC. Calculate the
magnitude and direction of the net force on the +4µC charge due to the other
two charges.
Solution
Step 1: Calculate the distance between the +4µC charge and the −3µC charge
using the Pythagorean theorem. The distance r1can be found by dividing the
equilateral triangle into two right triangles.
r1=v
u
u
td
22
+ √3d
2+d!2
=r(1)2+√3+12
9
Step 2: Calculate the distance between the +4µC charge and the +5µC
charge. The distance r2can be found in the same manner.
r2=v
u
u
td
22
+ √3d
2!2
=r(1)2+√32
Step 3: Calculate the magnitude of the force F1between the +4µC and
−3µC charges using Coulomb’s law.
F1=k|q1||q2|
r2
1
Step 4: Calculate the magnitude of the force F2between the +4µC and
+5µC charges in the same way.
F2=k|q1||q3|
r2
2
Step 5: Determine the direction of the net force on the +4µC charge. Since
the forces are vectors, we need to consider their directions. The force F1will be
directed away from the −3µC charge, and the force F2will be directed towards
the +5µC charge. Calculate the net force by considering the vector sum of F1
and F2.
Question 13
Question
Three point charges are arranged as shown in the diagram below. Each charge
has a magnitude of 3.00 µC. Calculate the magnitude and direction of the net
force on the charge q3.
q1q2
q3
Solution
Step 1: Calculate the distances between the charges using the Pythagorean
theorem. Let the length of each side of the equilateral triangle be L.
L=p22+ 1.7322
L=√4+3
L=√7
10
Step 2: Calculate the electric force between charges using Coulomb’s law:
F=k|q1q2|
r2where - kis the Coulomb constant (8.99 ×109N·m2/C2), - q1
and q2are the magnitudes of the charges, and - ris the distance between the
charges.
Step 3: Calculate the force between q1and q3:
F13 = 8.99 ×109(3 ×10−6)(3 ×10−6)
(√7)2
Step 4: Calculate the force between q2and q3:
F23 = 8.99 ×109(3 ×10−6)(3 ×10−6)
(√7)2
Step 5: The net force on q3is the vector sum of the forces F13 and F23.
Since the forces are acting at 120° angles to each other, the net force magnitude
can be calculated using the law of cosines:
Net force =qF2
13 +F2
23 + 2F13F23 cos(120◦)
Step 6: Now, plug in the values to find the magnitude and direction of the
net force on charge q3.
Question 14
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length a. The magnitudes of the charges are q,2q, and 3q. Calculate
the net electric force on the charge qdue to the other two charges.
Solution
Let’s denote the charges as follows: q1=q,q2= 2q, and q3= 3q. The distances
from q1to q2and q3are both a.
By Coulomb’s law, the magnitude of the electric force between two charges
qiand qjis Fij =k|qi|·|qj|
r2, where kis Coulomb’s constant (8.99 ×109N m2/C2)
and ris the distance between the charges.
Step 1: We start by finding the electric force on charge q1due to charge q2.
Given that |q1|=q,|q2|= 2q, and r=a, we have:
F12 =k|q1|·|q2|
r2= 8.99 ×109q·2q
a2
Step 2: Next, we find the electric force on charge q1due to charge q3. Given
that |q1|=q,|q3|= 3q, and r=a, we have:
F13 =k|q1|·|q3|
r2= 8.99 ×109q·3q
a2
11
Step 3: Now, we calculate the net electric force on charge q1by summing
the forces due to q2and q3.
Fnet =F12 +F13 = 8.99 ×1092q2
a2+ 8.99 ×1093q2
a2
Therefore, the net electric force on the charge qdue to the other two charges
is:
Fnet = 8.99 ×1095q2
a2N
Question 15
Question
Two point charges, q1= +4.0µC and q2=−6.0µC, are placed 0.10 m apart in
a vacuum. Calculate the magnitude of the electric force that each charge exerts
on the other.
Solution
Let’s denote the magnitude of the electric force between two point charges as F
and their separation distance as r. We can use Coulomb’s law to calculate the
electric force between these two charges. Coulomb’s law is given by:
F=k|q1q2|
r2,
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Step 1: Calculate the electric force that charge q1exerts on charge q2. This
force will be attractive since the charges are of opposite signs.
F1→2=k|q1q2|
r2.
Plugging in the given values:
F1→2= (8.99 ×109N m2/C2)|(4.0×10−6C)(−6.0×10−6C)|
(0.10 m)2.
Step 2: Calculate the magnitude of the electric force between q1and q2.
F1→2= 8.99 ×109×(4.0×10−6)(6.0×10−6)
0.01 .
F1→2= 8.99 ×109×2.4×10−11.
F1→2= 21.576 ×10−2N.
Therefore, the magnitude of the electric force that q1exerts on q2is 2.16 ×
10−1N.
12
Step 3: The force that charge q2exerts on charge q1will be equal in mag-
nitude but opposite in direction to the force calculated in Step 2.
Hence, the magnitude of the electric force that each charge exerts on the
other is 2.16 ×10−1N.
Question 16
Question
Three point charges are placed at the vertices of an equilateral triangle with
side length aas shown below. The charges are +q,−2q, and +4q. Calculate the
magnitude of the electric force on the +qcharge due to the other two charges.
+4q
−2q+q
Solution
Step 1: Calculate the electric force on the +qcharge due to the +4qcharge.
The electric force Fbetween two charges q1and q2separated by a distance ris
given by Coulomb’s law:
F=kq1q2
r2
where kis the Coulomb constant (8.9875 ×109N m2/C2).
In this case, the charge q1= +q,q2= +4q, and the distance between them is
the side length of the equilateral triangle a. Since the charges are located at the
vertices of an equilateral triangle, the distance between them is a
2. Therefore,
the electric force on the +qcharge due to the +4qcharge is:
F=kq(+4q)
(a
2)2=k4q2
a2
4
=k16q2
a2=72kq2
a2
Step 2: Calculate the electric force on the +qcharge due to the −2qcharge.
Similarly, the electric force on the +qcharge due to the −2qcharge can be
calculated:
F=kq(−2q)
a2=−2kq2
a2=−18kq2
a2
Step 3: Combine the two forces. The total force on the +qcharge is the
vector sum of the forces due to the +4qand −2qcharges. Since the forces are
in opposite directions along the same line:
Ftotal =72kq2
a2−18kq2
a2=54kq2
a2
Therefore, the magnitude of the electric force on the +qcharge due to the
other two charges is 54kq2
a2.
13
Question 17
Question
Three point charges are arranged as follows: a charge of +2.0µC at the origin,
a charge of −4.0µC at (0,3m), and a charge of +6.0µC at (4 m,0). Calculate
the net electric force on the charge at the origin.
Solution
To find the net electric force on the charge at the origin, we need to calculate
the individual electric forces due to each of the other charges and then sum
them up vectorially.
Step 1: Calculate the electric force due to the charge of +2.0µC at the
origin.
The electric force F1due to this charge can be calculated using Coulomb’s
law:
F1=kq1q
r2,
where kis the Coulomb constant, q1and qare the magnitudes of the two charges,
and ris the distance between the charges.
The distance between the charges is 0since the charge creating the force is
at the origin. Hence, r= 0.
Therefore, the force due to the charge of +2.0µC is 0.
Step 2: Calculate the electric force due to the charge of −4.0µC at (0,3m).
The electric force F2due to this charge can be calculated using Coulomb’s
law:
F2=kq2q
r2,
where q2and qare the magnitudes of the two charges, and ris the distance
between the charges.
The distance between the charges is √02+ 32= 3 m.
Plugging in the values and accounting for the direction of the force, which
would be along the negative y-axis, we get:
F2=−6.75 Nˆ
j.
Step 3: Calculate the electric force due to the charge of +6.0µC at (4 m,0).
The electric force F3due to this charge can be calculated using Coulomb’s
law:
F3=kq3q
r2,
where q3and qare the magnitudes of the two charges, and ris the distance
between the charges.
The distance between the charges is √42+ 02= 4 m.
Plugging in the values and accounting for the direction of the force, which
would be along the positive x-axis, we get:
F3= 6.75 Nˆ
i.
14
Step 4: Find the net electric force on the charge at the origin.
The net force on the charge at the origin is the vector sum of the individual
forces:
Fnet =F1+F2+F3.
Substitute the calculated values and summing the forces vectorially, we get:
Fnet =−6.75 Nˆ
j+ 6.75 Nˆ
i.
Therefore, the net electric force on the charge at the origin is −6.75 Nˆ
j+
6.75 Nˆ
i.
Question 18
Question
Two point charges, Q1=−8.0µC and Q2= 4.0µC, are placed at points A
and B, respectively, with a distance of 40 cm between them. Calculate the
magnitude and direction of the electric force that charge Q1exerts on charge
Q2. Given that the Coulomb constant is k= 8.99 ×109N m2/C2.
Solution
Step 1: Find the magnitude of the electric force using the equation:
F=k· |Q1|·|Q2|
r2
where Fis the electric force, kis the Coulomb constant, |Q1|and |Q2|are the
magnitudes of the charges, and ris the distance between the charges.
Plugging in the given values:
F=(8.99 ×109N m2/C2)·(8.0×10−6C)·(4.0×10−6C)
(0.40 m)2
F=8.99 ×8.0×4.0
0.16 ×103N
F= 8.99 ×8.0×4.0×103N
F= 2.8768 ×105N
Therefore, the magnitude of the electric force that charge Q1exerts on charge
Q2is 2.88 ×105N.
Step 2: Find the direction of the electric force. Since charge Q1is negative,
the force on charge Q2will be attractive, pulling Q2towards Q1. Thus, the
direction of the electric force is from charge Q2towards charge Q1.
15
Question 19
Question
Two point charges, q1= 8.0�C and q2=−10.0�C, are placed 40.0cm apart.
Determine the magnitude and direction of the electric force that q2exerts on
q1.
Solution
Step 1: First, we need to calculate the electric force between the two charges
using Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Plug in the given values:
F=(8.99 ×109)· |8.0×10−6·(−10.0×10−6)|
(0.40)2
Step 3: Calculate the magnitude of the electric force:
F=(8.99 ×109)·(8.0×10−6·10.0×10−6)
0.16
F=8.99 ×109·80 ×10−12
0.16
F=7.192 ×10−3
0.16 = 0.04495 N
Step 4: Now, to determine the direction of the force, we consider the fact
that q2exerts an attractive force on q1(since the charges have opposite signs).
Therefore, the force on q1points towards q2.
Step 5: So, the electric force that q2exerts on q1has a magnitude of 0.04495 N
and direction towards q2.
Question 20
Question
Two point charges, q1=−3.0µC and q2= 6.0µC, are placed 10.0cm apart.
Calculate the magnitude of the electric force that each charge exerts on the
other.
16
Solution
Step 1: Convert the given charges to coulombs.
q1=−3.0×10−6C
q2= 6.0×10−6C
Step 2: Calculate the distance between the charges in meters. Given: r=
10.0cm = 0.10 m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k|q1||q2|
r2
where k= 8.99 ×109N·m2/C2.
Step 4: Substitute the known values and calculate the magnitude of the
electric force.
F=8.99 ×109N·m2/C2×3.0×10−6C×6.0×10−6C
(0.10 m)2
F=161.82
0.01 = 16182 N
Therefore, the magnitude of the electric force that each charge exerts on the
other is 16182 N.
Question 21
Question
Two point charges, q1= 3.0×10−6C and q2=−4.0×10−6C, are placed 10
cm apart in a vacuum. Calculate the magnitude of the electric force between
the charges.
Solution
Step 1: Convert all given values to SI units. Given: q1= 3.0×10−6C, q2=
−4.0×10−6C, r= 10 cm = 0.10 m.
Step 2: Calculate the magnitude of the electric force between the charges
using Coulomb’s Law: The magnitude of the electric force between two point
charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2,
where k= 8.99 ×109N m2/C2is the electrostatic constant.
17
Substitute the given values into the formula:
F=8.99 ×109×3.0×10−6×4.0×10−6
(0.10)2.
Step 3: Calculate the electric force.
F=8.99 ×109×3.0×10−6×4.0×10−6
(0.10)2=107.88
0.01 = 10788 N.
Therefore, the magnitude of the electric force between the charges is 10788
N.
Question 22
Question
Three point charges are located at the corners of an equilateral triangle with
side length a. Each charge has a magnitude of q. Calculate the magnitude of
the net electric force on one of the charges due to the other two charges.
Solution
Let’s label the charges q1,q2, and q3at the vertices of the equilateral triangle.
We will find the magnitude of the net electric force on q1due to q2and q3.
Step 1: Calculate the electric force from q2on q1. The magnitude of the
electric force between two point charges can be calculated using Coulomb’s Law:
F=k|q1q2|
r2
where kis the Coulomb constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
In this case, the distance between q1and q2is a. The direction of the force
will be along the line connecting q1and q2.
Step 2: Calculate the electric force from q3on q1. Similarly, we can calculate
the electric force from q3on q1using Coulomb’s Law. The distance between q1
and q3is also ain this equilateral triangle configuration.
Step 3: Find the angle between the forces. The forces from q2and q3on q1
are equal in magnitude but opposite in direction. They form a 120◦angle with
respect to each other.
Step 4: Calculate the net electric force on q1. To find the net electric force
on q1, we can treat the forces from q2and q3as vectors. Since they are at 120◦
to each other, we can treat them as components of a single force vector.
The net force magnitude will be the vector sum of the forces from q2and q3:
Fnet =qF2
2+F2
3+ 2F2F3cos(120◦)
Solving this expression will give us the magnitude of the net electric force
on q1due to q2and q3.
18
Question 23
Question
Two point charges, q1= +3.00 nC and q2=−5.00 nC are located on the x-axis
at positions x= 2.00 m and x=−4.00 m respectively. Calculate the magnitude
and direction of the electric force on q1at x= 2.00 m due to the presence of
charge q2.
Solution
Step 1: Calculate the distance between the charges. Given that x1= 2.00 m and
x2=−4.00 m, we can find the distance between the charges using the formula:
r=|x2−x1|
r=| − 4.00 m−2.00 m|
r=| − 6.00 m|
r= 6.00 m
Therefore, the distance between the charges is 6.00 m.
Step 2: Calculate the electric force. The magnitude of the electric force
between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
where - kis the electrostatic constant 8.99 ×109N m2/C2-q1= +3.00 nC =
3.00 ×10−9C - q2=−5.00 nC =−5.00 ×10−9C - r= 6.00 m
Plugging in the values:
F= (8.99 ×109)|3.00 ×10−9× −5.00 ×10−9|
(6.00)2
F= (8.99 ×109)15.00 ×10−18
36
F= (8.99 ×109)×0.4167
F≈3.75 ×109N
Step 3: Determine the direction of the force. The force on q1is towards q2
because the charges are opposite in sign. Thus, the force is attractive.
Therefore, the magnitude of the electric force on q1at x= 2.00 m due to
the presence of charge q2at x=−4.00 m is approximately 3.75 ×109N and is
attractive.
19
Question 24
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 9.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between these two
charges.
Solution
Step 1: Convert the charges to Coulombs. Given that 1µC= 10−6C, we have:
q1=−4.0µC=−4.0×10−6C and q2= 6.0µC= 6.0×10−6C.
Step 2: Calculate the electric force using Coulomb’s Law. 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 Coulomb’s constant with a value of 8.9875 ×109N m2/C2,q1and q2
are the magnitudes of the charges, and ris the separation distance.
Step 3: Substitute the given values into Coulomb’s Law and solve for the
electric force. Substitute k= 8.9875 ×109N m2/C2,q1=−4.0×10−6C,
q2= 6.0×10−6C, and r= 9.0cm = 0.09 m into Coulomb’s Law:
F= 8.9875 ×109×|(−4.0×10−6)(6.0×10−6)|
(0.09)2
Step 4: Calculate the magnitude of the electric force. Performing the calcu-
lations:
F= 8.9875 ×109×24 ×10−12
0.0081
F= 8.9875 ×109×2.96 ×10−9
F= 26.6N
Therefore, the magnitude of the electric force between the two charges is
26.6N.
Question 25
Question
Two point charges q1=−3.0µC and q2= 5.0µC are placed 25 cm apart.
Calculate the magnitude of the electric force between the charges.
20
Solution
Step 1: Convert the charges to Coulombs.
Given: q1=−3.0µC=−3.0×10−6C
q2= 5.0µC= 5.0×10−6C
Step 2: Calculate the distance between the charges.
Given: r= 25 cm = 0.25 m
Step 3: Calculate the magnitude of the electric force.
The electric force between two point charges is given by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the electrostatic constant with a value of 8.99 ×109Nm2/C2.
Substitute the given values into the formula:
F= (8.99 ×109)·|(−3.0×10−6)·(5.0×10−6)|
(0.25)2
Step 4: Calculate the magnitude of the electric force.
F= 8.99 ×109·|(−3.0×10−6)·(5.0×10−6)|
(0.25)2
F= 8.99 ×109·15 ×10−12
0.0625
F= 8.99 ×109·2.4×10−10
F= 2.16 ×10−1N
Therefore, the magnitude of the electric force between the charges is 2.16 ×
10−1N.
21
Step 2: Determine the direction of the electric force on charge q1. Since
charge q2is negative, the force on q1will be attractive. Therefore, the direction
of the electric force on charge q1is towards charge q2.
Step 3: Calculate the magnitude of the electric force on charge q2. Since the
magnitudes of the charges are the same, the magnitude of the force on q2will
be the same as on q1, which is 21.576 N.
Step 4: Determine the direction of the electric force on charge q2. Since
charge q1is positive, the force on q2will be attractive also. Therefore, the
direction of the electric force on charge q2is towards charge q1.
Question 3
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length 2. The charges are +2µC at the top vertex, −3µC at the bottom
left vertex, and +5µC at the bottom right vertex. Calculate the magnitude and
direction of the net electrostatic force acting on the +2µC charge due to the
other charges.
Solution
To find the net electrostatic force acting on the +2µC charge, we need to cal-
culate the individual forces due to each of the other charges and then add them
as vectors.
Step 1: Calculate the force due to the −3µC charge at the bottom left ver-
tex. The magnitude of the force between two point charges q1and q2separated
by distance ris given by Coulomb’s Law:
F=k|q1q2|
r2
where kis the Coulomb’s constant (k≈8.99 ×109Nm2/C2).
For the +2µC charge and the −3µC charge, the distance between them is the
height of the equilateral triangle, h=√3. So, the force between these charges
is:
Fleft =(8.99 ×109)|2×10−6×3×10−6|
(√3)2
Step 2: Calculate the direction of the force due to the −3µC charge. Since
the −3µC charge is negative, the force on the +2µC charge will be attractive,
i.e., towards the −3µC charge. Therefore, the direction is towards the bottom
left vertex.
Step 3: Calculate the force due to the +5µC charge at the bottom right
vertex. Using Coulomb’s Law, the force between the +2µC charge and the
2
+5µC charge is:
Fright =(8.99 ×109)|2×10−6×5×10−6|
(√3)2
Step 4: Calculate the direction of the force due to the +5µC charge. Since
both charges are positive, the force on the +2µC charge will be repulsive, i.e.,
away from the +5µC charge. Therefore, the direction is away from the bottom
right vertex.
Step 5: Find the net force on the +2µC charge. To find the net force, we
need to add these forces as vectors. Let the force due to the −3µC charge be
Fleft and the force due to the +5µC charge be
Fright. The net force is:
Fnet =
Fleft +
Fright
Now, calculate the magnitude and direction of the net force.
Question 4
Question
Two point charges, q1= +2 µC and q2=−3µC, are placed 10 cm apart.
Calculate the magnitude and direction of the electric force each charge exerts
on the other. Assume the charges are located on the x-axis, with q1at the origin
and q2at x= 10 cm.
Solution
Step 1: Calculate the distance between the charges in meters. Given that the
charges are 10 cm apart, the distance rin meters is:
r= 10 cm = 0.10 m
Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force between two point charges is given by Coulomb’s law:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging in the given values:
F=(8.99 ×109)(2 ×10−6)(3 ×10−6)
(0.10)2
Calculating:
F=53.94
0.01 = 5394 N
3
Therefore, the magnitude of the electric force between the charges is 5394
N.
Step 3: Determine the direction of the force. Since q1is positive and q2is
negative, the force between them is attractive and directed from q2towards q1.
Thus, the electric force Fon q1points towards q1. The electric force Fon
q2points towards q2.
In conclusion, the magnitude of the electric force each charge exerts on the
other is 5394 N, and the direction of the force on q1is towards q1, while the
direction of the force on q2is towards q2.
Question 5
Question
Two point charges, q1 = +2.0 µC and q2 = -4.0 µC, are placed 10.0 cm apart
in air. Calculate the magnitude and direction of the electric force that q1 exerts
on q2.
Solution
Step 1: Convert µC to Coulombs. Given that 1µC = 10−6C, we can convert the
charges as follows: q1 = 2.0µC = 2.0×10−6C q2 = −4.0µC =−4.0×10−6C
Step 2: Calculate the distance between the two charges. Given that the
charges are placed 10.0 cm apart, we convert this distance to meters: d=
10.0cm = 10.0×10−2m= 0.10m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
The electric force between the charges is given by Coulomb’s Law: F=k·|q1·q2|
d2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the values: F= (8.99 ×109)·|2.0×10−6·−4.0×10−6|
0.102
F= 71.92 ×10−12/0.01
F= 7.192 ×10−10N
Step 4: Find the direction of the force. Since q1 is positive and q2 is negative,
the force between them will be attractive. Therefore, the force exerted by q1
on q2 is directed towards q1.
The magnitude of the electric force that q1 exerts on q2 is 7.192 ×10−10N
and the direction is towards q1.
Question 6
Question
Two point charges, +2 µC and −3µC, are placed 6 cm apart in air. Calculate
the magnitude and direction of the electric force that one charge exerts on the
other.
4
Solution
Step 1: Convert the charges to Coulombs. We have +2 µC= 2 ×10−6C and
−3µC=−3×10−6C.
Step 2: Calculate the electric force using Coulomb’s Law:
The magnitude of the electric force between two point charges is given by:
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 charges, - ris the distance between the
charges.
Plugging in the values, we get:
F=8.99 ×109· |2×10−6· −3×10−6|
(0.06)2
Step 3: Calculate the magnitude of the electric force:
F=8.99 ×109·6×10−12
0.0036
F=53.94 ×10−3
0.0036
F= 14950 N
Step 4: Determine the direction of the electric force: The force is attractive
since one charge is positive and the other is negative. It acts along the line
joining the charges, from the positive charge to the negative charge.
Therefore, the magnitude of the electric force that the positive charge exerts
on the negative charge is 14950 N, directed from the positive charge to the
negative charge.
Question 7
Question
Three point charges are arranged as shown: - Charge q1= +2.0µC is located at
the origin. - Charge q2=−3.0µC is located at coordinates (0,0.6m). - Charge
q3= +1.0µC is located at coordinates (0.8m,0).
Calculate the net electric force acting on q1due to q2and q3.
Solution
Step 1: Find the electric force between q1and q2: The electric force F12 between
two point charges is given by Coulomb’s law:
F12 =k|q1||q2|
r2
12
5
where - kis Coulomb’s constant (8.99 ×109N m2/C2), - q1and q2are the
magnitudes of the point charges, and - r12 is the distance between the two
charges.
Given that q1= 2.0µC,q2=−3.0µC, and r12 = 0.6m, we have:
F12 =(8.99 ×109)×(2.0×10−6)×(3.0×10−6)
(0.6)2
Step 2: Calculate F12:
F12 =8.99 ×2.0×3.0
(0.6)2×103N
Step 3: Simplify the expression:
F12 =53.94
0.36 ×103N
F12 = 149.83 ×103N
F12 = 1.50 ×105N
Step 4: Find the direction of F12: The direction of F12 is repulsive since the
charges q1and q2are of opposite sign.
Step 5: Repeat steps 1-4 for the electric force F13 between q1and q3and
then find the net electric force acting on q1.
Question 8
Question
Two point charges, q1= +2.0µC and q2=−4.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the two
charges.
Solution
Step 1: Convert the charges to the standard unit of Coulombs. Given: q1=
+2.0µC= 2.0×10−6C, and q2=−4.0µC=−4.0×10−6C.
Step 2: Recall the formula for the magnitude of electric force between two
point charges:
F=k·|q1·q2|
r2
where kis the Coulomb constant (k= 8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the separation distance between the charges.
Step 3: Substitute the given values into the formula and solve for the electric
force.
F= (8.99 ×109)·|2.0×10−6· −4.0×10−6|
(0.10)2
6
F= (8.99 ×109)·8.0×10−12
0.01
F= 8.99 ×109·8.0×10−10
F= 71.92 ×10−1
F= 7.192 N
Step 4: Therefore, the magnitude of the electric force between the two
charges is 7.192 N.
Question 9
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed on the x-axis at
positions x=−2.0m and x= 2.0m, respectively. Calculate the magnitude and
direction of the electric force on q1due to q2.
Solution
Step 1: Calculate the distance between the two charges. Given: q1=−4.0µC=
−4.0×10−6Cq2= 6.0µC= 6.0×10−6Cx1=−2.0mx2= 2.0m
The distance rbetween the two charges is:
r=|x2−x1|=|2.0−(−2.0)|= 4.0m
Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force Fbetween two charges is given by Coulomb’s law:
F=k·|q1·q2|
r2
where k= 8.99 ×109N m2C−2is the Coulomb’s constant.
Plugging in the values:
F= 8.99 ×109·| − 4.0×10−6·6.0×10−6|
(4.0)2
F= 8.99 ×109·24 ×10−12
16
F= 8.99 ×109·1.5×10−12
F≈13.485 N
Step 3: Determine the direction of the force. The force on q1due to q2
is attractive since the charges have opposite signs. Therefore, the force acts
towards the positive xdirection.
So, the magnitude of the electric force on q1due to q2is approximately
13.485 N, and it acts in the positive xdirection.
7
Question 10
Question
Two point charges, q1=−3.5µC and q2= 1.8µC, are placed 12 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to SI units.
q1=−3.5µC =−3.5×10−6C
q2= 1.8µC = 1.8×10−6C
Step 2: Determine the distance between the charges in meters.
d= 12 cm = 0.12 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
F=k|q1||q2|
d2
Where: k= 8.99 ×109Nm2/C2is the Coulomb constant
Step 4: Substitute the given values into the formula and calculate.
F=(8.99 ×109)(3.5×10−6)(1.8×10−6)
0.122
F=56.735 ×10−15
0.0144
F= 56.735 ×10−15 ×69.444
F= 3938 ×10−15 N
F= 3.938 ×10−12 N
Therefore, the magnitude of the electric force between the charges is 3.938 ×
10−12 N.
Question 11
Question
Two point charges, Q1=−2.0µC and Q2= 4.0µC, are placed 0.10 meters
apart in a vacuum. Calculate the magnitude and direction of the net electric
force on Q1.
8
Solution
Given: Q1=−2.0µC
Q2= 4.0µC
Distance between charges, r= 0.10 m
k= 8.99 ×109N·m2/C2(Coulomb’s constant)
Step 1: Calculate the electric force between the two charges using Coulomb’s
law:
F=k·|Q1·Q2|
r2
F= 8.99 ×109·2.0×10−6·4.0×10−6
0.12
Step 2: Calculate the magnitude of electric force:
F= 8.99 ×109·8.0×10−12
0.01
F= 8.99 ×109·8.0×10−10
F= 7.192 ×10−1N
Step 3: Determine the direction of the net electric force: Since Q1and Q2
have opposite charges, the electric force on Q1will be attractive towards Q2.
Therefore, the net electric force on Q1is directed towards Q2.
Thus, the magnitude of the net electric force on Q1is 7.192 ×10−1N and
the direction is towards Q2.
Question 12
Question
Three point charges are placed at the corners of an equilateral triangle with
side length d= 2 m. The charges are +4µC,−3µC and +5µC. Calculate the
magnitude and direction of the net force on the +4µC charge due to the other
two charges.
Solution
Step 1: Calculate the distance between the +4µC charge and the −3µC charge
using the Pythagorean theorem. The distance r1can be found by dividing the
equilateral triangle into two right triangles.
r1=v
u
u
td
22
+ √3d
2+d!2
=r(1)2+√3+12
9
Step 2: Calculate the distance between the +4µC charge and the +5µC
charge. The distance r2can be found in the same manner.
r2=v
u
u
td
22
+ √3d
2!2
=r(1)2+√32
Step 3: Calculate the magnitude of the force F1between the +4µC and
−3µC charges using Coulomb’s law.
F1=k|q1||q2|
r2
1
Step 4: Calculate the magnitude of the force F2between the +4µC and
+5µC charges in the same way.
F2=k|q1||q3|
r2
2
Step 5: Determine the direction of the net force on the +4µC charge. Since
the forces are vectors, we need to consider their directions. The force F1will be
directed away from the −3µC charge, and the force F2will be directed towards
the +5µC charge. Calculate the net force by considering the vector sum of F1
and F2.
Question 13
Question
Three point charges are arranged as shown in the diagram below. Each charge
has a magnitude of 3.00 µC. Calculate the magnitude and direction of the net
force on the charge q3.
q1q2
q3
Solution
Step 1: Calculate the distances between the charges using the Pythagorean
theorem. Let the length of each side of the equilateral triangle be L.
L=p22+ 1.7322
L=√4+3
L=√7
10
Step 2: Calculate the electric force between charges using Coulomb’s law:
F=k|q1q2|
r2where - kis the Coulomb constant (8.99 ×109N·m2/C2), - q1
and q2are the magnitudes of the charges, and - ris the distance between the
charges.
Step 3: Calculate the force between q1and q3:
F13 = 8.99 ×109(3 ×10−6)(3 ×10−6)
(√7)2
Step 4: Calculate the force between q2and q3:
F23 = 8.99 ×109(3 ×10−6)(3 ×10−6)
(√7)2
Step 5: The net force on q3is the vector sum of the forces F13 and F23.
Since the forces are acting at 120° angles to each other, the net force magnitude
can be calculated using the law of cosines:
Net force =qF2
13 +F2
23 + 2F13F23 cos(120◦)
Step 6: Now, plug in the values to find the magnitude and direction of the
net force on charge q3.
Question 14
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length a. The magnitudes of the charges are q,2q, and 3q. Calculate
the net electric force on the charge qdue to the other two charges.
Solution
Let’s denote the charges as follows: q1=q,q2= 2q, and q3= 3q. The distances
from q1to q2and q3are both a.
By Coulomb’s law, the magnitude of the electric force between two charges
qiand qjis Fij =k|qi|·|qj|
r2, where kis Coulomb’s constant (8.99 ×109N m2/C2)
and ris the distance between the charges.
Step 1: We start by finding the electric force on charge q1due to charge q2.
Given that |q1|=q,|q2|= 2q, and r=a, we have:
F12 =k|q1|·|q2|
r2= 8.99 ×109q·2q
a2
Step 2: Next, we find the electric force on charge q1due to charge q3. Given
that |q1|=q,|q3|= 3q, and r=a, we have:
F13 =k|q1|·|q3|
r2= 8.99 ×109q·3q
a2
11
Step 3: Now, we calculate the net electric force on charge q1by summing
the forces due to q2and q3.
Fnet =F12 +F13 = 8.99 ×1092q2
a2+ 8.99 ×1093q2
a2
Therefore, the net electric force on the charge qdue to the other two charges
is:
Fnet = 8.99 ×1095q2
a2N
Question 15
Question
Two point charges, q1= +4.0µC and q2=−6.0µC, are placed 0.10 m apart in
a vacuum. Calculate the magnitude of the electric force that each charge exerts
on the other.
Solution
Let’s denote the magnitude of the electric force between two point charges as F
and their separation distance as r. We can use Coulomb’s law to calculate the
electric force between these two charges. Coulomb’s law is given by:
F=k|q1q2|
r2,
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Step 1: Calculate the electric force that charge q1exerts on charge q2. This
force will be attractive since the charges are of opposite signs.
F1→2=k|q1q2|
r2.
Plugging in the given values:
F1→2= (8.99 ×109N m2/C2)|(4.0×10−6C)(−6.0×10−6C)|
(0.10 m)2.
Step 2: Calculate the magnitude of the electric force between q1and q2.
F1→2= 8.99 ×109×(4.0×10−6)(6.0×10−6)
0.01 .
F1→2= 8.99 ×109×2.4×10−11.
F1→2= 21.576 ×10−2N.
Therefore, the magnitude of the electric force that q1exerts on q2is 2.16 ×
10−1N.
12
Step 3: The force that charge q2exerts on charge q1will be equal in mag-
nitude but opposite in direction to the force calculated in Step 2.
Hence, the magnitude of the electric force that each charge exerts on the
other is 2.16 ×10−1N.
Question 16
Question
Three point charges are placed at the vertices of an equilateral triangle with
side length aas shown below. The charges are +q,−2q, and +4q. Calculate the
magnitude of the electric force on the +qcharge due to the other two charges.
+4q
−2q+q
Solution
Step 1: Calculate the electric force on the +qcharge due to the +4qcharge.
The electric force Fbetween two charges q1and q2separated by a distance ris
given by Coulomb’s law:
F=kq1q2
r2
where kis the Coulomb constant (8.9875 ×109N m2/C2).
In this case, the charge q1= +q,q2= +4q, and the distance between them is
the side length of the equilateral triangle a. Since the charges are located at the
vertices of an equilateral triangle, the distance between them is a
2. Therefore,
the electric force on the +qcharge due to the +4qcharge is:
F=kq(+4q)
(a
2)2=k4q2
a2
4
=k16q2
a2=72kq2
a2
Step 2: Calculate the electric force on the +qcharge due to the −2qcharge.
Similarly, the electric force on the +qcharge due to the −2qcharge can be
calculated:
F=kq(−2q)
a2=−2kq2
a2=−18kq2
a2
Step 3: Combine the two forces. The total force on the +qcharge is the
vector sum of the forces due to the +4qand −2qcharges. Since the forces are
in opposite directions along the same line:
Ftotal =72kq2
a2−18kq2
a2=54kq2
a2
Therefore, the magnitude of the electric force on the +qcharge due to the
other two charges is 54kq2
a2.
13
Question 17
Question
Three point charges are arranged as follows: a charge of +2.0µC at the origin,
a charge of −4.0µC at (0,3m), and a charge of +6.0µC at (4 m,0). Calculate
the net electric force on the charge at the origin.
Solution
To find the net electric force on the charge at the origin, we need to calculate
the individual electric forces due to each of the other charges and then sum
them up vectorially.
Step 1: Calculate the electric force due to the charge of +2.0µC at the
origin.
The electric force F1due to this charge can be calculated using Coulomb’s
law:
F1=kq1q
r2,
where kis the Coulomb constant, q1and qare the magnitudes of the two charges,
and ris the distance between the charges.
The distance between the charges is 0since the charge creating the force is
at the origin. Hence, r= 0.
Therefore, the force due to the charge of +2.0µC is 0.
Step 2: Calculate the electric force due to the charge of −4.0µC at (0,3m).
The electric force F2due to this charge can be calculated using Coulomb’s
law:
F2=kq2q
r2,
where q2and qare the magnitudes of the two charges, and ris the distance
between the charges.
The distance between the charges is √02+ 32= 3 m.
Plugging in the values and accounting for the direction of the force, which
would be along the negative y-axis, we get:
F2=−6.75 Nˆ
j.
Step 3: Calculate the electric force due to the charge of +6.0µC at (4 m,0).
The electric force F3due to this charge can be calculated using Coulomb’s
law:
F3=kq3q
r2,
where q3and qare the magnitudes of the two charges, and ris the distance
between the charges.
The distance between the charges is √42+ 02= 4 m.
Plugging in the values and accounting for the direction of the force, which
would be along the positive x-axis, we get:
F3= 6.75 Nˆ
i.
14
Step 4: Find the net electric force on the charge at the origin.
The net force on the charge at the origin is the vector sum of the individual
forces:
Fnet =F1+F2+F3.
Substitute the calculated values and summing the forces vectorially, we get:
Fnet =−6.75 Nˆ
j+ 6.75 Nˆ
i.
Therefore, the net electric force on the charge at the origin is −6.75 Nˆ
j+
6.75 Nˆ
i.
Question 18
Question
Two point charges, Q1=−8.0µC and Q2= 4.0µC, are placed at points A
and B, respectively, with a distance of 40 cm between them. Calculate the
magnitude and direction of the electric force that charge Q1exerts on charge
Q2. Given that the Coulomb constant is k= 8.99 ×109N m2/C2.
Solution
Step 1: Find the magnitude of the electric force using the equation:
F=k· |Q1|·|Q2|
r2
where Fis the electric force, kis the Coulomb constant, |Q1|and |Q2|are the
magnitudes of the charges, and ris the distance between the charges.
Plugging in the given values:
F=(8.99 ×109N m2/C2)·(8.0×10−6C)·(4.0×10−6C)
(0.40 m)2
F=8.99 ×8.0×4.0
0.16 ×103N
F= 8.99 ×8.0×4.0×103N
F= 2.8768 ×105N
Therefore, the magnitude of the electric force that charge Q1exerts on charge
Q2is 2.88 ×105N.
Step 2: Find the direction of the electric force. Since charge Q1is negative,
the force on charge Q2will be attractive, pulling Q2towards Q1. Thus, the
direction of the electric force is from charge Q2towards charge Q1.
15
Question 19
Question
Two point charges, q1= 8.0�C and q2=−10.0�C, are placed 40.0cm apart.
Determine the magnitude and direction of the electric force that q2exerts on
q1.
Solution
Step 1: First, we need to calculate the electric force between the two charges
using Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Plug in the given values:
F=(8.99 ×109)· |8.0×10−6·(−10.0×10−6)|
(0.40)2
Step 3: Calculate the magnitude of the electric force:
F=(8.99 ×109)·(8.0×10−6·10.0×10−6)
0.16
F=8.99 ×109·80 ×10−12
0.16
F=7.192 ×10−3
0.16 = 0.04495 N
Step 4: Now, to determine the direction of the force, we consider the fact
that q2exerts an attractive force on q1(since the charges have opposite signs).
Therefore, the force on q1points towards q2.
Step 5: So, the electric force that q2exerts on q1has a magnitude of 0.04495 N
and direction towards q2.
Question 20
Question
Two point charges, q1=−3.0µC and q2= 6.0µC, are placed 10.0cm apart.
Calculate the magnitude of the electric force that each charge exerts on the
other.
16
Solution
Step 1: Convert the given charges to coulombs.
q1=−3.0×10−6C
q2= 6.0×10−6C
Step 2: Calculate the distance between the charges in meters. Given: r=
10.0cm = 0.10 m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k|q1||q2|
r2
where k= 8.99 ×109N·m2/C2.
Step 4: Substitute the known values and calculate the magnitude of the
electric force.
F=8.99 ×109N·m2/C2×3.0×10−6C×6.0×10−6C
(0.10 m)2
F=161.82
0.01 = 16182 N
Therefore, the magnitude of the electric force that each charge exerts on the
other is 16182 N.
Question 21
Question
Two point charges, q1= 3.0×10−6C and q2=−4.0×10−6C, are placed 10
cm apart in a vacuum. Calculate the magnitude of the electric force between
the charges.
Solution
Step 1: Convert all given values to SI units. Given: q1= 3.0×10−6C, q2=
−4.0×10−6C, r= 10 cm = 0.10 m.
Step 2: Calculate the magnitude of the electric force between the charges
using Coulomb’s Law: The magnitude of the electric force between two point
charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2,
where k= 8.99 ×109N m2/C2is the electrostatic constant.
17
Substitute the given values into the formula:
F=8.99 ×109×3.0×10−6×4.0×10−6
(0.10)2.
Step 3: Calculate the electric force.
F=8.99 ×109×3.0×10−6×4.0×10−6
(0.10)2=107.88
0.01 = 10788 N.
Therefore, the magnitude of the electric force between the charges is 10788
N.
Question 22
Question
Three point charges are located at the corners of an equilateral triangle with
side length a. Each charge has a magnitude of q. Calculate the magnitude of
the net electric force on one of the charges due to the other two charges.
Solution
Let’s label the charges q1,q2, and q3at the vertices of the equilateral triangle.
We will find the magnitude of the net electric force on q1due to q2and q3.
Step 1: Calculate the electric force from q2on q1. The magnitude of the
electric force between two point charges can be calculated using Coulomb’s Law:
F=k|q1q2|
r2
where kis the Coulomb constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
In this case, the distance between q1and q2is a. The direction of the force
will be along the line connecting q1and q2.
Step 2: Calculate the electric force from q3on q1. Similarly, we can calculate
the electric force from q3on q1using Coulomb’s Law. The distance between q1
and q3is also ain this equilateral triangle configuration.
Step 3: Find the angle between the forces. The forces from q2and q3on q1
are equal in magnitude but opposite in direction. They form a 120◦angle with
respect to each other.
Step 4: Calculate the net electric force on q1. To find the net electric force
on q1, we can treat the forces from q2and q3as vectors. Since they are at 120◦
to each other, we can treat them as components of a single force vector.
The net force magnitude will be the vector sum of the forces from q2and q3:
Fnet =qF2
2+F2
3+ 2F2F3cos(120◦)
Solving this expression will give us the magnitude of the net electric force
on q1due to q2and q3.
18
Question 23
Question
Two point charges, q1= +3.00 nC and q2=−5.00 nC are located on the x-axis
at positions x= 2.00 m and x=−4.00 m respectively. Calculate the magnitude
and direction of the electric force on q1at x= 2.00 m due to the presence of
charge q2.
Solution
Step 1: Calculate the distance between the charges. Given that x1= 2.00 m and
x2=−4.00 m, we can find the distance between the charges using the formula:
r=|x2−x1|
r=| − 4.00 m−2.00 m|
r=| − 6.00 m|
r= 6.00 m
Therefore, the distance between the charges is 6.00 m.
Step 2: Calculate the electric force. The magnitude of the electric force
between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
where - kis the electrostatic constant 8.99 ×109N m2/C2-q1= +3.00 nC =
3.00 ×10−9C - q2=−5.00 nC =−5.00 ×10−9C - r= 6.00 m
Plugging in the values:
F= (8.99 ×109)|3.00 ×10−9× −5.00 ×10−9|
(6.00)2
F= (8.99 ×109)15.00 ×10−18
36
F= (8.99 ×109)×0.4167
F≈3.75 ×109N
Step 3: Determine the direction of the force. The force on q1is towards q2
because the charges are opposite in sign. Thus, the force is attractive.
Therefore, the magnitude of the electric force on q1at x= 2.00 m due to
the presence of charge q2at x=−4.00 m is approximately 3.75 ×109N and is
attractive.
19
Question 24
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 9.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between these two
charges.
Solution
Step 1: Convert the charges to Coulombs. Given that 1µC= 10−6C, we have:
q1=−4.0µC=−4.0×10−6C and q2= 6.0µC= 6.0×10−6C.
Step 2: Calculate the electric force using Coulomb’s Law. 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 Coulomb’s constant with a value of 8.9875 ×109N m2/C2,q1and q2
are the magnitudes of the charges, and ris the separation distance.
Step 3: Substitute the given values into Coulomb’s Law and solve for the
electric force. Substitute k= 8.9875 ×109N m2/C2,q1=−4.0×10−6C,
q2= 6.0×10−6C, and r= 9.0cm = 0.09 m into Coulomb’s Law:
F= 8.9875 ×109×|(−4.0×10−6)(6.0×10−6)|
(0.09)2
Step 4: Calculate the magnitude of the electric force. Performing the calcu-
lations:
F= 8.9875 ×109×24 ×10−12
0.0081
F= 8.9875 ×109×2.96 ×10−9
F= 26.6N
Therefore, the magnitude of the electric force between the two charges is
26.6N.
Question 25
Question
Two point charges q1=−3.0µC and q2= 5.0µC are placed 25 cm apart.
Calculate the magnitude of the electric force between the charges.
20
Solution
Step 1: Convert the charges to Coulombs.
Given: q1=−3.0µC=−3.0×10−6C
q2= 5.0µC= 5.0×10−6C
Step 2: Calculate the distance between the charges.
Given: r= 25 cm = 0.25 m
Step 3: Calculate the magnitude of the electric force.
The electric force between two point charges is given by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the electrostatic constant with a value of 8.99 ×109Nm2/C2.
Substitute the given values into the formula:
F= (8.99 ×109)·|(−3.0×10−6)·(5.0×10−6)|
(0.25)2
Step 4: Calculate the magnitude of the electric force.
F= 8.99 ×109·|(−3.0×10−6)·(5.0×10−6)|
(0.25)2
F= 8.99 ×109·15 ×10−12
0.0625
F= 8.99 ×109·2.4×10−10
F= 2.16 ×10−1N
Therefore, the magnitude of the electric force between the charges is 2.16 ×
10−1N.
21
Step 2: Determine the direction of the electric force on charge q1. Since
charge q2is negative, the force on q1will be attractive. Therefore, the direction
of the electric force on charge q1is towards charge q2.
Step 3: Calculate the magnitude of the electric force on charge q2. Since the
magnitudes of the charges are the same, the magnitude of the force on q2will
be the same as on q1, which is 21.576 N.
Step 4: Determine the direction of the electric force on charge q2. Since
charge q1is positive, the force on q2will be attractive also. Therefore, the
direction of the electric force on charge q2is towards charge q1.
Question 3
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length 2. The charges are +2µC at the top vertex, −3µC at the bottom
left vertex, and +5µC at the bottom right vertex. Calculate the magnitude and
direction of the net electrostatic force acting on the +2µC charge due to the
other charges.
Solution
To find the net electrostatic force acting on the +2µC charge, we need to cal-
culate the individual forces due to each of the other charges and then add them
as vectors.
Step 1: Calculate the force due to the −3µC charge at the bottom left ver-
tex. The magnitude of the force between two point charges q1and q2separated
by distance ris given by Coulomb’s Law:
F=k|q1q2|
r2
where kis the Coulomb’s constant (k≈8.99 ×109Nm2/C2).
For the +2µC charge and the −3µC charge, the distance between them is the
height of the equilateral triangle, h=√3. So, the force between these charges
is:
Fleft =(8.99 ×109)|2×10−6×3×10−6|
(√3)2
Step 2: Calculate the direction of the force due to the −3µC charge. Since
the −3µC charge is negative, the force on the +2µC charge will be attractive,
i.e., towards the −3µC charge. Therefore, the direction is towards the bottom
left vertex.
Step 3: Calculate the force due to the +5µC charge at the bottom right
vertex. Using Coulomb’s Law, the force between the +2µC charge and the
2
+5µC charge is:
Fright =(8.99 ×109)|2×10−6×5×10−6|
(√3)2
Step 4: Calculate the direction of the force due to the +5µC charge. Since
both charges are positive, the force on the +2µC charge will be repulsive, i.e.,
away from the +5µC charge. Therefore, the direction is away from the bottom
right vertex.
Step 5: Find the net force on the +2µC charge. To find the net force, we
need to add these forces as vectors. Let the force due to the −3µC charge be
Fleft and the force due to the +5µC charge be
Fright. The net force is:
Fnet =
Fleft +
Fright
Now, calculate the magnitude and direction of the net force.
Question 4
Question
Two point charges, q1= +2 µC and q2=−3µC, are placed 10 cm apart.
Calculate the magnitude and direction of the electric force each charge exerts
on the other. Assume the charges are located on the x-axis, with q1at the origin
and q2at x= 10 cm.
Solution
Step 1: Calculate the distance between the charges in meters. Given that the
charges are 10 cm apart, the distance rin meters is:
r= 10 cm = 0.10 m
Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force between two point charges is given by Coulomb’s law:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging in the given values:
F=(8.99 ×109)(2 ×10−6)(3 ×10−6)
(0.10)2
Calculating:
F=53.94
0.01 = 5394 N
3
Therefore, the magnitude of the electric force between the charges is 5394
N.
Step 3: Determine the direction of the force. Since q1is positive and q2is
negative, the force between them is attractive and directed from q2towards q1.
Thus, the electric force Fon q1points towards q1. The electric force Fon
q2points towards q2.
In conclusion, the magnitude of the electric force each charge exerts on the
other is 5394 N, and the direction of the force on q1is towards q1, while the
direction of the force on q2is towards q2.
Question 5
Question
Two point charges, q1 = +2.0 µC and q2 = -4.0 µC, are placed 10.0 cm apart
in air. Calculate the magnitude and direction of the electric force that q1 exerts
on q2.
Solution
Step 1: Convert µC to Coulombs. Given that 1µC = 10−6C, we can convert the
charges as follows: q1 = 2.0µC = 2.0×10−6C q2 = −4.0µC =−4.0×10−6C
Step 2: Calculate the distance between the two charges. Given that the
charges are placed 10.0 cm apart, we convert this distance to meters: d=
10.0cm = 10.0×10−2m= 0.10m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
The electric force between the charges is given by Coulomb’s Law: F=k·|q1·q2|
d2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the values: F= (8.99 ×109)·|2.0×10−6·−4.0×10−6|
0.102
F= 71.92 ×10−12/0.01
F= 7.192 ×10−10N
Step 4: Find the direction of the force. Since q1 is positive and q2 is negative,
the force between them will be attractive. Therefore, the force exerted by q1
on q2 is directed towards q1.
The magnitude of the electric force that q1 exerts on q2 is 7.192 ×10−10N
and the direction is towards q1.
Question 6
Question
Two point charges, +2 µC and −3µC, are placed 6 cm apart in air. Calculate
the magnitude and direction of the electric force that one charge exerts on the
other.
4
Solution
Step 1: Convert the charges to Coulombs. We have +2 µC= 2 ×10−6C and
−3µC=−3×10−6C.
Step 2: Calculate the electric force using Coulomb’s Law:
The magnitude of the electric force between two point charges is given by:
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 charges, - ris the distance between the
charges.
Plugging in the values, we get:
F=8.99 ×109· |2×10−6· −3×10−6|
(0.06)2
Step 3: Calculate the magnitude of the electric force:
F=8.99 ×109·6×10−12
0.0036
F=53.94 ×10−3
0.0036
F= 14950 N
Step 4: Determine the direction of the electric force: The force is attractive
since one charge is positive and the other is negative. It acts along the line
joining the charges, from the positive charge to the negative charge.
Therefore, the magnitude of the electric force that the positive charge exerts
on the negative charge is 14950 N, directed from the positive charge to the
negative charge.
Question 7
Question
Three point charges are arranged as shown: - Charge q1= +2.0µC is located at
the origin. - Charge q2=−3.0µC is located at coordinates (0,0.6m). - Charge
q3= +1.0µC is located at coordinates (0.8m,0).
Calculate the net electric force acting on q1due to q2and q3.
Solution
Step 1: Find the electric force between q1and q2: The electric force F12 between
two point charges is given by Coulomb’s law:
F12 =k|q1||q2|
r2
12
5
where - kis Coulomb’s constant (8.99 ×109N m2/C2), - q1and q2are the
magnitudes of the point charges, and - r12 is the distance between the two
charges.
Given that q1= 2.0µC,q2=−3.0µC, and r12 = 0.6m, we have:
F12 =(8.99 ×109)×(2.0×10−6)×(3.0×10−6)
(0.6)2
Step 2: Calculate F12:
F12 =8.99 ×2.0×3.0
(0.6)2×103N
Step 3: Simplify the expression:
F12 =53.94
0.36 ×103N
F12 = 149.83 ×103N
F12 = 1.50 ×105N
Step 4: Find the direction of F12: The direction of F12 is repulsive since the
charges q1and q2are of opposite sign.
Step 5: Repeat steps 1-4 for the electric force F13 between q1and q3and
then find the net electric force acting on q1.
Question 8
Question
Two point charges, q1= +2.0µC and q2=−4.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the two
charges.
Solution
Step 1: Convert the charges to the standard unit of Coulombs. Given: q1=
+2.0µC= 2.0×10−6C, and q2=−4.0µC=−4.0×10−6C.
Step 2: Recall the formula for the magnitude of electric force between two
point charges:
F=k·|q1·q2|
r2
where kis the Coulomb constant (k= 8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the separation distance between the charges.
Step 3: Substitute the given values into the formula and solve for the electric
force.
F= (8.99 ×109)·|2.0×10−6· −4.0×10−6|
(0.10)2
6
F= (8.99 ×109)·8.0×10−12
0.01
F= 8.99 ×109·8.0×10−10
F= 71.92 ×10−1
F= 7.192 N
Step 4: Therefore, the magnitude of the electric force between the two
charges is 7.192 N.
Question 9
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed on the x-axis at
positions x=−2.0m and x= 2.0m, respectively. Calculate the magnitude and
direction of the electric force on q1due to q2.
Solution
Step 1: Calculate the distance between the two charges. Given: q1=−4.0µC=
−4.0×10−6Cq2= 6.0µC= 6.0×10−6Cx1=−2.0mx2= 2.0m
The distance rbetween the two charges is:
r=|x2−x1|=|2.0−(−2.0)|= 4.0m
Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force Fbetween two charges is given by Coulomb’s law:
F=k·|q1·q2|
r2
where k= 8.99 ×109N m2C−2is the Coulomb’s constant.
Plugging in the values:
F= 8.99 ×109·| − 4.0×10−6·6.0×10−6|
(4.0)2
F= 8.99 ×109·24 ×10−12
16
F= 8.99 ×109·1.5×10−12
F≈13.485 N
Step 3: Determine the direction of the force. The force on q1due to q2
is attractive since the charges have opposite signs. Therefore, the force acts
towards the positive xdirection.
So, the magnitude of the electric force on q1due to q2is approximately
13.485 N, and it acts in the positive xdirection.
7
Question 10
Question
Two point charges, q1=−3.5µC and q2= 1.8µC, are placed 12 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to SI units.
q1=−3.5µC =−3.5×10−6C
q2= 1.8µC = 1.8×10−6C
Step 2: Determine the distance between the charges in meters.
d= 12 cm = 0.12 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
F=k|q1||q2|
d2
Where: k= 8.99 ×109Nm2/C2is the Coulomb constant
Step 4: Substitute the given values into the formula and calculate.
F=(8.99 ×109)(3.5×10−6)(1.8×10−6)
0.122
F=56.735 ×10−15
0.0144
F= 56.735 ×10−15 ×69.444
F= 3938 ×10−15 N
F= 3.938 ×10−12 N
Therefore, the magnitude of the electric force between the charges is 3.938 ×
10−12 N.
Question 11
Question
Two point charges, Q1=−2.0µC and Q2= 4.0µC, are placed 0.10 meters
apart in a vacuum. Calculate the magnitude and direction of the net electric
force on Q1.
8
Solution
Given: Q1=−2.0µC
Q2= 4.0µC
Distance between charges, r= 0.10 m
k= 8.99 ×109N·m2/C2(Coulomb’s constant)
Step 1: Calculate the electric force between the two charges using Coulomb’s
law:
F=k·|Q1·Q2|
r2
F= 8.99 ×109·2.0×10−6·4.0×10−6
0.12
Step 2: Calculate the magnitude of electric force:
F= 8.99 ×109·8.0×10−12
0.01
F= 8.99 ×109·8.0×10−10
F= 7.192 ×10−1N
Step 3: Determine the direction of the net electric force: Since Q1and Q2
have opposite charges, the electric force on Q1will be attractive towards Q2.
Therefore, the net electric force on Q1is directed towards Q2.
Thus, the magnitude of the net electric force on Q1is 7.192 ×10−1N and
the direction is towards Q2.
Question 12
Question
Three point charges are placed at the corners of an equilateral triangle with
side length d= 2 m. The charges are +4µC,−3µC and +5µC. Calculate the
magnitude and direction of the net force on the +4µC charge due to the other
two charges.
Solution
Step 1: Calculate the distance between the +4µC charge and the −3µC charge
using the Pythagorean theorem. The distance r1can be found by dividing the
equilateral triangle into two right triangles.
r1=v
u
u
td
22
+ √3d
2+d!2
=r(1)2+√3+12
9
Step 2: Calculate the distance between the +4µC charge and the +5µC
charge. The distance r2can be found in the same manner.
r2=v
u
u
td
22
+ √3d
2!2
=r(1)2+√32
Step 3: Calculate the magnitude of the force F1between the +4µC and
−3µC charges using Coulomb’s law.
F1=k|q1||q2|
r2
1
Step 4: Calculate the magnitude of the force F2between the +4µC and
+5µC charges in the same way.
F2=k|q1||q3|
r2
2
Step 5: Determine the direction of the net force on the +4µC charge. Since
the forces are vectors, we need to consider their directions. The force F1will be
directed away from the −3µC charge, and the force F2will be directed towards
the +5µC charge. Calculate the net force by considering the vector sum of F1
and F2.
Question 13
Question
Three point charges are arranged as shown in the diagram below. Each charge
has a magnitude of 3.00 µC. Calculate the magnitude and direction of the net
force on the charge q3.
q1q2
q3
Solution
Step 1: Calculate the distances between the charges using the Pythagorean
theorem. Let the length of each side of the equilateral triangle be L.
L=p22+ 1.7322
L=√4+3
L=√7
10
Step 2: Calculate the electric force between charges using Coulomb’s law:
F=k|q1q2|
r2where - kis the Coulomb constant (8.99 ×109N·m2/C2), - q1
and q2are the magnitudes of the charges, and - ris the distance between the
charges.
Step 3: Calculate the force between q1and q3:
F13 = 8.99 ×109(3 ×10−6)(3 ×10−6)
(√7)2
Step 4: Calculate the force between q2and q3:
F23 = 8.99 ×109(3 ×10−6)(3 ×10−6)
(√7)2
Step 5: The net force on q3is the vector sum of the forces F13 and F23.
Since the forces are acting at 120° angles to each other, the net force magnitude
can be calculated using the law of cosines:
Net force =qF2
13 +F2
23 + 2F13F23 cos(120◦)
Step 6: Now, plug in the values to find the magnitude and direction of the
net force on charge q3.
Question 14
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length a. The magnitudes of the charges are q,2q, and 3q. Calculate
the net electric force on the charge qdue to the other two charges.
Solution
Let’s denote the charges as follows: q1=q,q2= 2q, and q3= 3q. The distances
from q1to q2and q3are both a.
By Coulomb’s law, the magnitude of the electric force between two charges
qiand qjis Fij =k|qi|·|qj|
r2, where kis Coulomb’s constant (8.99 ×109N m2/C2)
and ris the distance between the charges.
Step 1: We start by finding the electric force on charge q1due to charge q2.
Given that |q1|=q,|q2|= 2q, and r=a, we have:
F12 =k|q1|·|q2|
r2= 8.99 ×109q·2q
a2
Step 2: Next, we find the electric force on charge q1due to charge q3. Given
that |q1|=q,|q3|= 3q, and r=a, we have:
F13 =k|q1|·|q3|
r2= 8.99 ×109q·3q
a2
11
Step 3: Now, we calculate the net electric force on charge q1by summing
the forces due to q2and q3.
Fnet =F12 +F13 = 8.99 ×1092q2
a2+ 8.99 ×1093q2
a2
Therefore, the net electric force on the charge qdue to the other two charges
is:
Fnet = 8.99 ×1095q2
a2N
Question 15
Question
Two point charges, q1= +4.0µC and q2=−6.0µC, are placed 0.10 m apart in
a vacuum. Calculate the magnitude of the electric force that each charge exerts
on the other.
Solution
Let’s denote the magnitude of the electric force between two point charges as F
and their separation distance as r. We can use Coulomb’s law to calculate the
electric force between these two charges. Coulomb’s law is given by:
F=k|q1q2|
r2,
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Step 1: Calculate the electric force that charge q1exerts on charge q2. This
force will be attractive since the charges are of opposite signs.
F1→2=k|q1q2|
r2.
Plugging in the given values:
F1→2= (8.99 ×109N m2/C2)|(4.0×10−6C)(−6.0×10−6C)|
(0.10 m)2.
Step 2: Calculate the magnitude of the electric force between q1and q2.
F1→2= 8.99 ×109×(4.0×10−6)(6.0×10−6)
0.01 .
F1→2= 8.99 ×109×2.4×10−11.
F1→2= 21.576 ×10−2N.
Therefore, the magnitude of the electric force that q1exerts on q2is 2.16 ×
10−1N.
12
Step 3: The force that charge q2exerts on charge q1will be equal in mag-
nitude but opposite in direction to the force calculated in Step 2.
Hence, the magnitude of the electric force that each charge exerts on the
other is 2.16 ×10−1N.
Question 16
Question
Three point charges are placed at the vertices of an equilateral triangle with
side length aas shown below. The charges are +q,−2q, and +4q. Calculate the
magnitude of the electric force on the +qcharge due to the other two charges.
+4q
−2q+q
Solution
Step 1: Calculate the electric force on the +qcharge due to the +4qcharge.
The electric force Fbetween two charges q1and q2separated by a distance ris
given by Coulomb’s law:
F=kq1q2
r2
where kis the Coulomb constant (8.9875 ×109N m2/C2).
In this case, the charge q1= +q,q2= +4q, and the distance between them is
the side length of the equilateral triangle a. Since the charges are located at the
vertices of an equilateral triangle, the distance between them is a
2. Therefore,
the electric force on the +qcharge due to the +4qcharge is:
F=kq(+4q)
(a
2)2=k4q2
a2
4
=k16q2
a2=72kq2
a2
Step 2: Calculate the electric force on the +qcharge due to the −2qcharge.
Similarly, the electric force on the +qcharge due to the −2qcharge can be
calculated:
F=kq(−2q)
a2=−2kq2
a2=−18kq2
a2
Step 3: Combine the two forces. The total force on the +qcharge is the
vector sum of the forces due to the +4qand −2qcharges. Since the forces are
in opposite directions along the same line:
Ftotal =72kq2
a2−18kq2
a2=54kq2
a2
Therefore, the magnitude of the electric force on the +qcharge due to the
other two charges is 54kq2
a2.
13
Question 17
Question
Three point charges are arranged as follows: a charge of +2.0µC at the origin,
a charge of −4.0µC at (0,3m), and a charge of +6.0µC at (4 m,0). Calculate
the net electric force on the charge at the origin.
Solution
To find the net electric force on the charge at the origin, we need to calculate
the individual electric forces due to each of the other charges and then sum
them up vectorially.
Step 1: Calculate the electric force due to the charge of +2.0µC at the
origin.
The electric force F1due to this charge can be calculated using Coulomb’s
law:
F1=kq1q
r2,
where kis the Coulomb constant, q1and qare the magnitudes of the two charges,
and ris the distance between the charges.
The distance between the charges is 0since the charge creating the force is
at the origin. Hence, r= 0.
Therefore, the force due to the charge of +2.0µC is 0.
Step 2: Calculate the electric force due to the charge of −4.0µC at (0,3m).
The electric force F2due to this charge can be calculated using Coulomb’s
law:
F2=kq2q
r2,
where q2and qare the magnitudes of the two charges, and ris the distance
between the charges.
The distance between the charges is √02+ 32= 3 m.
Plugging in the values and accounting for the direction of the force, which
would be along the negative y-axis, we get:
F2=−6.75 Nˆ
j.
Step 3: Calculate the electric force due to the charge of +6.0µC at (4 m,0).
The electric force F3due to this charge can be calculated using Coulomb’s
law:
F3=kq3q
r2,
where q3and qare the magnitudes of the two charges, and ris the distance
between the charges.
The distance between the charges is √42+ 02= 4 m.
Plugging in the values and accounting for the direction of the force, which
would be along the positive x-axis, we get:
F3= 6.75 Nˆ
i.
14
Step 4: Find the net electric force on the charge at the origin.
The net force on the charge at the origin is the vector sum of the individual
forces:
Fnet =F1+F2+F3.
Substitute the calculated values and summing the forces vectorially, we get:
Fnet =−6.75 Nˆ
j+ 6.75 Nˆ
i.
Therefore, the net electric force on the charge at the origin is −6.75 Nˆ
j+
6.75 Nˆ
i.
Question 18
Question
Two point charges, Q1=−8.0µC and Q2= 4.0µC, are placed at points A
and B, respectively, with a distance of 40 cm between them. Calculate the
magnitude and direction of the electric force that charge Q1exerts on charge
Q2. Given that the Coulomb constant is k= 8.99 ×109N m2/C2.
Solution
Step 1: Find the magnitude of the electric force using the equation:
F=k· |Q1|·|Q2|
r2
where Fis the electric force, kis the Coulomb constant, |Q1|and |Q2|are the
magnitudes of the charges, and ris the distance between the charges.
Plugging in the given values:
F=(8.99 ×109N m2/C2)·(8.0×10−6C)·(4.0×10−6C)
(0.40 m)2
F=8.99 ×8.0×4.0
0.16 ×103N
F= 8.99 ×8.0×4.0×103N
F= 2.8768 ×105N
Therefore, the magnitude of the electric force that charge Q1exerts on charge
Q2is 2.88 ×105N.
Step 2: Find the direction of the electric force. Since charge Q1is negative,
the force on charge Q2will be attractive, pulling Q2towards Q1. Thus, the
direction of the electric force is from charge Q2towards charge Q1.
15
Question 19
Question
Two point charges, q1= 8.0�C and q2=−10.0�C, are placed 40.0cm apart.
Determine the magnitude and direction of the electric force that q2exerts on
q1.
Solution
Step 1: First, we need to calculate the electric force between the two charges
using Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Plug in the given values:
F=(8.99 ×109)· |8.0×10−6·(−10.0×10−6)|
(0.40)2
Step 3: Calculate the magnitude of the electric force:
F=(8.99 ×109)·(8.0×10−6·10.0×10−6)
0.16
F=8.99 ×109·80 ×10−12
0.16
F=7.192 ×10−3
0.16 = 0.04495 N
Step 4: Now, to determine the direction of the force, we consider the fact
that q2exerts an attractive force on q1(since the charges have opposite signs).
Therefore, the force on q1points towards q2.
Step 5: So, the electric force that q2exerts on q1has a magnitude of 0.04495 N
and direction towards q2.
Question 20
Question
Two point charges, q1=−3.0µC and q2= 6.0µC, are placed 10.0cm apart.
Calculate the magnitude of the electric force that each charge exerts on the
other.
16
Solution
Step 1: Convert the given charges to coulombs.
q1=−3.0×10−6C
q2= 6.0×10−6C
Step 2: Calculate the distance between the charges in meters. Given: r=
10.0cm = 0.10 m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k|q1||q2|
r2
where k= 8.99 ×109N·m2/C2.
Step 4: Substitute the known values and calculate the magnitude of the
electric force.
F=8.99 ×109N·m2/C2×3.0×10−6C×6.0×10−6C
(0.10 m)2
F=161.82
0.01 = 16182 N
Therefore, the magnitude of the electric force that each charge exerts on the
other is 16182 N.
Question 21
Question
Two point charges, q1= 3.0×10−6C and q2=−4.0×10−6C, are placed 10
cm apart in a vacuum. Calculate the magnitude of the electric force between
the charges.
Solution
Step 1: Convert all given values to SI units. Given: q1= 3.0×10−6C, q2=
−4.0×10−6C, r= 10 cm = 0.10 m.
Step 2: Calculate the magnitude of the electric force between the charges
using Coulomb’s Law: The magnitude of the electric force between two point
charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2,
where k= 8.99 ×109N m2/C2is the electrostatic constant.
17
Substitute the given values into the formula:
F=8.99 ×109×3.0×10−6×4.0×10−6
(0.10)2.
Step 3: Calculate the electric force.
F=8.99 ×109×3.0×10−6×4.0×10−6
(0.10)2=107.88
0.01 = 10788 N.
Therefore, the magnitude of the electric force between the charges is 10788
N.
Question 22
Question
Three point charges are located at the corners of an equilateral triangle with
side length a. Each charge has a magnitude of q. Calculate the magnitude of
the net electric force on one of the charges due to the other two charges.
Solution
Let’s label the charges q1,q2, and q3at the vertices of the equilateral triangle.
We will find the magnitude of the net electric force on q1due to q2and q3.
Step 1: Calculate the electric force from q2on q1. The magnitude of the
electric force between two point charges can be calculated using Coulomb’s Law:
F=k|q1q2|
r2
where kis the Coulomb constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
In this case, the distance between q1and q2is a. The direction of the force
will be along the line connecting q1and q2.
Step 2: Calculate the electric force from q3on q1. Similarly, we can calculate
the electric force from q3on q1using Coulomb’s Law. The distance between q1
and q3is also ain this equilateral triangle configuration.
Step 3: Find the angle between the forces. The forces from q2and q3on q1
are equal in magnitude but opposite in direction. They form a 120◦angle with
respect to each other.
Step 4: Calculate the net electric force on q1. To find the net electric force
on q1, we can treat the forces from q2and q3as vectors. Since they are at 120◦
to each other, we can treat them as components of a single force vector.
The net force magnitude will be the vector sum of the forces from q2and q3:
Fnet =qF2
2+F2
3+ 2F2F3cos(120◦)
Solving this expression will give us the magnitude of the net electric force
on q1due to q2and q3.
18
Question 23
Question
Two point charges, q1= +3.00 nC and q2=−5.00 nC are located on the x-axis
at positions x= 2.00 m and x=−4.00 m respectively. Calculate the magnitude
and direction of the electric force on q1at x= 2.00 m due to the presence of
charge q2.
Solution
Step 1: Calculate the distance between the charges. Given that x1= 2.00 m and
x2=−4.00 m, we can find the distance between the charges using the formula:
r=|x2−x1|
r=| − 4.00 m−2.00 m|
r=| − 6.00 m|
r= 6.00 m
Therefore, the distance between the charges is 6.00 m.
Step 2: Calculate the electric force. The magnitude of the electric force
between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
where - kis the electrostatic constant 8.99 ×109N m2/C2-q1= +3.00 nC =
3.00 ×10−9C - q2=−5.00 nC =−5.00 ×10−9C - r= 6.00 m
Plugging in the values:
F= (8.99 ×109)|3.00 ×10−9× −5.00 ×10−9|
(6.00)2
F= (8.99 ×109)15.00 ×10−18
36
F= (8.99 ×109)×0.4167
F≈3.75 ×109N
Step 3: Determine the direction of the force. The force on q1is towards q2
because the charges are opposite in sign. Thus, the force is attractive.
Therefore, the magnitude of the electric force on q1at x= 2.00 m due to
the presence of charge q2at x=−4.00 m is approximately 3.75 ×109N and is
attractive.
19
Question 24
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 9.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between these two
charges.
Solution
Step 1: Convert the charges to Coulombs. Given that 1µC= 10−6C, we have:
q1=−4.0µC=−4.0×10−6C and q2= 6.0µC= 6.0×10−6C.
Step 2: Calculate the electric force using Coulomb’s Law. 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 Coulomb’s constant with a value of 8.9875 ×109N m2/C2,q1and q2
are the magnitudes of the charges, and ris the separation distance.
Step 3: Substitute the given values into Coulomb’s Law and solve for the
electric force. Substitute k= 8.9875 ×109N m2/C2,q1=−4.0×10−6C,
q2= 6.0×10−6C, and r= 9.0cm = 0.09 m into Coulomb’s Law:
F= 8.9875 ×109×|(−4.0×10−6)(6.0×10−6)|
(0.09)2
Step 4: Calculate the magnitude of the electric force. Performing the calcu-
lations:
F= 8.9875 ×109×24 ×10−12
0.0081
F= 8.9875 ×109×2.96 ×10−9
F= 26.6N
Therefore, the magnitude of the electric force between the two charges is
26.6N.
Question 25
Question
Two point charges q1=−3.0µC and q2= 5.0µC are placed 25 cm apart.
Calculate the magnitude of the electric force between the charges.
20
Solution
Step 1: Convert the charges to Coulombs.
Given: q1=−3.0µC=−3.0×10−6C
q2= 5.0µC= 5.0×10−6C
Step 2: Calculate the distance between the charges.
Given: r= 25 cm = 0.25 m
Step 3: Calculate the magnitude of the electric force.
The electric force between two point charges is given by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the electrostatic constant with a value of 8.99 ×109Nm2/C2.
Substitute the given values into the formula:
F= (8.99 ×109)·|(−3.0×10−6)·(5.0×10−6)|
(0.25)2
Step 4: Calculate the magnitude of the electric force.
F= 8.99 ×109·|(−3.0×10−6)·(5.0×10−6)|
(0.25)2
F= 8.99 ×109·15 ×10−12
0.0625
F= 8.99 ×109·2.4×10−10
F= 2.16 ×10−1N
Therefore, the magnitude of the electric force between the charges is 2.16 ×
10−1N.
21
Step 2: Determine the direction of the electric force on charge q1. Since
charge q2is negative, the force on q1will be attractive. Therefore, the direction
of the electric force on charge q1is towards charge q2.
Step 3: Calculate the magnitude of the electric force on charge q2. Since the
magnitudes of the charges are the same, the magnitude of the force on q2will
be the same as on q1, which is 21.576 N.
Step 4: Determine the direction of the electric force on charge q2. Since
charge q1is positive, the force on q2will be attractive also. Therefore, the
direction of the electric force on charge q2is towards charge q1.
Question 3
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length 2. The charges are +2µC at the top vertex, −3µC at the bottom
left vertex, and +5µC at the bottom right vertex. Calculate the magnitude and
direction of the net electrostatic force acting on the +2µC charge due to the
other charges.
Solution
To find the net electrostatic force acting on the +2µC charge, we need to cal-
culate the individual forces due to each of the other charges and then add them
as vectors.
Step 1: Calculate the force due to the −3µC charge at the bottom left ver-
tex. The magnitude of the force between two point charges q1and q2separated
by distance ris given by Coulomb’s Law:
F=k|q1q2|
r2
where kis the Coulomb’s constant (k≈8.99 ×109Nm2/C2).
For the +2µC charge and the −3µC charge, the distance between them is the
height of the equilateral triangle, h=√3. So, the force between these charges
is:
Fleft =(8.99 ×109)|2×10−6×3×10−6|
(√3)2
Step 2: Calculate the direction of the force due to the −3µC charge. Since
the −3µC charge is negative, the force on the +2µC charge will be attractive,
i.e., towards the −3µC charge. Therefore, the direction is towards the bottom
left vertex.
Step 3: Calculate the force due to the +5µC charge at the bottom right
vertex. Using Coulomb’s Law, the force between the +2µC charge and the
2
+5µC charge is:
Fright =(8.99 ×109)|2×10−6×5×10−6|
(√3)2
Step 4: Calculate the direction of the force due to the +5µC charge. Since
both charges are positive, the force on the +2µC charge will be repulsive, i.e.,
away from the +5µC charge. Therefore, the direction is away from the bottom
right vertex.
Step 5: Find the net force on the +2µC charge. To find the net force, we
need to add these forces as vectors. Let the force due to the −3µC charge be
Fleft and the force due to the +5µC charge be
Fright. The net force is:
Fnet =
Fleft +
Fright
Now, calculate the magnitude and direction of the net force.
Question 4
Question
Two point charges, q1= +2 µC and q2=−3µC, are placed 10 cm apart.
Calculate the magnitude and direction of the electric force each charge exerts
on the other. Assume the charges are located on the x-axis, with q1at the origin
and q2at x= 10 cm.
Solution
Step 1: Calculate the distance between the charges in meters. Given that the
charges are 10 cm apart, the distance rin meters is:
r= 10 cm = 0.10 m
Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force between two point charges is given by Coulomb’s law:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging in the given values:
F=(8.99 ×109)(2 ×10−6)(3 ×10−6)
(0.10)2
Calculating:
F=53.94
0.01 = 5394 N
3
Therefore, the magnitude of the electric force between the charges is 5394
N.
Step 3: Determine the direction of the force. Since q1is positive and q2is
negative, the force between them is attractive and directed from q2towards q1.
Thus, the electric force Fon q1points towards q1. The electric force Fon
q2points towards q2.
In conclusion, the magnitude of the electric force each charge exerts on the
other is 5394 N, and the direction of the force on q1is towards q1, while the
direction of the force on q2is towards q2.
Question 5
Question
Two point charges, q1 = +2.0 µC and q2 = -4.0 µC, are placed 10.0 cm apart
in air. Calculate the magnitude and direction of the electric force that q1 exerts
on q2.
Solution
Step 1: Convert µC to Coulombs. Given that 1µC = 10−6C, we can convert the
charges as follows: q1 = 2.0µC = 2.0×10−6C q2 = −4.0µC =−4.0×10−6C
Step 2: Calculate the distance between the two charges. Given that the
charges are placed 10.0 cm apart, we convert this distance to meters: d=
10.0cm = 10.0×10−2m= 0.10m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
The electric force between the charges is given by Coulomb’s Law: F=k·|q1·q2|
d2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the values: F= (8.99 ×109)·|2.0×10−6·−4.0×10−6|
0.102
F= 71.92 ×10−12/0.01
F= 7.192 ×10−10N
Step 4: Find the direction of the force. Since q1 is positive and q2 is negative,
the force between them will be attractive. Therefore, the force exerted by q1
on q2 is directed towards q1.
The magnitude of the electric force that q1 exerts on q2 is 7.192 ×10−10N
and the direction is towards q1.
Question 6
Question
Two point charges, +2 µC and −3µC, are placed 6 cm apart in air. Calculate
the magnitude and direction of the electric force that one charge exerts on the
other.
4
Solution
Step 1: Convert the charges to Coulombs. We have +2 µC= 2 ×10−6C and
−3µC=−3×10−6C.
Step 2: Calculate the electric force using Coulomb’s Law:
The magnitude of the electric force between two point charges is given by:
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 charges, - ris the distance between the
charges.
Plugging in the values, we get:
F=8.99 ×109· |2×10−6· −3×10−6|
(0.06)2
Step 3: Calculate the magnitude of the electric force:
F=8.99 ×109·6×10−12
0.0036
F=53.94 ×10−3
0.0036
F= 14950 N
Step 4: Determine the direction of the electric force: The force is attractive
since one charge is positive and the other is negative. It acts along the line
joining the charges, from the positive charge to the negative charge.
Therefore, the magnitude of the electric force that the positive charge exerts
on the negative charge is 14950 N, directed from the positive charge to the
negative charge.
Question 7
Question
Three point charges are arranged as shown: - Charge q1= +2.0µC is located at
the origin. - Charge q2=−3.0µC is located at coordinates (0,0.6m). - Charge
q3= +1.0µC is located at coordinates (0.8m,0).
Calculate the net electric force acting on q1due to q2and q3.
Solution
Step 1: Find the electric force between q1and q2: The electric force F12 between
two point charges is given by Coulomb’s law:
F12 =k|q1||q2|
r2
12
5
where - kis Coulomb’s constant (8.99 ×109N m2/C2), - q1and q2are the
magnitudes of the point charges, and - r12 is the distance between the two
charges.
Given that q1= 2.0µC,q2=−3.0µC, and r12 = 0.6m, we have:
F12 =(8.99 ×109)×(2.0×10−6)×(3.0×10−6)
(0.6)2
Step 2: Calculate F12:
F12 =8.99 ×2.0×3.0
(0.6)2×103N
Step 3: Simplify the expression:
F12 =53.94
0.36 ×103N
F12 = 149.83 ×103N
F12 = 1.50 ×105N
Step 4: Find the direction of F12: The direction of F12 is repulsive since the
charges q1and q2are of opposite sign.
Step 5: Repeat steps 1-4 for the electric force F13 between q1and q3and
then find the net electric force acting on q1.
Question 8
Question
Two point charges, q1= +2.0µC and q2=−4.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the two
charges.
Solution
Step 1: Convert the charges to the standard unit of Coulombs. Given: q1=
+2.0µC= 2.0×10−6C, and q2=−4.0µC=−4.0×10−6C.
Step 2: Recall the formula for the magnitude of electric force between two
point charges:
F=k·|q1·q2|
r2
where kis the Coulomb constant (k= 8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the separation distance between the charges.
Step 3: Substitute the given values into the formula and solve for the electric
force.
F= (8.99 ×109)·|2.0×10−6· −4.0×10−6|
(0.10)2
6
F= (8.99 ×109)·8.0×10−12
0.01
F= 8.99 ×109·8.0×10−10
F= 71.92 ×10−1
F= 7.192 N
Step 4: Therefore, the magnitude of the electric force between the two
charges is 7.192 N.
Question 9
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed on the x-axis at
positions x=−2.0m and x= 2.0m, respectively. Calculate the magnitude and
direction of the electric force on q1due to q2.
Solution
Step 1: Calculate the distance between the two charges. Given: q1=−4.0µC=
−4.0×10−6Cq2= 6.0µC= 6.0×10−6Cx1=−2.0mx2= 2.0m
The distance rbetween the two charges is:
r=|x2−x1|=|2.0−(−2.0)|= 4.0m
Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force Fbetween two charges is given by Coulomb’s law:
F=k·|q1·q2|
r2
where k= 8.99 ×109N m2C−2is the Coulomb’s constant.
Plugging in the values:
F= 8.99 ×109·| − 4.0×10−6·6.0×10−6|
(4.0)2
F= 8.99 ×109·24 ×10−12
16
F= 8.99 ×109·1.5×10−12
F≈13.485 N
Step 3: Determine the direction of the force. The force on q1due to q2
is attractive since the charges have opposite signs. Therefore, the force acts
towards the positive xdirection.
So, the magnitude of the electric force on q1due to q2is approximately
13.485 N, and it acts in the positive xdirection.
7
Question 10
Question
Two point charges, q1=−3.5µC and q2= 1.8µC, are placed 12 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to SI units.
q1=−3.5µC =−3.5×10−6C
q2= 1.8µC = 1.8×10−6C
Step 2: Determine the distance between the charges in meters.
d= 12 cm = 0.12 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
F=k|q1||q2|
d2
Where: k= 8.99 ×109Nm2/C2is the Coulomb constant
Step 4: Substitute the given values into the formula and calculate.
F=(8.99 ×109)(3.5×10−6)(1.8×10−6)
0.122
F=56.735 ×10−15
0.0144
F= 56.735 ×10−15 ×69.444
F= 3938 ×10−15 N
F= 3.938 ×10−12 N
Therefore, the magnitude of the electric force between the charges is 3.938 ×
10−12 N.
Question 11
Question
Two point charges, Q1=−2.0µC and Q2= 4.0µC, are placed 0.10 meters
apart in a vacuum. Calculate the magnitude and direction of the net electric
force on Q1.
8
Solution
Given: Q1=−2.0µC
Q2= 4.0µC
Distance between charges, r= 0.10 m
k= 8.99 ×109N·m2/C2(Coulomb’s constant)
Step 1: Calculate the electric force between the two charges using Coulomb’s
law:
F=k·|Q1·Q2|
r2
F= 8.99 ×109·2.0×10−6·4.0×10−6
0.12
Step 2: Calculate the magnitude of electric force:
F= 8.99 ×109·8.0×10−12
0.01
F= 8.99 ×109·8.0×10−10
F= 7.192 ×10−1N
Step 3: Determine the direction of the net electric force: Since Q1and Q2
have opposite charges, the electric force on Q1will be attractive towards Q2.
Therefore, the net electric force on Q1is directed towards Q2.
Thus, the magnitude of the net electric force on Q1is 7.192 ×10−1N and
the direction is towards Q2.
Question 12
Question
Three point charges are placed at the corners of an equilateral triangle with
side length d= 2 m. The charges are +4µC,−3µC and +5µC. Calculate the
magnitude and direction of the net force on the +4µC charge due to the other
two charges.
Solution
Step 1: Calculate the distance between the +4µC charge and the −3µC charge
using the Pythagorean theorem. The distance r1can be found by dividing the
equilateral triangle into two right triangles.
r1=v
u
u
td
22
+ √3d
2+d!2
=r(1)2+√3+12
9
Step 2: Calculate the distance between the +4µC charge and the +5µC
charge. The distance r2can be found in the same manner.
r2=v
u
u
td
22
+ √3d
2!2
=r(1)2+√32
Step 3: Calculate the magnitude of the force F1between the +4µC and
−3µC charges using Coulomb’s law.
F1=k|q1||q2|
r2
1
Step 4: Calculate the magnitude of the force F2between the +4µC and
+5µC charges in the same way.
F2=k|q1||q3|
r2
2
Step 5: Determine the direction of the net force on the +4µC charge. Since
the forces are vectors, we need to consider their directions. The force F1will be
directed away from the −3µC charge, and the force F2will be directed towards
the +5µC charge. Calculate the net force by considering the vector sum of F1
and F2.
Question 13
Question
Three point charges are arranged as shown in the diagram below. Each charge
has a magnitude of 3.00 µC. Calculate the magnitude and direction of the net
force on the charge q3.
q1q2
q3
Solution
Step 1: Calculate the distances between the charges using the Pythagorean
theorem. Let the length of each side of the equilateral triangle be L.
L=p22+ 1.7322
L=√4+3
L=√7
10
Step 2: Calculate the electric force between charges using Coulomb’s law:
F=k|q1q2|
r2where - kis the Coulomb constant (8.99 ×109N·m2/C2), - q1
and q2are the magnitudes of the charges, and - ris the distance between the
charges.
Step 3: Calculate the force between q1and q3:
F13 = 8.99 ×109(3 ×10−6)(3 ×10−6)
(√7)2
Step 4: Calculate the force between q2and q3:
F23 = 8.99 ×109(3 ×10−6)(3 ×10−6)
(√7)2
Step 5: The net force on q3is the vector sum of the forces F13 and F23.
Since the forces are acting at 120° angles to each other, the net force magnitude
can be calculated using the law of cosines:
Net force =qF2
13 +F2
23 + 2F13F23 cos(120◦)
Step 6: Now, plug in the values to find the magnitude and direction of the
net force on charge q3.
Question 14
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length a. The magnitudes of the charges are q,2q, and 3q. Calculate
the net electric force on the charge qdue to the other two charges.
Solution
Let’s denote the charges as follows: q1=q,q2= 2q, and q3= 3q. The distances
from q1to q2and q3are both a.
By Coulomb’s law, the magnitude of the electric force between two charges
qiand qjis Fij =k|qi|·|qj|
r2, where kis Coulomb’s constant (8.99 ×109N m2/C2)
and ris the distance between the charges.
Step 1: We start by finding the electric force on charge q1due to charge q2.
Given that |q1|=q,|q2|= 2q, and r=a, we have:
F12 =k|q1|·|q2|
r2= 8.99 ×109q·2q
a2
Step 2: Next, we find the electric force on charge q1due to charge q3. Given
that |q1|=q,|q3|= 3q, and r=a, we have:
F13 =k|q1|·|q3|
r2= 8.99 ×109q·3q
a2
11
Step 3: Now, we calculate the net electric force on charge q1by summing
the forces due to q2and q3.
Fnet =F12 +F13 = 8.99 ×1092q2
a2+ 8.99 ×1093q2
a2
Therefore, the net electric force on the charge qdue to the other two charges
is:
Fnet = 8.99 ×1095q2
a2N
Question 15
Question
Two point charges, q1= +4.0µC and q2=−6.0µC, are placed 0.10 m apart in
a vacuum. Calculate the magnitude of the electric force that each charge exerts
on the other.
Solution
Let’s denote the magnitude of the electric force between two point charges as F
and their separation distance as r. We can use Coulomb’s law to calculate the
electric force between these two charges. Coulomb’s law is given by:
F=k|q1q2|
r2,
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Step 1: Calculate the electric force that charge q1exerts on charge q2. This
force will be attractive since the charges are of opposite signs.
F1→2=k|q1q2|
r2.
Plugging in the given values:
F1→2= (8.99 ×109N m2/C2)|(4.0×10−6C)(−6.0×10−6C)|
(0.10 m)2.
Step 2: Calculate the magnitude of the electric force between q1and q2.
F1→2= 8.99 ×109×(4.0×10−6)(6.0×10−6)
0.01 .
F1→2= 8.99 ×109×2.4×10−11.
F1→2= 21.576 ×10−2N.
Therefore, the magnitude of the electric force that q1exerts on q2is 2.16 ×
10−1N.
12
Step 3: The force that charge q2exerts on charge q1will be equal in mag-
nitude but opposite in direction to the force calculated in Step 2.
Hence, the magnitude of the electric force that each charge exerts on the
other is 2.16 ×10−1N.
Question 16
Question
Three point charges are placed at the vertices of an equilateral triangle with
side length aas shown below. The charges are +q,−2q, and +4q. Calculate the
magnitude of the electric force on the +qcharge due to the other two charges.
+4q
−2q+q
Solution
Step 1: Calculate the electric force on the +qcharge due to the +4qcharge.
The electric force Fbetween two charges q1and q2separated by a distance ris
given by Coulomb’s law:
F=kq1q2
r2
where kis the Coulomb constant (8.9875 ×109N m2/C2).
In this case, the charge q1= +q,q2= +4q, and the distance between them is
the side length of the equilateral triangle a. Since the charges are located at the
vertices of an equilateral triangle, the distance between them is a
2. Therefore,
the electric force on the +qcharge due to the +4qcharge is:
F=kq(+4q)
(a
2)2=k4q2
a2
4
=k16q2
a2=72kq2
a2
Step 2: Calculate the electric force on the +qcharge due to the −2qcharge.
Similarly, the electric force on the +qcharge due to the −2qcharge can be
calculated:
F=kq(−2q)
a2=−2kq2
a2=−18kq2
a2
Step 3: Combine the two forces. The total force on the +qcharge is the
vector sum of the forces due to the +4qand −2qcharges. Since the forces are
in opposite directions along the same line:
Ftotal =72kq2
a2−18kq2
a2=54kq2
a2
Therefore, the magnitude of the electric force on the +qcharge due to the
other two charges is 54kq2
a2.
13
Question 17
Question
Three point charges are arranged as follows: a charge of +2.0µC at the origin,
a charge of −4.0µC at (0,3m), and a charge of +6.0µC at (4 m,0). Calculate
the net electric force on the charge at the origin.
Solution
To find the net electric force on the charge at the origin, we need to calculate
the individual electric forces due to each of the other charges and then sum
them up vectorially.
Step 1: Calculate the electric force due to the charge of +2.0µC at the
origin.
The electric force F1due to this charge can be calculated using Coulomb’s
law:
F1=kq1q
r2,
where kis the Coulomb constant, q1and qare the magnitudes of the two charges,
and ris the distance between the charges.
The distance between the charges is 0since the charge creating the force is
at the origin. Hence, r= 0.
Therefore, the force due to the charge of +2.0µC is 0.
Step 2: Calculate the electric force due to the charge of −4.0µC at (0,3m).
The electric force F2due to this charge can be calculated using Coulomb’s
law:
F2=kq2q
r2,
where q2and qare the magnitudes of the two charges, and ris the distance
between the charges.
The distance between the charges is √02+ 32= 3 m.
Plugging in the values and accounting for the direction of the force, which
would be along the negative y-axis, we get:
F2=−6.75 Nˆ
j.
Step 3: Calculate the electric force due to the charge of +6.0µC at (4 m,0).
The electric force F3due to this charge can be calculated using Coulomb’s
law:
F3=kq3q
r2,
where q3and qare the magnitudes of the two charges, and ris the distance
between the charges.
The distance between the charges is √42+ 02= 4 m.
Plugging in the values and accounting for the direction of the force, which
would be along the positive x-axis, we get:
F3= 6.75 Nˆ
i.
14
Step 4: Find the net electric force on the charge at the origin.
The net force on the charge at the origin is the vector sum of the individual
forces:
Fnet =F1+F2+F3.
Substitute the calculated values and summing the forces vectorially, we get:
Fnet =−6.75 Nˆ
j+ 6.75 Nˆ
i.
Therefore, the net electric force on the charge at the origin is −6.75 Nˆ
j+
6.75 Nˆ
i.
Question 18
Question
Two point charges, Q1=−8.0µC and Q2= 4.0µC, are placed at points A
and B, respectively, with a distance of 40 cm between them. Calculate the
magnitude and direction of the electric force that charge Q1exerts on charge
Q2. Given that the Coulomb constant is k= 8.99 ×109N m2/C2.
Solution
Step 1: Find the magnitude of the electric force using the equation:
F=k· |Q1|·|Q2|
r2
where Fis the electric force, kis the Coulomb constant, |Q1|and |Q2|are the
magnitudes of the charges, and ris the distance between the charges.
Plugging in the given values:
F=(8.99 ×109N m2/C2)·(8.0×10−6C)·(4.0×10−6C)
(0.40 m)2
F=8.99 ×8.0×4.0
0.16 ×103N
F= 8.99 ×8.0×4.0×103N
F= 2.8768 ×105N
Therefore, the magnitude of the electric force that charge Q1exerts on charge
Q2is 2.88 ×105N.
Step 2: Find the direction of the electric force. Since charge Q1is negative,
the force on charge Q2will be attractive, pulling Q2towards Q1. Thus, the
direction of the electric force is from charge Q2towards charge Q1.
15
Question 19
Question
Two point charges, q1= 8.0�C and q2=−10.0�C, are placed 40.0cm apart.
Determine the magnitude and direction of the electric force that q2exerts on
q1.
Solution
Step 1: First, we need to calculate the electric force between the two charges
using Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Plug in the given values:
F=(8.99 ×109)· |8.0×10−6·(−10.0×10−6)|
(0.40)2
Step 3: Calculate the magnitude of the electric force:
F=(8.99 ×109)·(8.0×10−6·10.0×10−6)
0.16
F=8.99 ×109·80 ×10−12
0.16
F=7.192 ×10−3
0.16 = 0.04495 N
Step 4: Now, to determine the direction of the force, we consider the fact
that q2exerts an attractive force on q1(since the charges have opposite signs).
Therefore, the force on q1points towards q2.
Step 5: So, the electric force that q2exerts on q1has a magnitude of 0.04495 N
and direction towards q2.
Question 20
Question
Two point charges, q1=−3.0µC and q2= 6.0µC, are placed 10.0cm apart.
Calculate the magnitude of the electric force that each charge exerts on the
other.
16
Solution
Step 1: Convert the given charges to coulombs.
q1=−3.0×10−6C
q2= 6.0×10−6C
Step 2: Calculate the distance between the charges in meters. Given: r=
10.0cm = 0.10 m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k|q1||q2|
r2
where k= 8.99 ×109N·m2/C2.
Step 4: Substitute the known values and calculate the magnitude of the
electric force.
F=8.99 ×109N·m2/C2×3.0×10−6C×6.0×10−6C
(0.10 m)2
F=161.82
0.01 = 16182 N
Therefore, the magnitude of the electric force that each charge exerts on the
other is 16182 N.
Question 21
Question
Two point charges, q1= 3.0×10−6C and q2=−4.0×10−6C, are placed 10
cm apart in a vacuum. Calculate the magnitude of the electric force between
the charges.
Solution
Step 1: Convert all given values to SI units. Given: q1= 3.0×10−6C, q2=
−4.0×10−6C, r= 10 cm = 0.10 m.
Step 2: Calculate the magnitude of the electric force between the charges
using Coulomb’s Law: The magnitude of the electric force between two point
charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2,
where k= 8.99 ×109N m2/C2is the electrostatic constant.
17
Substitute the given values into the formula:
F=8.99 ×109×3.0×10−6×4.0×10−6
(0.10)2.
Step 3: Calculate the electric force.
F=8.99 ×109×3.0×10−6×4.0×10−6
(0.10)2=107.88
0.01 = 10788 N.
Therefore, the magnitude of the electric force between the charges is 10788
N.
Question 22
Question
Three point charges are located at the corners of an equilateral triangle with
side length a. Each charge has a magnitude of q. Calculate the magnitude of
the net electric force on one of the charges due to the other two charges.
Solution
Let’s label the charges q1,q2, and q3at the vertices of the equilateral triangle.
We will find the magnitude of the net electric force on q1due to q2and q3.
Step 1: Calculate the electric force from q2on q1. The magnitude of the
electric force between two point charges can be calculated using Coulomb’s Law:
F=k|q1q2|
r2
where kis the Coulomb constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
In this case, the distance between q1and q2is a. The direction of the force
will be along the line connecting q1and q2.
Step 2: Calculate the electric force from q3on q1. Similarly, we can calculate
the electric force from q3on q1using Coulomb’s Law. The distance between q1
and q3is also ain this equilateral triangle configuration.
Step 3: Find the angle between the forces. The forces from q2and q3on q1
are equal in magnitude but opposite in direction. They form a 120◦angle with
respect to each other.
Step 4: Calculate the net electric force on q1. To find the net electric force
on q1, we can treat the forces from q2and q3as vectors. Since they are at 120◦
to each other, we can treat them as components of a single force vector.
The net force magnitude will be the vector sum of the forces from q2and q3:
Fnet =qF2
2+F2
3+ 2F2F3cos(120◦)
Solving this expression will give us the magnitude of the net electric force
on q1due to q2and q3.
18
Question 23
Question
Two point charges, q1= +3.00 nC and q2=−5.00 nC are located on the x-axis
at positions x= 2.00 m and x=−4.00 m respectively. Calculate the magnitude
and direction of the electric force on q1at x= 2.00 m due to the presence of
charge q2.
Solution
Step 1: Calculate the distance between the charges. Given that x1= 2.00 m and
x2=−4.00 m, we can find the distance between the charges using the formula:
r=|x2−x1|
r=| − 4.00 m−2.00 m|
r=| − 6.00 m|
r= 6.00 m
Therefore, the distance between the charges is 6.00 m.
Step 2: Calculate the electric force. The magnitude of the electric force
between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
where - kis the electrostatic constant 8.99 ×109N m2/C2-q1= +3.00 nC =
3.00 ×10−9C - q2=−5.00 nC =−5.00 ×10−9C - r= 6.00 m
Plugging in the values:
F= (8.99 ×109)|3.00 ×10−9× −5.00 ×10−9|
(6.00)2
F= (8.99 ×109)15.00 ×10−18
36
F= (8.99 ×109)×0.4167
F≈3.75 ×109N
Step 3: Determine the direction of the force. The force on q1is towards q2
because the charges are opposite in sign. Thus, the force is attractive.
Therefore, the magnitude of the electric force on q1at x= 2.00 m due to
the presence of charge q2at x=−4.00 m is approximately 3.75 ×109N and is
attractive.
19
Question 24
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 9.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between these two
charges.
Solution
Step 1: Convert the charges to Coulombs. Given that 1µC= 10−6C, we have:
q1=−4.0µC=−4.0×10−6C and q2= 6.0µC= 6.0×10−6C.
Step 2: Calculate the electric force using Coulomb’s Law. 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 Coulomb’s constant with a value of 8.9875 ×109N m2/C2,q1and q2
are the magnitudes of the charges, and ris the separation distance.
Step 3: Substitute the given values into Coulomb’s Law and solve for the
electric force. Substitute k= 8.9875 ×109N m2/C2,q1=−4.0×10−6C,
q2= 6.0×10−6C, and r= 9.0cm = 0.09 m into Coulomb’s Law:
F= 8.9875 ×109×|(−4.0×10−6)(6.0×10−6)|
(0.09)2
Step 4: Calculate the magnitude of the electric force. Performing the calcu-
lations:
F= 8.9875 ×109×24 ×10−12
0.0081
F= 8.9875 ×109×2.96 ×10−9
F= 26.6N
Therefore, the magnitude of the electric force between the two charges is
26.6N.
Question 25
Question
Two point charges q1=−3.0µC and q2= 5.0µC are placed 25 cm apart.
Calculate the magnitude of the electric force between the charges.
20
Solution
Step 1: Convert the charges to Coulombs.
Given: q1=−3.0µC=−3.0×10−6C
q2= 5.0µC= 5.0×10−6C
Step 2: Calculate the distance between the charges.
Given: r= 25 cm = 0.25 m
Step 3: Calculate the magnitude of the electric force.
The electric force between two point charges is given by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the electrostatic constant with a value of 8.99 ×109Nm2/C2.
Substitute the given values into the formula:
F= (8.99 ×109)·|(−3.0×10−6)·(5.0×10−6)|
(0.25)2
Step 4: Calculate the magnitude of the electric force.
F= 8.99 ×109·|(−3.0×10−6)·(5.0×10−6)|
(0.25)2
F= 8.99 ×109·15 ×10−12
0.0625
F= 8.99 ×109·2.4×10−10
F= 2.16 ×10−1N
Therefore, the magnitude of the electric force between the charges is 2.16 ×
10−1N.
21
Step 2: Determine the direction of the electric force on charge q1. Since
charge q2is negative, the force on q1will be attractive. Therefore, the direction
of the electric force on charge q1is towards charge q2.
Step 3: Calculate the magnitude of the electric force on charge q2. Since the
magnitudes of the charges are the same, the magnitude of the force on q2will
be the same as on q1, which is 21.576 N.
Step 4: Determine the direction of the electric force on charge q2. Since
charge q1is positive, the force on q2will be attractive also. Therefore, the
direction of the electric force on charge q2is towards charge q1.
Question 3
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length 2. The charges are +2µC at the top vertex, −3µC at the bottom
left vertex, and +5µC at the bottom right vertex. Calculate the magnitude and
direction of the net electrostatic force acting on the +2µC charge due to the
other charges.
Solution
To find the net electrostatic force acting on the +2µC charge, we need to cal-
culate the individual forces due to each of the other charges and then add them
as vectors.
Step 1: Calculate the force due to the −3µC charge at the bottom left ver-
tex. The magnitude of the force between two point charges q1and q2separated
by distance ris given by Coulomb’s Law:
F=k|q1q2|
r2
where kis the Coulomb’s constant (k≈8.99 ×109Nm2/C2).
For the +2µC charge and the −3µC charge, the distance between them is the
height of the equilateral triangle, h=√3. So, the force between these charges
is:
Fleft =(8.99 ×109)|2×10−6×3×10−6|
(√3)2
Step 2: Calculate the direction of the force due to the −3µC charge. Since
the −3µC charge is negative, the force on the +2µC charge will be attractive,
i.e., towards the −3µC charge. Therefore, the direction is towards the bottom
left vertex.
Step 3: Calculate the force due to the +5µC charge at the bottom right
vertex. Using Coulomb’s Law, the force between the +2µC charge and the
2
+5µC charge is:
Fright =(8.99 ×109)|2×10−6×5×10−6|
(√3)2
Step 4: Calculate the direction of the force due to the +5µC charge. Since
both charges are positive, the force on the +2µC charge will be repulsive, i.e.,
away from the +5µC charge. Therefore, the direction is away from the bottom
right vertex.
Step 5: Find the net force on the +2µC charge. To find the net force, we
need to add these forces as vectors. Let the force due to the −3µC charge be
Fleft and the force due to the +5µC charge be
Fright. The net force is:
Fnet =
Fleft +
Fright
Now, calculate the magnitude and direction of the net force.
Question 4
Question
Two point charges, q1= +2 µC and q2=−3µC, are placed 10 cm apart.
Calculate the magnitude and direction of the electric force each charge exerts
on the other. Assume the charges are located on the x-axis, with q1at the origin
and q2at x= 10 cm.
Solution
Step 1: Calculate the distance between the charges in meters. Given that the
charges are 10 cm apart, the distance rin meters is:
r= 10 cm = 0.10 m
Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force between two point charges is given by Coulomb’s law:
F=k|q1q2|
r2
where k= 8.99 ×109N m2/C2is the electrostatic constant.
Plugging in the given values:
F=(8.99 ×109)(2 ×10−6)(3 ×10−6)
(0.10)2
Calculating:
F=53.94
0.01 = 5394 N
3
Therefore, the magnitude of the electric force between the charges is 5394
N.
Step 3: Determine the direction of the force. Since q1is positive and q2is
negative, the force between them is attractive and directed from q2towards q1.
Thus, the electric force Fon q1points towards q1. The electric force Fon
q2points towards q2.
In conclusion, the magnitude of the electric force each charge exerts on the
other is 5394 N, and the direction of the force on q1is towards q1, while the
direction of the force on q2is towards q2.
Question 5
Question
Two point charges, q1 = +2.0 µC and q2 = -4.0 µC, are placed 10.0 cm apart
in air. Calculate the magnitude and direction of the electric force that q1 exerts
on q2.
Solution
Step 1: Convert µC to Coulombs. Given that 1µC = 10−6C, we can convert the
charges as follows: q1 = 2.0µC = 2.0×10−6C q2 = −4.0µC =−4.0×10−6C
Step 2: Calculate the distance between the two charges. Given that the
charges are placed 10.0 cm apart, we convert this distance to meters: d=
10.0cm = 10.0×10−2m= 0.10m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
The electric force between the charges is given by Coulomb’s Law: F=k·|q1·q2|
d2
where kis the electrostatic constant (8.99 ×109N m2/C2).
Substitute the values: F= (8.99 ×109)·|2.0×10−6·−4.0×10−6|
0.102
F= 71.92 ×10−12/0.01
F= 7.192 ×10−10N
Step 4: Find the direction of the force. Since q1 is positive and q2 is negative,
the force between them will be attractive. Therefore, the force exerted by q1
on q2 is directed towards q1.
The magnitude of the electric force that q1 exerts on q2 is 7.192 ×10−10N
and the direction is towards q1.
Question 6
Question
Two point charges, +2 µC and −3µC, are placed 6 cm apart in air. Calculate
the magnitude and direction of the electric force that one charge exerts on the
other.
4
Solution
Step 1: Convert the charges to Coulombs. We have +2 µC= 2 ×10−6C and
−3µC=−3×10−6C.
Step 2: Calculate the electric force using Coulomb’s Law:
The magnitude of the electric force between two point charges is given by:
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 charges, - ris the distance between the
charges.
Plugging in the values, we get:
F=8.99 ×109· |2×10−6· −3×10−6|
(0.06)2
Step 3: Calculate the magnitude of the electric force:
F=8.99 ×109·6×10−12
0.0036
F=53.94 ×10−3
0.0036
F= 14950 N
Step 4: Determine the direction of the electric force: The force is attractive
since one charge is positive and the other is negative. It acts along the line
joining the charges, from the positive charge to the negative charge.
Therefore, the magnitude of the electric force that the positive charge exerts
on the negative charge is 14950 N, directed from the positive charge to the
negative charge.
Question 7
Question
Three point charges are arranged as shown: - Charge q1= +2.0µC is located at
the origin. - Charge q2=−3.0µC is located at coordinates (0,0.6m). - Charge
q3= +1.0µC is located at coordinates (0.8m,0).
Calculate the net electric force acting on q1due to q2and q3.
Solution
Step 1: Find the electric force between q1and q2: The electric force F12 between
two point charges is given by Coulomb’s law:
F12 =k|q1||q2|
r2
12
5
where - kis Coulomb’s constant (8.99 ×109N m2/C2), - q1and q2are the
magnitudes of the point charges, and - r12 is the distance between the two
charges.
Given that q1= 2.0µC,q2=−3.0µC, and r12 = 0.6m, we have:
F12 =(8.99 ×109)×(2.0×10−6)×(3.0×10−6)
(0.6)2
Step 2: Calculate F12:
F12 =8.99 ×2.0×3.0
(0.6)2×103N
Step 3: Simplify the expression:
F12 =53.94
0.36 ×103N
F12 = 149.83 ×103N
F12 = 1.50 ×105N
Step 4: Find the direction of F12: The direction of F12 is repulsive since the
charges q1and q2are of opposite sign.
Step 5: Repeat steps 1-4 for the electric force F13 between q1and q3and
then find the net electric force acting on q1.
Question 8
Question
Two point charges, q1= +2.0µC and q2=−4.0µC, are placed 10.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between the two
charges.
Solution
Step 1: Convert the charges to the standard unit of Coulombs. Given: q1=
+2.0µC= 2.0×10−6C, and q2=−4.0µC=−4.0×10−6C.
Step 2: Recall the formula for the magnitude of electric force between two
point charges:
F=k·|q1·q2|
r2
where kis the Coulomb constant (k= 8.99 ×109N m2/C2), q1and q2are the
magnitudes of the charges, and ris the separation distance between the charges.
Step 3: Substitute the given values into the formula and solve for the electric
force.
F= (8.99 ×109)·|2.0×10−6· −4.0×10−6|
(0.10)2
6
F= (8.99 ×109)·8.0×10−12
0.01
F= 8.99 ×109·8.0×10−10
F= 71.92 ×10−1
F= 7.192 N
Step 4: Therefore, the magnitude of the electric force between the two
charges is 7.192 N.
Question 9
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed on the x-axis at
positions x=−2.0m and x= 2.0m, respectively. Calculate the magnitude and
direction of the electric force on q1due to q2.
Solution
Step 1: Calculate the distance between the two charges. Given: q1=−4.0µC=
−4.0×10−6Cq2= 6.0µC= 6.0×10−6Cx1=−2.0mx2= 2.0m
The distance rbetween the two charges is:
r=|x2−x1|=|2.0−(−2.0)|= 4.0m
Step 2: Calculate the magnitude of the electric force. The magnitude of the
electric force Fbetween two charges is given by Coulomb’s law:
F=k·|q1·q2|
r2
where k= 8.99 ×109N m2C−2is the Coulomb’s constant.
Plugging in the values:
F= 8.99 ×109·| − 4.0×10−6·6.0×10−6|
(4.0)2
F= 8.99 ×109·24 ×10−12
16
F= 8.99 ×109·1.5×10−12
F≈13.485 N
Step 3: Determine the direction of the force. The force on q1due to q2
is attractive since the charges have opposite signs. Therefore, the force acts
towards the positive xdirection.
So, the magnitude of the electric force on q1due to q2is approximately
13.485 N, and it acts in the positive xdirection.
7
Question 10
Question
Two point charges, q1=−3.5µC and q2= 1.8µC, are placed 12 cm apart in a
vacuum. Calculate the magnitude of the electric force between the charges.
Solution
Step 1: Convert the charges to SI units.
q1=−3.5µC =−3.5×10−6C
q2= 1.8µC = 1.8×10−6C
Step 2: Determine the distance between the charges in meters.
d= 12 cm = 0.12 m
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.
F=k|q1||q2|
d2
Where: k= 8.99 ×109Nm2/C2is the Coulomb constant
Step 4: Substitute the given values into the formula and calculate.
F=(8.99 ×109)(3.5×10−6)(1.8×10−6)
0.122
F=56.735 ×10−15
0.0144
F= 56.735 ×10−15 ×69.444
F= 3938 ×10−15 N
F= 3.938 ×10−12 N
Therefore, the magnitude of the electric force between the charges is 3.938 ×
10−12 N.
Question 11
Question
Two point charges, Q1=−2.0µC and Q2= 4.0µC, are placed 0.10 meters
apart in a vacuum. Calculate the magnitude and direction of the net electric
force on Q1.
8
Solution
Given: Q1=−2.0µC
Q2= 4.0µC
Distance between charges, r= 0.10 m
k= 8.99 ×109N·m2/C2(Coulomb’s constant)
Step 1: Calculate the electric force between the two charges using Coulomb’s
law:
F=k·|Q1·Q2|
r2
F= 8.99 ×109·2.0×10−6·4.0×10−6
0.12
Step 2: Calculate the magnitude of electric force:
F= 8.99 ×109·8.0×10−12
0.01
F= 8.99 ×109·8.0×10−10
F= 7.192 ×10−1N
Step 3: Determine the direction of the net electric force: Since Q1and Q2
have opposite charges, the electric force on Q1will be attractive towards Q2.
Therefore, the net electric force on Q1is directed towards Q2.
Thus, the magnitude of the net electric force on Q1is 7.192 ×10−1N and
the direction is towards Q2.
Question 12
Question
Three point charges are placed at the corners of an equilateral triangle with
side length d= 2 m. The charges are +4µC,−3µC and +5µC. Calculate the
magnitude and direction of the net force on the +4µC charge due to the other
two charges.
Solution
Step 1: Calculate the distance between the +4µC charge and the −3µC charge
using the Pythagorean theorem. The distance r1can be found by dividing the
equilateral triangle into two right triangles.
r1=v
u
u
td
22
+ √3d
2+d!2
=r(1)2+√3+12
9
Step 2: Calculate the distance between the +4µC charge and the +5µC
charge. The distance r2can be found in the same manner.
r2=v
u
u
td
22
+ √3d
2!2
=r(1)2+√32
Step 3: Calculate the magnitude of the force F1between the +4µC and
−3µC charges using Coulomb’s law.
F1=k|q1||q2|
r2
1
Step 4: Calculate the magnitude of the force F2between the +4µC and
+5µC charges in the same way.
F2=k|q1||q3|
r2
2
Step 5: Determine the direction of the net force on the +4µC charge. Since
the forces are vectors, we need to consider their directions. The force F1will be
directed away from the −3µC charge, and the force F2will be directed towards
the +5µC charge. Calculate the net force by considering the vector sum of F1
and F2.
Question 13
Question
Three point charges are arranged as shown in the diagram below. Each charge
has a magnitude of 3.00 µC. Calculate the magnitude and direction of the net
force on the charge q3.
q1q2
q3
Solution
Step 1: Calculate the distances between the charges using the Pythagorean
theorem. Let the length of each side of the equilateral triangle be L.
L=p22+ 1.7322
L=√4+3
L=√7
10
Step 2: Calculate the electric force between charges using Coulomb’s law:
F=k|q1q2|
r2where - kis the Coulomb constant (8.99 ×109N·m2/C2), - q1
and q2are the magnitudes of the charges, and - ris the distance between the
charges.
Step 3: Calculate the force between q1and q3:
F13 = 8.99 ×109(3 ×10−6)(3 ×10−6)
(√7)2
Step 4: Calculate the force between q2and q3:
F23 = 8.99 ×109(3 ×10−6)(3 ×10−6)
(√7)2
Step 5: The net force on q3is the vector sum of the forces F13 and F23.
Since the forces are acting at 120° angles to each other, the net force magnitude
can be calculated using the law of cosines:
Net force =qF2
13 +F2
23 + 2F13F23 cos(120◦)
Step 6: Now, plug in the values to find the magnitude and direction of the
net force on charge q3.
Question 14
Question
Three point charges are placed at the vertices of an equilateral triangle with
sides of length a. The magnitudes of the charges are q,2q, and 3q. Calculate
the net electric force on the charge qdue to the other two charges.
Solution
Let’s denote the charges as follows: q1=q,q2= 2q, and q3= 3q. The distances
from q1to q2and q3are both a.
By Coulomb’s law, the magnitude of the electric force between two charges
qiand qjis Fij =k|qi|·|qj|
r2, where kis Coulomb’s constant (8.99 ×109N m2/C2)
and ris the distance between the charges.
Step 1: We start by finding the electric force on charge q1due to charge q2.
Given that |q1|=q,|q2|= 2q, and r=a, we have:
F12 =k|q1|·|q2|
r2= 8.99 ×109q·2q
a2
Step 2: Next, we find the electric force on charge q1due to charge q3. Given
that |q1|=q,|q3|= 3q, and r=a, we have:
F13 =k|q1|·|q3|
r2= 8.99 ×109q·3q
a2
11
Step 3: Now, we calculate the net electric force on charge q1by summing
the forces due to q2and q3.
Fnet =F12 +F13 = 8.99 ×1092q2
a2+ 8.99 ×1093q2
a2
Therefore, the net electric force on the charge qdue to the other two charges
is:
Fnet = 8.99 ×1095q2
a2N
Question 15
Question
Two point charges, q1= +4.0µC and q2=−6.0µC, are placed 0.10 m apart in
a vacuum. Calculate the magnitude of the electric force that each charge exerts
on the other.
Solution
Let’s denote the magnitude of the electric force between two point charges as F
and their separation distance as r. We can use Coulomb’s law to calculate the
electric force between these two charges. Coulomb’s law is given by:
F=k|q1q2|
r2,
where k= 8.99 ×109N m2/C2is the Coulomb constant.
Step 1: Calculate the electric force that charge q1exerts on charge q2. This
force will be attractive since the charges are of opposite signs.
F1→2=k|q1q2|
r2.
Plugging in the given values:
F1→2= (8.99 ×109N m2/C2)|(4.0×10−6C)(−6.0×10−6C)|
(0.10 m)2.
Step 2: Calculate the magnitude of the electric force between q1and q2.
F1→2= 8.99 ×109×(4.0×10−6)(6.0×10−6)
0.01 .
F1→2= 8.99 ×109×2.4×10−11.
F1→2= 21.576 ×10−2N.
Therefore, the magnitude of the electric force that q1exerts on q2is 2.16 ×
10−1N.
12
Step 3: The force that charge q2exerts on charge q1will be equal in mag-
nitude but opposite in direction to the force calculated in Step 2.
Hence, the magnitude of the electric force that each charge exerts on the
other is 2.16 ×10−1N.
Question 16
Question
Three point charges are placed at the vertices of an equilateral triangle with
side length aas shown below. The charges are +q,−2q, and +4q. Calculate the
magnitude of the electric force on the +qcharge due to the other two charges.
+4q
−2q+q
Solution
Step 1: Calculate the electric force on the +qcharge due to the +4qcharge.
The electric force Fbetween two charges q1and q2separated by a distance ris
given by Coulomb’s law:
F=kq1q2
r2
where kis the Coulomb constant (8.9875 ×109N m2/C2).
In this case, the charge q1= +q,q2= +4q, and the distance between them is
the side length of the equilateral triangle a. Since the charges are located at the
vertices of an equilateral triangle, the distance between them is a
2. Therefore,
the electric force on the +qcharge due to the +4qcharge is:
F=kq(+4q)
(a
2)2=k4q2
a2
4
=k16q2
a2=72kq2
a2
Step 2: Calculate the electric force on the +qcharge due to the −2qcharge.
Similarly, the electric force on the +qcharge due to the −2qcharge can be
calculated:
F=kq(−2q)
a2=−2kq2
a2=−18kq2
a2
Step 3: Combine the two forces. The total force on the +qcharge is the
vector sum of the forces due to the +4qand −2qcharges. Since the forces are
in opposite directions along the same line:
Ftotal =72kq2
a2−18kq2
a2=54kq2
a2
Therefore, the magnitude of the electric force on the +qcharge due to the
other two charges is 54kq2
a2.
13
Question 17
Question
Three point charges are arranged as follows: a charge of +2.0µC at the origin,
a charge of −4.0µC at (0,3m), and a charge of +6.0µC at (4 m,0). Calculate
the net electric force on the charge at the origin.
Solution
To find the net electric force on the charge at the origin, we need to calculate
the individual electric forces due to each of the other charges and then sum
them up vectorially.
Step 1: Calculate the electric force due to the charge of +2.0µC at the
origin.
The electric force F1due to this charge can be calculated using Coulomb’s
law:
F1=kq1q
r2,
where kis the Coulomb constant, q1and qare the magnitudes of the two charges,
and ris the distance between the charges.
The distance between the charges is 0since the charge creating the force is
at the origin. Hence, r= 0.
Therefore, the force due to the charge of +2.0µC is 0.
Step 2: Calculate the electric force due to the charge of −4.0µC at (0,3m).
The electric force F2due to this charge can be calculated using Coulomb’s
law:
F2=kq2q
r2,
where q2and qare the magnitudes of the two charges, and ris the distance
between the charges.
The distance between the charges is √02+ 32= 3 m.
Plugging in the values and accounting for the direction of the force, which
would be along the negative y-axis, we get:
F2=−6.75 Nˆ
j.
Step 3: Calculate the electric force due to the charge of +6.0µC at (4 m,0).
The electric force F3due to this charge can be calculated using Coulomb’s
law:
F3=kq3q
r2,
where q3and qare the magnitudes of the two charges, and ris the distance
between the charges.
The distance between the charges is √42+ 02= 4 m.
Plugging in the values and accounting for the direction of the force, which
would be along the positive x-axis, we get:
F3= 6.75 Nˆ
i.
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Step 4: Find the net electric force on the charge at the origin.
The net force on the charge at the origin is the vector sum of the individual
forces:
Fnet =F1+F2+F3.
Substitute the calculated values and summing the forces vectorially, we get:
Fnet =−6.75 Nˆ
j+ 6.75 Nˆ
i.
Therefore, the net electric force on the charge at the origin is −6.75 Nˆ
j+
6.75 Nˆ
i.
Question 18
Question
Two point charges, Q1=−8.0µC and Q2= 4.0µC, are placed at points A
and B, respectively, with a distance of 40 cm between them. Calculate the
magnitude and direction of the electric force that charge Q1exerts on charge
Q2. Given that the Coulomb constant is k= 8.99 ×109N m2/C2.
Solution
Step 1: Find the magnitude of the electric force using the equation:
F=k· |Q1|·|Q2|
r2
where Fis the electric force, kis the Coulomb constant, |Q1|and |Q2|are the
magnitudes of the charges, and ris the distance between the charges.
Plugging in the given values:
F=(8.99 ×109N m2/C2)·(8.0×10−6C)·(4.0×10−6C)
(0.40 m)2
F=8.99 ×8.0×4.0
0.16 ×103N
F= 8.99 ×8.0×4.0×103N
F= 2.8768 ×105N
Therefore, the magnitude of the electric force that charge Q1exerts on charge
Q2is 2.88 ×105N.
Step 2: Find the direction of the electric force. Since charge Q1is negative,
the force on charge Q2will be attractive, pulling Q2towards Q1. Thus, the
direction of the electric force is from charge Q2towards charge Q1.
15
Question 19
Question
Two point charges, q1= 8.0�C and q2=−10.0�C, are placed 40.0cm apart.
Determine the magnitude and direction of the electric force that q2exerts on
q1.
Solution
Step 1: First, we need to calculate the electric force between the two charges
using Coulomb’s Law:
F=k· |q1·q2|
r2
where kis the Coulomb constant (8.99 ×109N·m2/C2), q1and q2are the
magnitudes of the charges, and ris the distance between the charges.
Step 2: Plug in the given values:
F=(8.99 ×109)· |8.0×10−6·(−10.0×10−6)|
(0.40)2
Step 3: Calculate the magnitude of the electric force:
F=(8.99 ×109)·(8.0×10−6·10.0×10−6)
0.16
F=8.99 ×109·80 ×10−12
0.16
F=7.192 ×10−3
0.16 = 0.04495 N
Step 4: Now, to determine the direction of the force, we consider the fact
that q2exerts an attractive force on q1(since the charges have opposite signs).
Therefore, the force on q1points towards q2.
Step 5: So, the electric force that q2exerts on q1has a magnitude of 0.04495 N
and direction towards q2.
Question 20
Question
Two point charges, q1=−3.0µC and q2= 6.0µC, are placed 10.0cm apart.
Calculate the magnitude of the electric force that each charge exerts on the
other.
16
Solution
Step 1: Convert the given charges to coulombs.
q1=−3.0×10−6C
q2= 6.0×10−6C
Step 2: Calculate the distance between the charges in meters. Given: r=
10.0cm = 0.10 m.
Step 3: Calculate the magnitude of the electric force using Coulomb’s Law:
F=k|q1||q2|
r2
where k= 8.99 ×109N·m2/C2.
Step 4: Substitute the known values and calculate the magnitude of the
electric force.
F=8.99 ×109N·m2/C2×3.0×10−6C×6.0×10−6C
(0.10 m)2
F=161.82
0.01 = 16182 N
Therefore, the magnitude of the electric force that each charge exerts on the
other is 16182 N.
Question 21
Question
Two point charges, q1= 3.0×10−6C and q2=−4.0×10−6C, are placed 10
cm apart in a vacuum. Calculate the magnitude of the electric force between
the charges.
Solution
Step 1: Convert all given values to SI units. Given: q1= 3.0×10−6C, q2=
−4.0×10−6C, r= 10 cm = 0.10 m.
Step 2: Calculate the magnitude of the electric force between the charges
using Coulomb’s Law: The magnitude of the electric force between two point
charges is given by Coulomb’s law:
F=k· |q1|·|q2|
r2,
where k= 8.99 ×109N m2/C2is the electrostatic constant.
17
Substitute the given values into the formula:
F=8.99 ×109×3.0×10−6×4.0×10−6
(0.10)2.
Step 3: Calculate the electric force.
F=8.99 ×109×3.0×10−6×4.0×10−6
(0.10)2=107.88
0.01 = 10788 N.
Therefore, the magnitude of the electric force between the charges is 10788
N.
Question 22
Question
Three point charges are located at the corners of an equilateral triangle with
side length a. Each charge has a magnitude of q. Calculate the magnitude of
the net electric force on one of the charges due to the other two charges.
Solution
Let’s label the charges q1,q2, and q3at the vertices of the equilateral triangle.
We will find the magnitude of the net electric force on q1due to q2and q3.
Step 1: Calculate the electric force from q2on q1. The magnitude of the
electric force between two point charges can be calculated using Coulomb’s Law:
F=k|q1q2|
r2
where kis the Coulomb constant, q1and q2are the magnitudes of the charges,
and ris the distance between the charges.
In this case, the distance between q1and q2is a. The direction of the force
will be along the line connecting q1and q2.
Step 2: Calculate the electric force from q3on q1. Similarly, we can calculate
the electric force from q3on q1using Coulomb’s Law. The distance between q1
and q3is also ain this equilateral triangle configuration.
Step 3: Find the angle between the forces. The forces from q2and q3on q1
are equal in magnitude but opposite in direction. They form a 120◦angle with
respect to each other.
Step 4: Calculate the net electric force on q1. To find the net electric force
on q1, we can treat the forces from q2and q3as vectors. Since they are at 120◦
to each other, we can treat them as components of a single force vector.
The net force magnitude will be the vector sum of the forces from q2and q3:
Fnet =qF2
2+F2
3+ 2F2F3cos(120◦)
Solving this expression will give us the magnitude of the net electric force
on q1due to q2and q3.
18
Question 23
Question
Two point charges, q1= +3.00 nC and q2=−5.00 nC are located on the x-axis
at positions x= 2.00 m and x=−4.00 m respectively. Calculate the magnitude
and direction of the electric force on q1at x= 2.00 m due to the presence of
charge q2.
Solution
Step 1: Calculate the distance between the charges. Given that x1= 2.00 m and
x2=−4.00 m, we can find the distance between the charges using the formula:
r=|x2−x1|
r=| − 4.00 m−2.00 m|
r=| − 6.00 m|
r= 6.00 m
Therefore, the distance between the charges is 6.00 m.
Step 2: Calculate the electric force. The magnitude of the electric force
between two point charges is given by Coulomb’s Law:
F=k|q1q2|
r2
where - kis the electrostatic constant 8.99 ×109N m2/C2-q1= +3.00 nC =
3.00 ×10−9C - q2=−5.00 nC =−5.00 ×10−9C - r= 6.00 m
Plugging in the values:
F= (8.99 ×109)|3.00 ×10−9× −5.00 ×10−9|
(6.00)2
F= (8.99 ×109)15.00 ×10−18
36
F= (8.99 ×109)×0.4167
F≈3.75 ×109N
Step 3: Determine the direction of the force. The force on q1is towards q2
because the charges are opposite in sign. Thus, the force is attractive.
Therefore, the magnitude of the electric force on q1at x= 2.00 m due to
the presence of charge q2at x=−4.00 m is approximately 3.75 ×109N and is
attractive.
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Question 24
Question
Two point charges, q1=−4.0µC and q2= 6.0µC, are placed 9.0 cm apart
in a vacuum. Calculate the magnitude of the electric force between these two
charges.
Solution
Step 1: Convert the charges to Coulombs. Given that 1µC= 10−6C, we have:
q1=−4.0µC=−4.0×10−6C and q2= 6.0µC= 6.0×10−6C.
Step 2: Calculate the electric force using Coulomb’s Law. 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 Coulomb’s constant with a value of 8.9875 ×109N m2/C2,q1and q2
are the magnitudes of the charges, and ris the separation distance.
Step 3: Substitute the given values into Coulomb’s Law and solve for the
electric force. Substitute k= 8.9875 ×109N m2/C2,q1=−4.0×10−6C,
q2= 6.0×10−6C, and r= 9.0cm = 0.09 m into Coulomb’s Law:
F= 8.9875 ×109×|(−4.0×10−6)(6.0×10−6)|
(0.09)2
Step 4: Calculate the magnitude of the electric force. Performing the calcu-
lations:
F= 8.9875 ×109×24 ×10−12
0.0081
F= 8.9875 ×109×2.96 ×10−9
F= 26.6N
Therefore, the magnitude of the electric force between the two charges is
26.6N.
Question 25
Question
Two point charges q1=−3.0µC and q2= 5.0µC are placed 25 cm apart.
Calculate the magnitude of the electric force between the charges.
20
Solution
Step 1: Convert the charges to Coulombs.
Given: q1=−3.0µC=−3.0×10−6C
q2= 5.0µC= 5.0×10−6C
Step 2: Calculate the distance between the charges.
Given: r= 25 cm = 0.25 m
Step 3: Calculate the magnitude of the electric force.
The electric force between two point charges is given by Coulomb’s Law:
F=k·|q1·q2|
r2
where kis the electrostatic constant with a value of 8.99 ×109Nm2/C2.
Substitute the given values into the formula:
F= (8.99 ×109)·|(−3.0×10−6)·(5.0×10−6)|
(0.25)2
Step 4: Calculate the magnitude of the electric force.
F= 8.99 ×109·|(−3.0×10−6)·(5.0×10−6)|
(0.25)2
F= 8.99 ×109·15 ×10−12
0.0625
F= 8.99 ×109·2.4×10−10
F= 2.16 ×10−1N
Therefore, the magnitude of the electric force between the charges is 2.16 ×
10−1N.
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