PHYS 232 - UNIVERSITY PHYSICS II Calculation of electric forces between point charges Question Bank Set 3

PHYS 232 - UNIVERSITY PHYSICS

II - Calculation of electric forces

between point charges

Question Bank - Set 3

Liberty University

Question 1

Question

Two point charges are placed at the corners of an equilateral triangle. The

charge at the top corner has a magnitude of +3µC and the charges at the bottom

corners have magnitudes of −4µC each. If the side length of the triangle is 2

m, calculate the magnitude of the net electric force on the charge at the top

corner.

Solution

Step 1: We first need to calculate the magnitude of the electric force between

the charge at the top corner and each of the charges at the bottom corners

using Coulomb’s law. The electric force between two point charges is given by

the equation:

F=kq1q2

where Fis the magnitude of the electric force, kis Coulomb’s constant (8.99 ×

109N m2/C2), q1and q2are the magnitudes of the two charges, and ris the

distance between the charges.

Step 2: Calculate the electric force between the top charge and one of the

bottom charges. Let q1= 3µC = 3 ×10−6C, q2=−4µC =−4×10−6C, and

r= 2 m. Plugging these values into the Coulomb’s law equation, we get:

F= (8.99 ×109)(3 ×10−6)(−4×10−6)

(2)2

F=−8.99 ×109×12 ×10−12

F=−26.97 ×10−3=−26.97 mN

So, the magnitude of the electric force between the top charge and one of

the bottom charges is 26.97 mN.

Step 3: Since the charges at the bottom corners are at the same distance

from the top charge, their electric forces will have the same magnitude (but

opposite direction due to the opposite signs of the charges).

Step 4: To find the net force on the top charge, we need to calculate the

vector sum of the electric forces from the charges at the bottom corners. Since

the forces are along the same line (due to the symmetry of the equilateral

triangle), the net force will be the sum of the magnitudes in the downward

direction. Thus, the net force is 2×26.97 = 53.94 mN pointing downward.

Therefore, the magnitude of the net electric force on the charge at the top

corner is 53.94 mN downward.

Question 2

Question

Two point charges, q1=−3.0µC and q2= 5.0µC, are separated by a distance

of 8.0cm. Calculate the magnitude of the electric force between the two charges.

Solution

Step 1: Convert the charges to Coulombs. We know that 1µC = 10−6C.

Therefore, the charges q1and q2in Coulombs are:

q1=−3.0µC =−3.0×10−6C

q2= 5.0µC = 5.0×10−6C

Step 2: Calculate the magnitude of the electric force using Coulomb’s Law,

which states that the magnitude of the electric force between two point charges

is given by:

F=k· |q1·q2|

where: - Fis the magnitude of the electric force, - kis Coulomb’s constant

(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the charges, - ris the

distance between the charges.

Step 3: Substitute the given values into the formula and calculate the electric

force:

F=(8.99 ×109N m2/C2)·|−3.0×10−6C·5.0×10−6C|

(0.08 m)2

F=(8.99 ×109)·(1.5×10−11)

0.0064

F=1.3485 ×10−1

0.0064

F= 21.08 N

Therefore, the magnitude of the electric force between the two charges is

21.08 N.

Question 3

Question

Three point charges are arranged at the vertices of an equilateral triangle as

shown below:

−2Q

The side length of the triangle is a.

What is the net electric force on the charge at the top vertex of the triangle

due to the other two charges?

Solution

Step 1: Calculate the electric force exerted on the charge at the top vertex by

the charge at the bottom left vertex. Let F1be the magnitude of the force. By

Coulomb’s Law, the magnitude of the force is given by:

F1=k|Q|2

Step 2: Determine the direction of the force on the charge at the top vertex

due to the charge at the bottom left vertex. The force between Qand Qwill be

repulsive since they are both positive charges. Therefore, the force on the top

charge from the bottom left charge will be directed along the line joining the

two charges (upwards).

Step 3: Calculate the electric force exerted on the charge at the top vertex

by the charge at the bottom right vertex. Let F2be the magnitude of the force.

By Coulomb’s Law, the magnitude of the force is given by:

F2=k|Q· −2Q|

a2=2k|Q|2

Here, the minus sign indicates that the force is attractive.

Step 4: Find the angle between the two forces. The angle between the two

forces is 60◦due to the equilateral triangle configuration.

Step 5: Calculate the net force. The net force is the vector sum of the two

forces F1and F2.

Fnet =√F2

1+F2

2+ 2F1F2cos(60◦)

Fnet =√(k|Q|2

a2)2

+(2k|Q|2

a2)2

+ 2 (k|Q|2

a2)(2k|Q|2

a2)cos(60◦)

Step 6: Simplify the expression.

Fnet =3√3k|Q|2

2a2

Therefore, the net electric force on the charge at the top vertex due to the

other two charges is 3√3k|Q|2

2a2directed upwards.

Question 4

Question

Two point charges, q1=−2µC and q2= 4µC, are placed 10 cm apart in air.

Calculate the magnitude of the electric force between the charges.

Solution

To calculate the magnitude of the electric force between the charges, we can use

Coulomb’s law:

F=k|q1q2|

where - Fis the magnitude of the electric force between the charges, - kis

Coulomb’s constant (8.99 ×109Nm2/C2), - q1and q2are the magnitudes of the

two charges, and - ris the distance between the charges.

Step 1: Identify the given values: - q1=−2µC, - q2= 4µC, - r= 0.10m

(10 cm converted to meters), - k= 8.99 ×109Nm2/C2.

Step 2: Substitute the values into Coulomb’s law:

F=(8.99 ×109Nm2/C2)|(−2×10−6C)(4 ×10−6C)|

(0.10m)2

F=(8.99 ×109)(8 ×10−12)

0.01

F=71.92 ×10−3

0.01

F= 7.192N

So, the magnitude of the electric force between the charges is 7.192 Newtons.

Question 5

Question

Two point charges, q1=−3.5µC and q2= 7.9µC, are placed 8.0 cm apart in a

vacuum. Calculate the magnitude of the electric force exerted on q2by q1.

Solution

Step 1: Convert the charges to coulombs.

q1=−3.5µC =−3.5×10−6C

q2= 7.9µC = 7.9×10−6C

Step 2: Calculate the distance between the charges.

r= 8.0cm = 8.0×10−2m

Step 3: Calculate the magnitude of the electric force using the formula:

F=k· |q1|·|q2|

where k= 8.99 ×109N m2/C2is the Coulomb constant.

Step 4: Substitute the given values into the formula to find the magnitude

of the electric force.

F=(8.99 ×109N m2/C2)·(3.5×10−6C)·(7.9×10−6C)

(8.0×10−2m)2

F=(8.99 ×109)·(3.5) ·(7.9)

64 ×10−4

F=250.415

0.0064

F= 39.088 ×103N

F= 39.1kN

Question 6

Question

Two point charges, q1=−4µC and q2= 5 µC, are placed 8cm apart on the

x-axis. What is the magnitude and direction of the electric force on q1due to

q2?

Solution

Step 1: Given that q1=−4µC,q2= 5 µC, and the distance between them is

r= 8 cm = 0.08 m, we can calculate the electric force using Coulomb’s law:

F=k·|q1|·|q2|

where k= 8.99 ×109N·m2/C2is the electrostatic constant.

Step 2: Substitute the given values:

F= (8.99 ×109)·4×10−6·5×10−6

(0.08)2

F= (8.99 ×109)·20 ×10−12

0.0064

F= 1429187.5N

Step 3: The direction of the electric force can be determined by using the

fact that like charges repel and opposite charges attract. Since q1and q2have

opposite signs, the force on q1due to q2will be attractive and directed towards

q2.

Therefore, the magnitude of the electric force on q1due to q2is 1429187.5N

and the direction is towards q2.

Question 7

Question

Two point charges with a magnitude of 4.5 µC and -3.0 µC are placed 8.0 cm

apart in a vacuum. Calculate the magnitude and direction of the electric force

exerted on the positive charge by the negative charge.

Solution

Step 1: Convert the distances and charges to SI units: Given: q1= 4.5×10−6

C, q2=−3.0×10−6C, r= 8.0×10−2m.

Step 2: Calculate the electric force using Coulomb’s Law: The electric force

between two point charges q1and q2separated by a distance ris given by

Coulomb’s Law:

F=k|q1q2|

where k= 8.99 ×109N m2/C2is the electrostatic constant.

Substitute the given values into the formula:

F=(8.99 ×109)(4.5×10−6)(3.0×10−6)

(8.0×10−2)2

Step 3: Calculate the magnitude of the electric force:

F=(8.99 ×109)(4.5×10−6)(3.0×10−6)

(8.0×10−2)2

F=121.455 ×10−15

64 ×10−4

F= 1898.05 ×10−11 = 1.89805 ×10−7N

Therefore, the magnitude of the electric force exerted on the positive charge

by the negative charge is 1.89805 ×10−7N.

Step 4: Determine the direction of the force: The force will be attractive

since the charges are of different signs. The negative charge will attract the

positive charge along the line joining the charges.

Question 8

Question

Two point charges, q1=−4.0µC and q2= 7.0µC, are placed 20 cm apart.

Calculate the magnitude of the electric force between the charges.

Solution

Step 1: Convert the charges to SI units.

Given: q1=−4.0µC=−4.0×10−6C

q2= 7.0µC= 7.0×10−6C

Step 2: Calculate the distance between the charges in meters.

Given: Distance r= 20 cm = 20 ×10−2m

Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.

The magnitude of the electric force between two point charges is given by:

F=k· |q1|·|q2|

where kis the electrostatic constant (k≈8.99 ×109N m2/C2).

Plugging in the given values:

F=8.99 ×109·|−4.0×10−6|·|7.0×10−6|

(20 ×10−2)2

Step 4: Calculate the magnitude of the electric force.

F=8.99 ×109·4.0×7.0×10−12

202×10−4

F=8.99 ×28.0×10−12

400 ×10−4

F=251.72 ×10−12

4×10−2

F=25.172 ×10−10

4×10−2

F= 6.293 ×10−8N

Therefore, the magnitude of the electric force between the charges is 6.293 ×

10−8N.

Question 9

Question

Two point charges, one with charge +3 µC and the other with charge −6µC,

are placed 10 cm apart in air. Calculate the magnitude of the electric force

between the charges.

Solution

Step 1: Convert the given charges into standard units. The charge of 3µC is

equal to 3×10−6C, and the charge of −6µC is equal to −6×10−6C.

Step 2: Express the distance between the charges in meters. Given that

the charges are 10 cm apart, we convert this to meters by dividing by 100:

10 cm = 0.1m.

Step 3: Calculate the magnitude of the electric force using Coulomb’s Law.

The magnitude of the electric force between two point charges q1and q2sepa-

rated by a distance ris given by

F=k|q1q2|

r2,

where kis Coulomb’s constant, which is approximately 8.99 ×109N m2/C2.

Substitute the given values into the formula:

F=(8.99 ×109)(3 ×10−6)(6 ×10−6)

(0.1)2.

Step 4: Calculate the magnitude of the electric force.

F=(8.99 ×109)(18 ×10−12)

0.01 ,

F=161.82 ×10−3

0.01 ,

F= 16.182 N.

So, the magnitude of the electric force between the charges is 16.182 N.

Question 10

Question

Two point charges, q1=−4µC and q2= 6 µC, are placed 8cm apart. Calculate

the magnitude of the electric force between the charges.

Solution

To calculate the magnitude of the electric force between the charges, we will use

Coulomb’s Law, which states that the magnitude of the electric force between

two point charges is given by:

F=k·|q1·q2|

where: F= magnitude of the electric force, k= Coulomb’s constant = 8.99 ×

109N m2/C2,q1and q2= magnitudes of the charges, and r= separation dis-

tance between the charges.

Step 1: Convert the charges to Coulombs: q1=−4µC=−4×10−6C,

q2= 6 µC= 6 ×10−6C.

Step 2: Substitute the given values into the formula:

F= (8.99 ×109N m2/C2)·| − 4×10−6·6×10−6|

(0.08 m)2

Step 3: Calculate the force:

F= 8.99 ×109×24 ×10−12

0.0064

F= 8.99 ×24

0.0064

F= 8.99 ×3750

F= 33.7125 ×109N

Therefore, the magnitude of the electric force between the charges is 33.7125×

109N.

Question 11

Question

Three point charges are placed at the vertices of an equilateral triangle of side

a. The charges are +q,−q, and +q. Calculate the net electric force on the

charge −q.

Solution

Let’s break down the solution into steps:

Step 1: Calculate the electric force on the charge −qdue to the charge +q.

The magnitude of the electric force between two charges q1and q2separated by

a distance ris given by Coulomb’s law:

F=k|q1q2|

where kis Coulomb’s constant, q1and q2are the charges, and ris the separation

distance.

Since the charges are placed at the vertices of an equilateral triangle of side

a, the distance between the charges +qand −qis a.

The direction of the force will be along the line joining the two charges,

which is from −qto +q. Thus, the force on −qis attractive towards +q.

Substitute q1=qand q2=−qinto Coulomb’s law:

F+q=k|q(−q)|

a2=kq2

Step 2: Calculate the electric force on the charge −qdue to the charge +q.

From symmetry, we know that this force will be equal in magnitude but opposite

in direction to the force calculated in Step 1. Therefore, the force on −qdue to

+qwill be:

F−q→+q=−kq2

Step 3: Calculate the electric force on the charge −qdue to the charge +q.

The net force on the charge −qwill be the vector sum of the forces calculated

in Step 1 and Step 2. Since the forces are along the same line but in opposite

directions, we subtract their magnitudes to find the net force:

Fnet =F+q−F−q→+q=kq2

a2+kq2

a2= 2kq2

Therefore, the net electric force on the charge −qis 2kq2

a2, directed towards

the charge +q.

Question 13

Question

Two point charges, q1=−2.0µC and q2= 4.0µC, are placed on the x-axis at

x=−1.0m and x= 1.0m, respectively. Calculate the electric force that each

charge exerts on the other.

Solution

Step 1: Calculate the distance between the charges. We have the charges placed

at x=−1.0m and x= 1.0m on the x-axis. The distance between the charges

can be found using the formula:

Distance between charges =|x2−x1|=|1.0m−(−1.0m)|= 2.0m

Step 2: Calculate the magnitude of the electric force between the charges.

The magnitude of the electric force between two point charges can be calculated

using Coulomb’s law:

F=k|q1·q2|

where kis Coulomb’s constant (8.99 ×109N m2/C2), q1and q2are the charges,

and ris the distance between them. Therefore, the force F12 on charge q2due

to q1is:

F12 = 8.99 ×109|(−2.0×10−6C)·(4.0×10−6C)|

(2.0m)2

F12 = 8.99 ×1098.0×10−12

4.0

F12 = 1.80 ×10−3N

By Newton’s third law, the force on charge q1due to q2is equal in magnitude

but opposite in direction. Therefore, the force F21 on charge q1due to q2is also

1.80 ×10−3N.

Question 15

Question

Two point charges, q1=−2µC and q2= 3µC, are placed 5 cm apart. Calculate

the magnitude of the electric force between the charges.

Solution

Step 1: Convert the given distances into meters. Given: q1=−2µC =−2×

10−6C

q2= 3µC = 3 ×10−6C

r= 5 cm = 5 ×10−2m

Step 2: Calculate the electric force using Coulomb’s Law: The electric force

between two charges is given by Coulomb’s Law:

F=k|q1·q2|

where k= 8.99 ×109N·m2/C2is the Coulomb’s constant.

Step 3: Plug in the given values and compute the electric force:

F= 8.99 ×109| − 2×10−6·3×10−6|

(5 ×10−2)2

Step 4: Simplify the expression:

F= 8.99 ×109| − 6×10−12|

2.5×10−4

F= 8.99 ×1096×10−12

2.5×10−4

F= 8.99 ×109×6

2.5×10−8

F= 8.99 ×109×2.4×10−8

F= 21.576 ×10 ×10−8

F= 2.1576 ×10−6N

Therefore, the magnitude of the electric force between the charges is 2.1576×

10−6N.

Question 16

Question

Two point charges are placed at the corners of an equilateral triangle with sides

of length a. One charge has a magnitude qand the other charge has a magnitude

3q. Calculate the magnitude of the electric force between these charges.

Solution

Let’s label the charges as q1=qand q2= 3q. The distance between these

charges is given by the side length of the equilateral triangle, a.

Step 1: Calculate the electric field due to q1at the position of q2. The

electric field Eat a distance raway from a point charge Qis given by the

formula:

E=k|Q|

where kis Coulomb’s constant 8.99 ×109N m2/C2.

For the electric field E1due to q1at the position of q2, we have:

E1=k|q|

Step 2: Calculate the electric force on q2due to q1. The electric force Fon

a charge Qin an electric field Eis given by:

F=Q·E

Thus, the force F1on q2due to q1is:

F1=q2·E1= 3q·k|q|

Step 3: Determine the direction of the force. Since both charges have the

same sign, the force between them is repulsive. The force vector will point away

from q1along the line connecting q1and q2.

Step 4: Calculate the magnitude of the electric force between the charges.

The magnitude of the force is:

|F1|= 3q·k|q|

a2=3kq2

Therefore, the magnitude of the electric force between the charges is 3kq2

a2.

Question 17

Question

Two point charges, q1=−3.0µC and q2= 5.0µC, are placed 8.0 cm apart.

Calculate the magnitude of the electric force between the charges.

Solution

Step 1: Convert the charges to Coulombs: The charges q1and q2are given

in microCoulombs (µC), so we need to convert them to Coulombs. Given:

q1=−3.0µC and q2= 5.0µC.

Step 2: Convert the charges to Coulombs: q1=−3.0µC =−3.0×10−6C

q2= 5.0µC = 5.0×10−6C

Step 3: Calculate the force: The magnitude of the electric force between two

point charges is given by Coulomb’s law:

F=k·|q1·q2|

where k= 8.99×109N·m2/C2is the Coulomb’s constant, |q1·q2|is the product

of the charges, and ris the distance between the charges.

Given: r= 8.0cm = 0.08 m

Step 4: Substitute the values and find the magnitude of the electric force:

F= (8.99 ×109)×|(−3.0×10−6)·(5.0×10−6)|

(0.08)2

Step 5: Calculate the magnitude of the electric force:

F= (8.99 ×109)×|(−1.5×10−11)|

0.0064

F= (8.99 ×109)×1.5×10−11

0.0064

F= (8.99 ×109)×2.34375 ×10−7

F= 2.109375 ×103N

Therefore, the magnitude of the electric force between the charges is 2.109375×

103N.

Question 18

Question

Three point charges are fixed in the xy-plane: a +3.0 µC charge at (0, 0) m, a

-5.0 µC charge at (4.0, 0) m, and a +4.0 µC charge at (0, 3.0) m. Calculate the

magnitude of the net force on the +4.0 µC charge.

Solution

Step 1: Determine the individual forces between each pair of charges.

The force between two point charges q1and q2separated by distance ris

given by Coulomb’s law:

F=k|q1·q2|

where k= 8.99 ×109N m2/C2is the Coulomb constant.

For the +3.0 µC and +4.0 µC charges, the distance between them is r1= 3.0

m. So, the force between them is:

F1= 8.99 ×109(3.0×10−6)(4.0×10−6)

(3.0)2

Step 2: Calculate the value of F1.

F1= (8.99 ×109)×(3.0×10−6)(4.0×10−6)

9.0

F1= 4.79 ×10−2N

For the -5.0 µC and +4.0 µC charges, the distance between them is r2= 5.0

m. So, the force between them is:

F2= 8.99 ×109(5.0×10−6)(4.0×10−6)

(5.0)2

Step 3: Calculate the value of F2.

F2= (8.99 ×109)×(5.0×10−6)(4.0×10−6)

25.0

F2= 7.19 ×10−3N

Step 4: Find the direction of each force by considering the signs of the

charges.

The force F1is repulsive because both charges are positive. The force F2is

attractive because one charge is positive and the other is negative.

Step 5: Calculate the net force on the +4.0 µC charge.

Since the forces F1and F2act in perpendicular directions along the x and

y-axes, their vector sum is the net force on the +4.0 µC charge. This magnitude

can be found using Pythagoras’ theorem.

Fnet =√F2

1+F2

Fnet =√(4.79 ×10−2)2+ (7.19 ×10−3)2

Step 6: Calculate the value of the net force Fnet.

Fnet =√(2.30 ×10−3)2+ (5.17 ×10−5)2

Fnet ≈2.30 ×10−3N

Therefore, the magnitude of the net force on the +4.0 µC charge is approx-

imately 2.30 ×10−3N.

Question 19

Question

Three point charges are arranged in a straight line as shown below:

Q1Q2Q3

The charges are Q1= +4.0µC,Q2=−2.0µC, and Q3=−6.0µC. The

distances between the charges are r12 = 0.50 mand r23 = 2.0m. Determine the

magnitude and direction of the net force on Q2due to the other two charges.

Solution

Step 1: Calculate the force on Q2due to Q1. The magnitude of the force F12

on Q2due to Q1is given by Coulomb’s Law:

F12 =k|Q1Q2|

where kis Coulomb’s constant (8.99 ×109N m2/C2).

Substitute the given values:

F12 =(8.99 ×109Nm2/C2)(4.0×10−6C)(2.0×10−6C)

(0.50 m)2

F12 ≈57504 N

Step 2: Determine the direction of the force F12. Since Q1is positive and

Q2is negative, the force F12 on Q2due to Q1will be attractive, towards Q1.

Step 3: Calculate the force on Q2due to Q3. The magnitude of the force

F23 on Q2due to Q3can be calculated using Coulomb’s Law:

F23 =k|Q2Q3|

Substitute the given values:

F23 =(8.99 ×109Nm2/C2)(2.0×10−6C)(6.0×10−6C)

(2.0m)2

F23 = 6742 N

Step 4: Determine the direction of the force F23. Since Q2and Q3are both

negative charges, the force F23 on Q2due to Q3will be repulsive, away from

Q3.

Step 5: Find the net force on Q2. To find the net force on Q2, we need to

consider the directions of the forces F12 and F23. Since F12 is to the left and

F23 is to the right, we need to subtract them to find the net force:

Net Force =F12 −F23

Net Force = 57504 N−6742 N

Net Force = 50762 Nto the left

Therefore, the magnitude of the net force on Q2due to the other two charges

is 50762 Nto the left.

Question 20

Question

Two point charges, Q1= 5 µC and Q2=−3µC, are placed 10 cm apart in a

vacuum. Calculate the magnitude of the electric force between them.

Solution

Step 1: Convert the charges to Coulombs. Given that 1µC= 10−6C,

Q1= 5 µC= 5 ×10−6C

Q2=−3µC=−3×10−6C

Step 2: Calculate the magnitude of the electric force using the formula for

the electric force between two point charges:

F=k·|Q1|·|Q2|

where k= 8.99 ×109N m2/C2is Coulomb’s constant, and r= 10 cm = 0.1m

is the separation between the charges.

Step 3: Substitute the given values into the formula to find the magnitude

of the electric force:

F= (8.99 ×109)·5×10−6·3×10−6

0.12

Step 4: Calculate the electric force:

F= 8.99 ×109·15 ×10−12

0.01

F= 8.99 ×109·1.5×10−10

F= 13.485 ×10−1

F= 1.35 N

Therefore, the magnitude of the electric force between the two charges is

1.35 N.

Question 21

Question

Two point charges, q1=−2.0µC and q2= 3.0µC, are placed 10.0 cm apart in

air. Calculate the magnitude of the electric force between them.

Solution

Step 1: Convert the charges to Coulombs. The charges are given in micro-

coulombs (µC). To convert to coulombs, we use the conversion factor 1µC=

10−6C. q1=−2.0µC=−2.0×10−6C=−2.0nC q2= 3.0µC= 3.0×10−6C=

3.0nC

Step 2: Calculate the electric force using Coulomb’s law. The magnitude of

the electric force between two point charges is given by Coulomb’s law:

F=k|q1q2|

where: - kis the Coulomb constant, 8.99 ×109N m2/C2, - q1and q2are the

magnitudes of the charges, - ris the separation distance between the charges.

Plugging in the values:

F=(8.99 ×109N m2/C2)· |(−2.0nC)(3.0nC)|

(0.10 m)2

Step 3: Calculate the electric force.

F=(8.99 ×109)·6.0×10−6

0.01

F=5.394 ×104

0.01

F= 5.394 ×106N

Therefore, the magnitude of the electric force between the charges is 5.394 ×

106N.

Question 22

Question

Three point charges are arranged along the x-axis as follows: +qat x=−a,

−2qat x= 0, and +3qat x= 2a. Calculate the net force on the charge at

x= 0.

Solution

1. Calculate the force on the charge at x= 0 due to the charge at x=−a.

The force between two charges is given by Coulomb’s law: F=kq1q2

r2, where k

is the Coulomb constant, q1and q2are the magnitudes of the charges, and ris

the distance between the charges. Since the charges at x= 0 and x=−ahave

opposite signs, the force is attractive. The distance between them is a.

So, the force on the charge at x= 0 due to the charge at x=−ais:

F1=k|+q|·|−2q|

a2=k2q2

2. Calculate the force on the charge at x= 0 due to the charge at x= 2a.

The force between these two charges is repulsive since they have the same sign.

The distance between them is 2a.

So, the force on the charge at x= 0 due to the charge at x= 2ais:

F2=k|+ 3q|·|−2q|

(2a)2=k6q2

4a2=k3q2

2a2

3. The net force on the charge at x= 0 is the vector sum of F1and F2.

Since they act in opposite directions, we need to subtract them:

Fnet =F2−F1=k(3q2

2a2−2q2

a2)=kq2

2a2

Therefore, the net force on the charge at x= 0 is kq2

2a2in the positive

x-direction.

Question 23

Question

Two point charges, q1= +4.0µC and q2=−2.0µC, are located 0.030 meters

apart. Calculate the magnitude of the electric force between these charges.

Solution

Step 1: Convert the charges to coulombs.

Given: q1= +4.0µC and q2=−2.0µC.

1µC= 10−6C. So, q1= +4.0×10−6C and q2=−2.0×10−6C.

Step 2: Calculate the electric force using Coulomb’s Law.

The formula for the electric force between two point charges is given by Coulomb’s

Law:

F=k· |q1·q2|

where Fis the magnitude of the electric force, q1and q2are the magnitudes

of the charges, kis the Coulomb constant (k= 8.99 ×109N m2/C2), ris the

separation between the charges.

Substitute the given values into the formula:

F=(8.99 ×109N m2/C2)· |4.0×10−6C· −2.0×10−6C|

(0.030 m)2

Step 3: Calculate the electric force.

F=(8.99 ×109)·8.0×10−12

(0.030)2

F=7.192 ×10−2

0.0009 = 79.91 N

Therefore, the magnitude of the electric force between the charges q1=

+4.0µC and q2=−2.0µC when they are 0.030 meters apart is 79.91 N.

Question 24

Question

Two point charges, q1=−3.0µC and q2= 5.0µC, are placed on the x-axis at

the points x=−1.0m and x= 1.0m, respectively. Calculate the magnitude

and direction of the force on q1due to q2.

Solution

Step 1: Calculate the distance between the charges. The distance between the

charges is the difference in their positions on the x-axis.

Distance =|1.0m−(−1.0m)|= 2.0m

Step 2: Calculate the force magnitude using Coulomb’s law. Coulomb’s law

states that the magnitude of the electrostatic force between two point charges

is given by:

F=k|q1||q2|

where k= 8.99 ×109N·m2/C2is the electrostatic constant, q1and q2are the

charges, and ris the distance between the charges.

Plugging in the values:

F= 8.99 ×109N·m2/C2·3.0×10−6C·5.0×10−6C

(2.0m)2

F= 8.99 ×109×15.0×10−12 N/4.0m2= 3.3735 mN

Step 3: Determine the direction of the force. Since q1is negative and q2is

positive, the force on q1is attractive towards q2.

Therefore, the magnitude of the force on q1due to q2is 3.3735 mN directed

towards q2.

Question 25

Question

Two point charges are placed on the x-axis. The first charge, q1= +4.0µC,

is located at x= 0.20 m and the second charge, q2=−2.0µC, is located at

x= 0.40 m. Calculate the total electric force experienced by a −3.0µC charge

placed at the origin.

Solution

Step 1: Calculate the electric force between the −3.0µC charge at the origin

and the +4.0µC charge at x= 0.20 m. The electric force between two point

charges is given by Coulomb’s law:

F=k|q1q2|

where - Fis the magnitude of the electric force, - kis Coulomb’s constant

(8.99 ×109N m2/C2), - q1and q2are the magnitudes of the charges, - ris the

distance between the charges.

Plugging in the values:

F1=(8.99 ×109N m2/C2)×(4.0×10−6C)×(3.0×10−6C)

(0.20 m)2

F1=35.96 ×3

0.04 = 2694 N

Therefore, the electric force between the −3.0µC charge at the origin and

the +4.0µC charge at x= 0.20 m is 2694 N in the positive x-direction.

Step 2: Calculate the electric force between the −3.0µC charge at the origin

and the −2.0µC charge at x= 0.40 m. Following the same steps as in Step 1,

we find:

F2=(8.99 ×109N m2/C2)×(2.0×10−6C)×(3.0×10−6C)

(0.40 m)2

F2=17.98 ×3

0.16 = 449.5N

Therefore, the electric force between the −3.0µC charge at the origin and

the −2.0µC charge at x= 0.40 m is 449.5N in the positive x-direction.

Step 3: Calculate the total electric force experienced by the −3.0µC charge

at the origin. The total force is the vector sum of the two forces:

Ftotal =F1+F2= 2694 N+ 449.5N= 3143.5N

Therefore, the total electric force experienced by the −3.0µC charge at the

origin is 3143.5N in the positive x-direction.