PHYS 232 - UNIVERSITY PHYSICS II Calculation of electric forces between point charges Question Bank Set 10

PHYS 232 - UNIVERSITY PHYSICS

II - Calculation of electric forces

between point charges

Question Bank - Set 10

Liberty University

Question 1

Question

Two point charges, q1= 3.2×10−6C and q2=−4.8×10−6C, are placed 6.0

cm apart in a vacuum. Calculate the magnitude of the electric force between

the charges.

Solution

Step 1: Convert the distance from cm to meters. Given that 1 cm = 1×10−2

m, the distance between the charges is:

d= 6.0cm ×1×10−2m/cm = 0.06 m

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

between two point charges is given by Coulomb’s Law:

F=k×|q1×q2|

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

Substitute the given values into the equation:

F= (8.99 ×109)×|3.2×10−6× −4.8×10−6|

(0.06)2

Step 3: Perform the calculations.

F= 8.99 ×109×3.2×4.8×10−12

0.0036

F= 8.99 ×109×15.36 ×10−12

0.0036

F= 8.99 ×15.36

0.0036 ×10−3

F= 8.99 ×4266.67

1×10−3

F= 3.84 ×104N

Therefore, the magnitude of the electric force between the charges is 3.84 ×

104N.

Question 2

Question

Two point charges q1= 3.0µC and q2=−4.0µC are placed 10.0 cm apart in

air. Calculate the magnitude and direction of the electric force that q1exerts

on q2.

Solution

Step 1: Convert the charges to coulombs. Given: q1= 3.0µC = 3.0×10−6C

q2=−4.0µC =−4.0×10−6C

Step 2: Find the distance between the charges in meters. Given: r= 10.0

cm = 10.0×10−2m= 0.10 m

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

Coulomb’s law states:

F=k|q1q2|

where kis Coulomb’s constant, k= 8.9875 ×109N·m2/C2.

Plugging in the values:

F= 8.9875 ×109×|3.0×10−6× −4.0×10−6|

(0.10)2

F= 8.9875 ×109×12 ×10−12

0.01

F= 8.9875 ×109×1.2×10−9

F= 10.785 ×10−1

F= 10.785 N

Step 4: Determine the direction of the force. The force is attractive since

one charge is positive and the other is negative.

Therefore, the magnitude of the electric force that q1exerts on q2is 10.785 N

directed towards q2.

Question 3

Question

Three point charges are arranged in an equilateral triangle. The charges are

+q,−2q, and +3qwith sides of length d. Calculate the magnitude of the net

electrostatic force on the +3qcharge due to the other two charges. Assume the

charges are free to move.

Solution

1. First, let’s calculate the electrostatic force between the +qand +3qcharges.

Since the charges are the same sign, this force will be repulsive. The magnitude

of the electrostatic force F1between a pair of charges qand Qis given by

Coulomb’s Law:

F1=k· |q|·|Q|

where kis the Coulomb constant (8.99 ×109N m2/C2), |q|is the magnitude of

the +qcharge, |Q|is the magnitude of the +3qcharge, and ris the distance

between the charges. Since this distance is din an equilateral triangle, r=d.

2. Substituting the given values into the formula, we get:

F1=(8.99 ×109)(q)(3q)

d2=26.97 ×109q2

So, the magnitude of the force F1between the +qand +3qcharges is 26.97×109q2

3. Next, let’s calculate the electrostatic force between the −2qand +3q

charges. Since the charges are opposite signs, this force will be attractive. Let’s

denote this force as F2.

4. Using Coulomb’s Law again, we have:

F2=k· |(−2q)|·|3q|

d2=6kq2

5. Now, to find the net force on the +3qcharge, we need to consider the

directions. Since F1and F2are along the same line, we simply subtract these

two forces:

Net force =F1−F2=26.97 ×109q2

d2−6kq2

=26.97 ×109−6×8.99 ×109q2

6. Therefore, the magnitude of the net electrostatic force on the +3qcharge

due to the other two charges is (26.97 −53.94) ×109q2

d2=−26.97 ×109q2

d2.

Question 4

Question

Three point charges are arranged along the x-axis: a charge of +3µC at the

origin, a charge of −5µC at x= 0.1m, and a charge of +2µC at x= 0.2m.

Calculate the net electric force on the charge of −5µC.

Solution

Step 1: Calculate the electric force between the charge of +3µC at the origin

and the charge of −5µC at x= 0.1m. The electric force F1between two charges

q1and q2separated by a distance ris given by Coulomb’s law:

F1=k·|q1q2|

where k= 8.99 ×109Nm²/C² is Coulomb’s constant.

Substitute the given values:

F1= 8.99 ×109·3×10−6×5×10−6

(0.1)2

Calculating F1gives:

F1= 1.35 ×10−3N

Step 2: Calculate the electric force between the charge of +2µC at x= 0.2

m and the charge of −5µC at x= 0.1m. Using Coulomb’s law again:

F2=k·|q1q2|

Substitute the given values:

F2= 8.99 ×109·2×10−6×5×10−6

(0.2−0.1)2

Calculating F2gives:

F2= 4.49 ×10−3N

Step 3: Find the net force on the charge of −5µC. Since the forces are in

opposite directions, we need to consider them as negative and positive values.

The net force is the sum of the forces:

Fnet =F1−F2= 1.35 ×10−3N−4.49 ×10−3N

Calculating Fnet gives:

Fnet =−3.14 ×10−3N

Therefore, the net electric force on the charge of −5µC is −3.14 ×10−3N,

pointing towards the charge of +2µC at x= 0.2m.

Question 5

Question

Two point charges, q1=−6.0µC and q2= 4.0µC, are placed 10.0 cm apart in

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

Solution

Given: q1=−6.0µC

q2= 4.0µC

r= 10.0cm

The electric force between two point charges q1and q2separated by a dis-

tance ris given by Coulomb’s law:

F=k·|q1·q2|

where kis the electrostatic constant equal to 8.99 ×109N·m2/C2.

Step 1: Convert the distance rfrom cm to m:

r= 10.0cm = 0.1m

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

F= (8.99 ×109)·| − 6.0×10−6·4.0×10−6|

(0.1)2

F= 8.99 ×109·24 ×10−12

0.01

F= 8.99 ×109·2.4×10−10

F= 2.1576 ×10−1N

The magnitude of the electric force between the charges q1and q2is 0.216 N.

Question 6

Question

Three point charges are located on the vertices of an equilateral triangle of side

length a. The charges are +q,−2q, and +q. Calculate the magnitude and

direction of the electric force on the positive charge +qdue to the other two

charges.

Solution

Given: - Charge at vertex A = +q, - Charge at vertex B = -2q, - Charge

at vertex C = +q, - Distance between charges = a(side length of equilateral

triangle), - Coulomb’s constant = k.

The direction of the electric force on the positive charge +qdue to the other

two charges will depend on the configuration of the charges. Since the charges

are symmetrical with respect to the positively charged particle, the net force on

the positive charge will be directed vertically downwards.

Step 1: Calculate the force on +qdue to −2qat vertex B: The electric force

FAB on the charge at A due to the charge at B is given by Coulomb’s law:

FAB =k· |q|·|−2q|

a2=2kq2

Step 2: Calculate the force on +qdue to +q at vertex C: The electric force

FAC on the charge at A due to the charge at C is given by Coulomb’s law:

FAC =k· |q|·|q|

a2=kq2

Step 3: Calculate the net force and its direction: The net force on the

charge at A will be the sum of the forces FAB and FAC in the vertical downward

direction:

Fnet =FAB +FAC =2kq2

a2+kq2

a2=3kq2

Therefore, the magnitude of the net force on the charge +qis 3kq2

a2and it is

directed vertically downwards.

Question 7

Question

Two point charges, q1= 3 µC and q2=−2µC, are placed 10 cm apart in air.

Calculate the magnitude of the electric force between these charges.

Solution

Step 1: Determine the distance between the charges in meters. Given that the

charges are 10 cm apart, we need to convert this distance to meters:

10 cm = 0.1m

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

using Coulomb’s Law: Coulomb’s Law states that the magnitude of the electric

force between two point charges is given by:

F=k· |q1·q2|

where: - Fis the magnitude of the electric force, - kis the electrostatic constant

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

the distance between the charges.

Plugging in the given values:

F=(8.99 ×109N m2/C2)· |3×10−6C· −2×10−6C|

(0.1m)2

Step 3: Calculate the electric force.

F=8.99 ×109·6×10−6·2×10−6

0.01

F=8.99 ×6×2

10 ×10−3

F=107.88

10 ×10−3

F= 10.788 ×10−3

F= 10.788 mN

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

10.788 mN.

Question 9

Question

Two point charges q1= +2.5µC and q2=−4.0µC are placed 10 cm apart.

Calculate the magnitude of the electric force between the charges.

Solution

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

Coulomb’s Law. Coulomb’s Law states that the magnitude of the electric force

between two point charges is given by:

F=k|q1q2|

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

(8.9875 ×109N·m2/C2), - q1and q2are the magnitudes of the charges, and -

ris the separation distance between the charges.

Step 1: Convert the given charges to coulombs:

q1= +2.5µC = 2.5×10−6C

q2=−4.0µC =−4.0×10−6C

Step 2: Determine the separation distance in meters:

r= 10 cm = 0.1m

Step 3: Calculate the magnitude of the electric force:

F= (8.9875 ×109)|2.5×10−6× −4.0×10−6|

(0.1)2

F= 8.9875 ×109×1.0×10−11 ×100

F= 8.9875 ×1013N

Therefore, the magnitude of the electric force between the charges is 8.9875×

1013 N.

Question 10

Question

Two point charges, q1=−3.0µC and q2= +6.0µC, are placed 0.20 m apart.

Calculate the magnitude and direction of the electric force that q2exerts on q1.

Solution

Step 1: Convert the given charges to standard units.

•q1=−3.0µC =−3.0×10−6C

•q2= +6.0µC = 6.0×10−6C

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

F=k·|q1·q2|

where kis the Coulomb constant (8.99 ×109N·m2/C2) and ris the distance

between the charges.

Step 3: Substitute the given values into the equation:

F= (8.99 ×109)·| − 3.0×10−6·6.0×10−6|

0.202

Step 4: Simplify the expression to find the magnitude of the electric force.

F= (8.99 ×109)·18 ×10−12

0.04 = 4.497 ×10−2N

Step 5: Determine the direction of the electric force. Since q2is positive, it

exerts a repulsive force on q1, which is negative. Thus, the direction of the force

is away from q2.

Therefore, the magnitude of the electric force that q2exerts on q1is 4.497 ×

10−2N, directed away from q2.

Question 11

Question

Three point charges are arranged in a line. Charge q1=−3.0µC is located

at x=−1.0m, charge q2= 6.0µC is located at the origin, and charge q3=

−4.0µC is located at x= 2.0m. Calculate the magnitude and direction of the

net electrostatic force on q1due to q2and q3.

Solution

Step 1: Calculate the force on q1due to q2. The electrostatic force between two

point charges is given by Coulomb’s Law:

F=k·|q1·q2|

where k= 8.99 ×109Nm2/C2is Coulomb’s constant, q1=−3.0µC =−3.0×

10−6C,q2= 6.0µC = 6.0×10−6C, and r= 1.0mis the distance between q1

and q2.

Plugging in these values, we get

Fq1,q2= 8.99 ×109·| − 3.0×10−6·6.0×10−6|

(1.0)2

Fq1,q2= 8.99 ×109·18 ×10−12

1.0

Fq1,q2= 161.82 ×10−3N

Step 2: Determine the direction of the force on q1due to q2. Since q1and

q2are oppositely charged, the force will be attractive. Therefore, the force will

act towards q2.

Step 3: Calculate the force on q1due to q3. Similarly, using Coulomb’s Law,

we find the force on q1due to q3:

Fq1,q3= 8.99 ×109·| − 3.0×10−6· −4.0×10−6|

(3.0)2

Fq1,q3= 8.99 ×109·12 ×10−12

9.0

Fq1,q3= 119.76 ×10−3N

Step 4: Determine the direction of the force on q1due to q3. As q1and q3

are both negative charges, the force between them will also be attractive, acting

towards q3.

Step 5: Calculate the net force on q1by summing the individual forces from

q2and q3. The net force is given by

Fnet =Fq1,q2+Fq1,q3

Fnet = 161.82 ×10−3+ 119.76 ×10−3

Fnet = 281.58 ×10−3N

The net force on q1is 281.58 ×10−3Ntowards the right.

Question 12

Question

Three point charges are placed at the following positions: q1= +3 nC at (0,0),

q2=−1nC at (4,0), and q3= +2 nC at (0,3). Calculate the net electric force

on q2due to q1and q3.

Solution

Step 1: Calculate the electric force on q2due to q1: The electric force between

two point charges q1and q2separated by a distance ris given by Coulomb’s

Law:

F1,2=k|q1||q2|

In this case, q1= +3 nC, q2=−1nC, and r= 4 (since q1and q2are 4 units

apart horizontally). Using k= 8.99 ×109N m2/C2, we have:

F1,2= (8.99 ×109)(3 ×10−9)(1 ×10−9)

(4)2

F1,2= (8.99 ×109)3

16 ×10−18

F1,2=2697

16 ×10−9

F1,2= 168.56 ×10−9

F1,2= 1.69 ×10−6N

Step 2: Calculate the electric force on q2due to q3: Similarly, using q2=

−1nC, q3= +2 nC, and r= 3 (since q2and q3are 3 units apart vertically), we

have:

F2,3=k|q2||q3|

F2,3= (8.99 ×109)(1 ×10−9)(2 ×10−9)

(3)2

F2,3=2×8.99

9×10−9

F2,3= 1.99 ×10−6N

Step 3: Calculate the net electric force on q2: The net force is the vector

sum of the two forces:

Net force on q2=F1,2+F2,3

Net force on q2= 1.69 ×10−6+ 1.99 ×10−6

Net force on q2= 3.68 ×10−6N

Therefore, the net electric force on q2due to q1and q3is 3.68 ×10−6N.

Question 13

Question

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

length d= 2.0m as shown below. The charges are +3.0µC at the upper corner,

−2.0µC at the lower left corner, and +6.0µC at the lower right corner. Calcu-

late the magnitude and direction of the net electric force on the charge at the

upper corner.

` `

U

3.0µC

−2.0µC6.0µC

Solution

Step 1: Calculate the electric force between the upper corner charge at +3.0µC

and the lower left corner charge at −2.0µC. The electric force F1between two

point charges q1and q2separated by a distance ris given by Coulomb’s Law:

F1=k· |q1|·|q2|

where kis the Coulomb constant 8.99 ×109N·m2/C2.

Substitute the given values into Coulomb’s Law:

F1=(8.99 ×109N·m2/C2)·(3.0×10−6C)·(2.0×10−6C)

(2.0m)2

F1=8.99 ×109×3.0×2.0

F1=53.94 ×109

F1= 13.485 ×109N

Step 2: Calculate the electric force between the upper corner charge at

+3.0µC and the lower right corner charge at +6.0µC. Similarly, use Coulomb’s

Law with the appropriate charges and distance:

F2=k· |3.0×10−6|·|6.0×10−6|

(2.0m)2

F2=8.99 ×109×3.0×6.0

F2= 13.485 ×109N

Step 3: Find the net force on the upper corner charge. Since the forces

between the upper corner charge and the two lower corner charges are along the

same line but in opposite directions, the net force will be the difference of these

two forces.

Net Force =F2−F1

Net Force = 13.485 ×109N−13.485 ×109N

Net Force = 0

Question 14

Question

Three point charges are arranged along the x-axis as follows: q1=−2.0µC at

x= 0.0m, q2= 4.0µC at x= 4.0m, and q3=−6.0µC at x= 8.0m. Calculate

the magnitude and direction of the net electrostatic force on q2.

Solution

Step 1: Calculate the force between q2and q1. The force between two charges

is given by Coulomb’s Law:

F=k·|q1·q2|

where Fis the magnitude of the force, kis Coulomb’s constant (8.99 ×109N·

m2/C2), q1, q2are the point charges, ris the distance between the charges.

The distance between q1and q2is r= 4.0m. Plugging in the values, we get:

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

(4.0)2= 3.59 N

Step 2: Calculate the force between q2and q3. The distance between q2and

q3is r= 4.0m. Plugging in the values, we get:

F23 = 8.99 ×109·|4.0×10−6·6.0×10−6|

(4.0)2= 6.74 N

Step 3: Calculate the net force on q2. The net force on q2is the vector sum

of F12 and F23. Since F12 is to the left (negative x-direction) and F23 is to the

right (positive x-direction), their magnitudes will be subtracted to find the net

force.

Fnet =|F23 −F12|=|6.74 −3.59|= 3.15 N

The direction of the net force will be towards the positive x-axis.

Question 15

Question

Three point charges are arranged in an equilateral triangle as shown below. The

charges are q1= +2.0µC, q2=−3.0µC, and q3= +4.0µC. Calculate the

magnitude of the net force on q2due to the other two charges.

q1q2

Solution

Step 1: Calculate the electric force between q2and q1. The force between two

charges q1and q2separated by a distance ris given by Coulomb’s law:

F=kq1q2

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

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

In our case, q1= +2.0µC, q2=−3.0µC, and r=[distance between q1and

q2in the equilateral triangle]. Let’s denote this distance as a.

The distance between q1and q2in an equilateral triangle can be calculated

using trigonometry. Since all sides of an equilateral triangle are equal, the

distance between q1and q2is a= 2lcos(30◦), where lis the side length of the

equilateral triangle. Given an equilateral triangle, l=[side length] and thus

a=[calculated distance].

Substitute the charges and distances into Coulomb’s law to find the force

F12 between q1and q2.

Step 2: Calculate the electric force between q2and q3(F23 ) and the angle

between F12 and F23.

Step 3: Using vector addition, calculate the net force on q2due to q1and

q3. The net force is the vector sum of F12 and F23.

Step 4: Calculate the magnitude of the net force on q2.

Question 16

Question

Two point charges, q1=−3µC and q2= 5 µC, are placed 10 cm apart in

a vacuum. Calculate the magnitude of the electric force between these two

charges.

Solution

Step 1: Convert all given quantities to standard SI units.

•q1=−3µC=−3×10−6C

•q2= 5 µC= 5 ×10−6C

•r= 10 cm = 10 ×10−2m

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

F=k·|q1|·|q2|

where kis the electrostatic constant with a value of 8.99 ×109N·m2/C2.

Step 3: Plug in the given values to find the magnitude of the electric force.

F= 8.99 ×109N·m2/C2·3×10−6C·5×10−6C

(10 ×10−2m)2

Step 4: Perform the calculations to find the numerical value of the electric

force.

F= 8.99 ×109×15 ×10−12 N

F= 134.85 ×10−3N

F= 0.13485 N

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

0.13485 N.

Question 17

Question

Three point charges are arranged as shown in the diagram below. The charge

q1=−5.0µC is located at point A at coordinates (0,0), the charge q2= 8.0µC

is located at point B at coordinates (2,0), and the charge q3=−3.0µC is located

at point C at coordinates (0,3).

Calculate the magnitude and direction of the net force on the charge at point

A due to the other two charges.

Solution

Step 1: Calculate the force exerted on the charge at point A due to the charge

at point B. Using Coulomb’s law, the magnitude of the force exerted by charge

q2on charge q1is given by:

FA,B =k|q1||q2|

A,B

Where k= 8.99 ×109Nm2/C2is the Coulomb constant, rA,B is the distance

between charge q1and charge q2which is 2 meters in this case. Plugging in the

values, we get:

FA,B =(8.99 ×109Nm2/C2)(5.0×10−6C)(8.0×10−6C)

(2 m)2

FA,B =35.96

4= 8.99 N

Step 2: Calculate the force exerted on the charge at point A due to the

charge at point C. Similarly, using Coulomb’s law, the magnitude of the force

exerted by charge q3on charge q1is given by:

FA,C =k|q1||q3|

A,C

Where rA,C is the distance between charge q1and charge q3which is 3 meters

in this case. Plugging in the values, we get:

FA,C =(8.99 ×109Nm2/C2)(5.0×10−6C)(3.0×10−6C)

(3 m)2

FA,C =13.485

9= 1.498 N

Step 3: Calculate the net force on the charge at point A. Since the charges

at B and C are at right angles to each other, the net force on charge q1is given

by the vector sum of the forces due to q2and q3. Using Pythagoras theorem,

the magnitude of the net force is:

Fnet =qF2

A,B +F2

A,C

Fnet =p(8.99 N)2+ (1.498 N)2

Fnet =√80.82 + 2.244

Fnet =√83.064

Fnet ≈9.12 N

The direction of the net force can be found by determining the angle between

the net force and the x-axis:

θ= tan−1FA,C

FA,B 

θ= tan−11.498

8.99 

θ≈

Question 18

Question

Three point charges, q1= +3 nC, q2=−2nC, and q3= +4 nC, are placed at

the vertices of an equilateral triangle with sides of length d= 0.10 m. What is

the magnitude of the net electric force on q1due to the other two charges?

Solution

Step 1: Find the electric force on q1due to q2. The formula to calculate the

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

Felec =k·|q1|·|q2|

where kis the Coulomb constant (8.99 ×109N m2/C2) and ris the distance

between the charges.

Substituting the given values, the force on q1due to q2is:

F1,2= (8.99 ×109)·3×10−9·2×10−9

(0.10)2

Step 2: Calculate the electric force on q1due to q3. Following the same steps

as in Step 1, the force on q1due to q3is:

F1,3= (8.99 ×109)·3×10−9·4×10−9

(0.10)2

Step 3: Find the net force on q1. The net force on q1is the vector sum of

the forces due to q2and q3, so

|

Fnet|=q(

F1,2+

F1,3)2

After calculating the individual forces using Coulomb’s Law, you can find

the total magnitude of the net electric force on q1. Remember to consider the

directions of the forces as well.

Question 19

Question

Two point charges, q1= 4.0µC and q2=−6.0µC, are placed 2.0 meters apart.

Determine the magnitude and direction of the electric force that each charge

exerts on the other.

Solution

Given: q1= 4.0µC= 4.0×10−6C, q2=−6.0µC=−6.0×10−6C, r= 2.0m.

We’ll first calculate the magnitude of the electric force between the two

charges using Coulomb’s Law:

F=k· |q1|·|q2|

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

Step 1: Calculate the magnitude of the force

F=8.99 ×109· |4.0×10−6|·|−6.0×10−6|

(2.0)2

F=8.99 ×109·4.0×10−6·6.0×10−6

4.0

F=2.1576 ×10−2

4.0

F= 5.394 ×10−3N

Step 2: Determine the direction of the force Since q1is positive and q2

is negative, the forces between them are attractive. Thus, the force that charge

q1exerts on charge q2is directed towards q2, and the force that charge q2exerts

on charge q1is directed towards q1(in the opposite direction).

Therefore, the magnitude of the force is 5.394 ×10−3N and the direction

of the force on q1by q2is attractive and directed towards q2. Similarly, the

direction of the force on q2by q1is attractive and directed towards q1.

Question 20

Question

Two point charges, q1= +2.0nC and q2=−3.0nC, are placed 6.0cm apart

in air. Calculate the magnitude and direction of the electric force between the

charges.

Solution

Let’s solve this problem step by step:

Step 1: Find the distance between the charges in meters. Given

that the charges are 6.0cm apart, we convert this to meters:

6.0cm = 6.0×10−2m= 0.06 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· |q1·q2|

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

and ris the distance between the charges in meters.

Plugging in the values:

F=(8.99 ×109N m2/C2)· |2.0×10−9C· −3.0×10−9C|

(0.06 m)2

F=(8.99 ×109)·(6 ×10−18)

0.0036

F=53.94 ×10−9

0.0036

F= 14.95 ×10−9N= 1.5×10−8N

Thus, the magnitude of the electric force between the charges is 1.5×10−8N.

Step 3: Determine the direction of the electric force. The electric

force between the charges is attractive since the charges have opposite signs.

Therefore, the direction of the force is from q1to q2or from −3.0nC to +2.0nC.

Question 21

Question

Two point charges, q1= +3.00 µC and q2=−5.00 µC, are placed 8.00 cm

apart in a vacuum. Calculate the magnitude of the electric force between these

charges.

Solution

Step 1: Convert the given charges to Coulombs for calculation:

q1= +3.00 µC= 3.00 ×10−6C

q2=−5.00 µC=−5.00 ×10−6C

Step 2: Convert the distance between the charges to meters:

d= 8.00 cm = 8.00 ×10−2m

Step 3: Calculate the magnitude of the electric force between these charges

using Coulomb’s law:

F=k· |q1·q2|

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

Step 4: Substitute the known values into the equation:

F=(8.99 ×109N m2/C2)· |(3.00 ×10−6C)·(−5.00 ×10−6C)|

(8.00 ×10−2m)2

=(8.99 ×109)·(3.00 ×10−6)·(5.00 ×10−6)

(8.00 ×10−2)2

=(8.99 ×109)·(15 ×10−12)

(0.08)2

=134.85 ×10−3

0.0064

= 21,132.81 N

Therefore, the magnitude of the electric force between the charges is 21,132.81 N.

Question 22

Question

Two point charges, q1= 3µC and q2=−5µC, 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. Step 2: Calculate the distance in

meters. Step 3: Use Coulomb’s Law to find the magnitude of the electric force

between the charges.

Step 1: Converting the charges to coulombs: q1= 3µC = 3×10−6C

q2=−5µC = −5×10−6C

Step 2: Calculating the distance in meters: The distance between the

charges is 10 cm = 10 ×10−2m = 0.1 m.

Step 3: Using Coulomb’s Law to find the magnitude of the electric force:

Coulomb’s Law states that the magnitude of the electric force between two point

charges is given by:

F=k· |q1·q2|

r2,

where kis the electrostatic constant (8.99 ×109Nm2/C2), q1and q2are the

charges, and ris the distance between the charges.

Substitute the given values into the formula:

F=(8.99 ×109)· |3×10−6·(−5×10−6)|

(0.1)2

F=(8.99 ×109)·15 ×10−12

0.01

F=1.3485 ×10−2

0.01

F= 1.3485 N

Therefore, the magnitude of the electric force between the charges q1and q2

is 1.3485 N.

Question 23

Question

Three point charges are arranged as shown in the diagram below. Calculate the

magnitude and direction of the electric force on the charge q1due to the other

two charges. Use q= 2.0µC and d= 2.0m.

•q1= +2.0µC

•q2=−4.0µC•q3= +6.0µC

Solution

Step 1: Calculate the electric force on q1due to q2. The electric force between

two point charges is given by Coulomb’s law:

F=kq1q2

where k≈8.99 ×109Nm2/C2is the Coulomb constant, q1and q2are the

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

Given that q1= 2.0µC, q2=−4.0µC, and d= 2.0m, we can substitute

these values into Coulomb’s law to find the force F12 on q1due to q2:

F12 =k|q1||q2|

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

(2.0m)2

F12 = (8.99 ×109)×(8.0×10−12)×(4) = 287.68 nN

The force F12 on q1due to q2is 287.68 nN.

Step 2: Determine the direction of the force F12 . Since q1and q2have

opposite charges, the force F12 will be attractive, i.e., it will act along the line

joining q1and q2from q2towards q1.

Step 3: Calculating the electric force on q1due to q3. Repeat the same

process as in Step 1 to find the force F13 on q1due to q3:

F13 =k|q1||q3|

F13 = (8.99 ×109)(2.0×10−6)(6.0×10−6)

(2.0)2= 674.04 nN

The force F13 on q1due to q3is 674.04 nN.

Step 4: Determine the direction of the force F13 . Since q1and q3have like

charges, the force F13 will be repulsive, i.e., it will act along the line joining q1

and q3from q1towards q3.

Step 5: Finding the net force on q1. In order to find the net force on q1, we

need to combine the forces F12

Question 24

Question

Two point charges, q1=−2.0µC and q2= 4.0µC, are placed 10.0 cm apart on

the x-axis. Calculate the magnitude and direction of the electric force that q1

exerts on q2.

Solution

Step 1: Calculate the distance between the charges in meters. Given that

1cm = 0.01 m, the distance between the charges is

r= 10.0cm ×0.01 m/cm = 0.10 m.

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

F=k



q1q2

r2



,

where k= 8.99×109N m2/C2is the Coulomb constant. Plugging in the values:

F= 8.99 ×109N m2

C2



−2.0×10−6C×4.0×10−6C

(0.10 m)2



F= 8.99 ×109N m2

C2



−8.0×10−12 C2

0.01 m2



F= 8.99 ×109N m2

C2×8.0×10−10 C= 7.2×10−1N.

Step 3: Determine the direction of the force. The force is attractive because

the charges have opposite signs. Thus, the force is directed from q1towards q2

along the x-axis.

Therefore, the magnitude of the electric force that q1exerts on q2is 7.2×

10−1N, directed from q1towards q2along the x-axis.

Question 25

Question

Two identical point charges, each with a charge of +2.0µC, are placed at a

distance of 5.0m apart. Determine the magnitude and direction of the electric

force between the charges.

Solution

Step 1: Convert the charges to standard units by using the conversion factor

1µC= 10−6C. Given that each charge is +2.0µC, the charges can be converted

to +2.0×10−6C.

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

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= distance between

the charges. Plugging in the values:

F=(8.99 ×109N m2/C2)· |(+2.0×10−6C)2|

(5.0m)2

Step 3: Calculate the magnitude of the electric force:

F=8.99 ×109·(2.0×10−6)2

5.02

Step 4: Simplify the expression further:

F=8.99 ×109·4.0×10−12

25.0

Step 5: Calculate the final answer:

F=35.96 ×10−3

25.0= 1.44 ×10−3N

Step 6: Determine the direction of the force. Since the charges are the same,

they will repel each other. Therefore, the direction of the force will be along

the line connecting the charges, pointing away from each charge.

Therefore, the magnitude of the electric force between the charges is 1.44 ×

10−3N, and the direction of the force is along the line connecting the charges,

pointing away from each charge.

A B

F= 8.99 ×109×15.36 ×10−12

0.0036

F= 8.99 ×15.36

0.0036 ×10−3

F= 8.99 ×4266.67

1×10−3

F= 3.84 ×104N

Therefore, the magnitude of the electric force between the charges is 3.84 ×

104N.

Question 2

Question

Two point charges q1= 3.0µC and q2=−4.0µC are placed 10.0 cm apart in

air. Calculate the magnitude and direction of the electric force that q1exerts

on q2.

Solution

Step 1: Convert the charges to coulombs. Given: q1= 3.0µC = 3.0×10−6C

q2=−4.0µC =−4.0×10−6C

Step 2: Find the distance between the charges in meters. Given: r= 10.0

cm = 10.0×10−2m= 0.10 m

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

Coulomb’s law states:

F=k|q1q2|

where kis Coulomb’s constant, k= 8.9875 ×109N·m2/C2.

Plugging in the values:

F= 8.9875 ×109×|3.0×10−6× −4.0×10−6|

(0.10)2

F= 8.9875 ×109×12 ×10−12

0.01

F= 8.9875 ×109×1.2×10−9

F= 10.785 ×10−1

F= 10.785 N

Step 4: Determine the direction of the force. The force is attractive since

one charge is positive and the other is negative.

Therefore, the magnitude of the electric force that q1exerts on q2is 10.785 N

directed towards q2.

Question 3

Question

Three point charges are arranged in an equilateral triangle. The charges are

+q,−2q, and +3qwith sides of length d. Calculate the magnitude of the net

electrostatic force on the +3qcharge due to the other two charges. Assume the

charges are free to move.

Solution

1. First, let’s calculate the electrostatic force between the +qand +3qcharges.

Since the charges are the same sign, this force will be repulsive. The magnitude

of the electrostatic force F1between a pair of charges qand Qis given by

Coulomb’s Law:

F1=k· |q|·|Q|

where kis the Coulomb constant (8.99 ×109N m2/C2), |q|is the magnitude of

the +qcharge, |Q|is the magnitude of the +3qcharge, and ris the distance

between the charges. Since this distance is din an equilateral triangle, r=d.

2. Substituting the given values into the formula, we get:

F1=(8.99 ×109)(q)(3q)

d2=26.97 ×109q2

So, the magnitude of the force F1between the +qand +3qcharges is 26.97×109q2

3. Next, let’s calculate the electrostatic force between the −2qand +3q

charges. Since the charges are opposite signs, this force will be attractive. Let’s

denote this force as F2.

4. Using Coulomb’s Law again, we have:

F2=k· |(−2q)|·|3q|

d2=6kq2

5. Now, to find the net force on the +3qcharge, we need to consider the

directions. Since F1and F2are along the same line, we simply subtract these

two forces:

Net force =F1−F2=26.97 ×109q2

d2−6kq2

=26.97 ×109−6×8.99 ×109q2

6. Therefore, the magnitude of the net electrostatic force on the +3qcharge

due to the other two charges is (26.97 −53.94) ×109q2

d2=−26.97 ×109q2

d2.

Question 4

Question

Three point charges are arranged along the x-axis: a charge of +3µC at the

origin, a charge of −5µC at x= 0.1m, and a charge of +2µC at x= 0.2m.

Calculate the net electric force on the charge of −5µC.

Solution

Step 1: Calculate the electric force between the charge of +3µC at the origin

and the charge of −5µC at x= 0.1m. The electric force F1between two charges

q1and q2separated by a distance ris given by Coulomb’s law:

F1=k·|q1q2|

where k= 8.99 ×109Nm²/C² is Coulomb’s constant.

Substitute the given values:

F1= 8.99 ×109·3×10−6×5×10−6

(0.1)2

Calculating F1gives:

F1= 1.35 ×10−3N

Step 2: Calculate the electric force between the charge of +2µC at x= 0.2

m and the charge of −5µC at x= 0.1m. Using Coulomb’s law again:

F2=k·|q1q2|

Substitute the given values:

F2= 8.99 ×109·2×10−6×5×10−6

(0.2−0.1)2

Calculating F2gives:

F2= 4.49 ×10−3N

Step 3: Find the net force on the charge of −5µC. Since the forces are in

opposite directions, we need to consider them as negative and positive values.

The net force is the sum of the forces:

Fnet =F1−F2= 1.35 ×10−3N−4.49 ×10−3N

Calculating Fnet gives:

Fnet =−3.14 ×10−3N

Therefore, the net electric force on the charge of −5µC is −3.14 ×10−3N,

pointing towards the charge of +2µC at x= 0.2m.

Question 5

Question

Two point charges, q1=−6.0µC and q2= 4.0µC, are placed 10.0 cm apart in

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

Solution

Given: q1=−6.0µC

q2= 4.0µC

r= 10.0cm

The electric force between two point charges q1and q2separated by a dis-

tance ris given by Coulomb’s law:

F=k·|q1·q2|

where kis the electrostatic constant equal to 8.99 ×109N·m2/C2.

Step 1: Convert the distance rfrom cm to m:

r= 10.0cm = 0.1m

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

F= (8.99 ×109)·| − 6.0×10−6·4.0×10−6|

(0.1)2

F= 8.99 ×109·24 ×10−12

0.01

F= 8.99 ×109·2.4×10−10

F= 2.1576 ×10−1N

The magnitude of the electric force between the charges q1and q2is 0.216 N.

Question 6

Question

Three point charges are located on the vertices of an equilateral triangle of side

length a. The charges are +q,−2q, and +q. Calculate the magnitude and

direction of the electric force on the positive charge +qdue to the other two

charges.

Solution

Given: - Charge at vertex A = +q, - Charge at vertex B = -2q, - Charge

at vertex C = +q, - Distance between charges = a(side length of equilateral

triangle), - Coulomb’s constant = k.

The direction of the electric force on the positive charge +qdue to the other

two charges will depend on the configuration of the charges. Since the charges

are symmetrical with respect to the positively charged particle, the net force on

the positive charge will be directed vertically downwards.

Step 1: Calculate the force on +qdue to −2qat vertex B: The electric force

FAB on the charge at A due to the charge at B is given by Coulomb’s law:

FAB =k· |q|·|−2q|

a2=2kq2

Step 2: Calculate the force on +qdue to +q at vertex C: The electric force

FAC on the charge at A due to the charge at C is given by Coulomb’s law:

FAC =k· |q|·|q|

a2=kq2

Step 3: Calculate the net force and its direction: The net force on the

charge at A will be the sum of the forces FAB and FAC in the vertical downward

direction:

Fnet =FAB +FAC =2kq2

a2+kq2

a2=3kq2

Therefore, the magnitude of the net force on the charge +qis 3kq2

a2and it is

directed vertically downwards.

Question 7

Question

Two point charges, q1= 3 µC and q2=−2µC, are placed 10 cm apart in air.

Calculate the magnitude of the electric force between these charges.

Solution

Step 1: Determine the distance between the charges in meters. Given that the

charges are 10 cm apart, we need to convert this distance to meters:

10 cm = 0.1m

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

using Coulomb’s Law: Coulomb’s Law states that the magnitude of the electric

force between two point charges is given by:

F=k· |q1·q2|

where: - Fis the magnitude of the electric force, - kis the electrostatic constant

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

the distance between the charges.

Plugging in the given values:

F=(8.99 ×109N m2/C2)· |3×10−6C· −2×10−6C|

(0.1m)2

Step 3: Calculate the electric force.

F=8.99 ×109·6×10−6·2×10−6

0.01

F=8.99 ×6×2

10 ×10−3

F=107.88

10 ×10−3

F= 10.788 ×10−3

F= 10.788 mN

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

10.788 mN.

Question 9

Question

Two point charges q1= +2.5µC and q2=−4.0µC are placed 10 cm apart.

Calculate the magnitude of the electric force between the charges.

Solution

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

Coulomb’s Law. Coulomb’s Law states that the magnitude of the electric force

between two point charges is given by:

F=k|q1q2|

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

(8.9875 ×109N·m2/C2), - q1and q2are the magnitudes of the charges, and -

ris the separation distance between the charges.

Step 1: Convert the given charges to coulombs:

q1= +2.5µC = 2.5×10−6C

q2=−4.0µC =−4.0×10−6C

Step 2: Determine the separation distance in meters:

r= 10 cm = 0.1m

Step 3: Calculate the magnitude of the electric force:

F= (8.9875 ×109)|2.5×10−6× −4.0×10−6|

(0.1)2

F= 8.9875 ×109×1.0×10−11 ×100

F= 8.9875 ×1013N

Therefore, the magnitude of the electric force between the charges is 8.9875×

1013 N.

Question 10

Question

Two point charges, q1=−3.0µC and q2= +6.0µC, are placed 0.20 m apart.

Calculate the magnitude and direction of the electric force that q2exerts on q1.

Solution

Step 1: Convert the given charges to standard units.

•q1=−3.0µC =−3.0×10−6C

•q2= +6.0µC = 6.0×10−6C

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

F=k·|q1·q2|

where kis the Coulomb constant (8.99 ×109N·m2/C2) and ris the distance

between the charges.

Step 3: Substitute the given values into the equation:

F= (8.99 ×109)·| − 3.0×10−6·6.0×10−6|

0.202

Step 4: Simplify the expression to find the magnitude of the electric force.

F= (8.99 ×109)·18 ×10−12

0.04 = 4.497 ×10−2N

Step 5: Determine the direction of the electric force. Since q2is positive, it

exerts a repulsive force on q1, which is negative. Thus, the direction of the force

is away from q2.

Therefore, the magnitude of the electric force that q2exerts on q1is 4.497 ×

10−2N, directed away from q2.

Question 11

Question

Three point charges are arranged in a line. Charge q1=−3.0µC is located

at x=−1.0m, charge q2= 6.0µC is located at the origin, and charge q3=

−4.0µC is located at x= 2.0m. Calculate the magnitude and direction of the

net electrostatic force on q1due to q2and q3.

Solution

Step 1: Calculate the force on q1due to q2. The electrostatic force between two

point charges is given by Coulomb’s Law:

F=k·|q1·q2|

where k= 8.99 ×109Nm2/C2is Coulomb’s constant, q1=−3.0µC =−3.0×

10−6C,q2= 6.0µC = 6.0×10−6C, and r= 1.0mis the distance between q1

and q2.

Plugging in these values, we get

Fq1,q2= 8.99 ×109·| − 3.0×10−6·6.0×10−6|

(1.0)2

Fq1,q2= 8.99 ×109·18 ×10−12

1.0

Fq1,q2= 161.82 ×10−3N

Step 2: Determine the direction of the force on q1due to q2. Since q1and

q2are oppositely charged, the force will be attractive. Therefore, the force will

act towards q2.

Step 3: Calculate the force on q1due to q3. Similarly, using Coulomb’s Law,

we find the force on q1due to q3:

Fq1,q3= 8.99 ×109·| − 3.0×10−6· −4.0×10−6|

(3.0)2

Fq1,q3= 8.99 ×109·12 ×10−12

9.0

Fq1,q3= 119.76 ×10−3N

Step 4: Determine the direction of the force on q1due to q3. As q1and q3

are both negative charges, the force between them will also be attractive, acting

towards q3.

Step 5: Calculate the net force on q1by summing the individual forces from

q2and q3. The net force is given by

Fnet =Fq1,q2+Fq1,q3

Fnet = 161.82 ×10−3+ 119.76 ×10−3

Fnet = 281.58 ×10−3N

The net force on q1is 281.58 ×10−3Ntowards the right.

Question 12

Question

Three point charges are placed at the following positions: q1= +3 nC at (0,0),

q2=−1nC at (4,0), and q3= +2 nC at (0,3). Calculate the net electric force

on q2due to q1and q3.

Solution

Step 1: Calculate the electric force on q2due to q1: The electric force between

two point charges q1and q2separated by a distance ris given by Coulomb’s

Law:

F1,2=k|q1||q2|

In this case, q1= +3 nC, q2=−1nC, and r= 4 (since q1and q2are 4 units

apart horizontally). Using k= 8.99 ×109N m2/C2, we have:

F1,2= (8.99 ×109)(3 ×10−9)(1 ×10−9)

(4)2

F1,2= (8.99 ×109)3

16 ×10−18

F1,2=2697

16 ×10−9

F1,2= 168.56 ×10−9

F1,2= 1.69 ×10−6N

Step 2: Calculate the electric force on q2due to q3: Similarly, using q2=

−1nC, q3= +2 nC, and r= 3 (since q2and q3are 3 units apart vertically), we

have:

F2,3=k|q2||q3|

F2,3= (8.99 ×109)(1 ×10−9)(2 ×10−9)

(3)2

F2,3=2×8.99

9×10−9

F2,3= 1.99 ×10−6N

Step 3: Calculate the net electric force on q2: The net force is the vector

sum of the two forces:

Net force on q2=F1,2+F2,3

Net force on q2= 1.69 ×10−6+ 1.99 ×10−6

Net force on q2= 3.68 ×10−6N

Therefore, the net electric force on q2due to q1and q3is 3.68 ×10−6N.

Question 13

Question

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

length d= 2.0m as shown below. The charges are +3.0µC at the upper corner,

−2.0µC at the lower left corner, and +6.0µC at the lower right corner. Calcu-

late the magnitude and direction of the net electric force on the charge at the

upper corner.

` `

U

3.0µC

−2.0µC6.0µC

Solution

Step 1: Calculate the electric force between the upper corner charge at +3.0µC

and the lower left corner charge at −2.0µC. The electric force F1between two

point charges q1and q2separated by a distance ris given by Coulomb’s Law:

F1=k· |q1|·|q2|

where kis the Coulomb constant 8.99 ×109N·m2/C2.

Substitute the given values into Coulomb’s Law:

F1=(8.99 ×109N·m2/C2)·(3.0×10−6C)·(2.0×10−6C)

(2.0m)2

F1=8.99 ×109×3.0×2.0

F1=53.94 ×109

F1= 13.485 ×109N

Step 2: Calculate the electric force between the upper corner charge at

+3.0µC and the lower right corner charge at +6.0µC. Similarly, use Coulomb’s

Law with the appropriate charges and distance:

F2=k· |3.0×10−6|·|6.0×10−6|

(2.0m)2

F2=8.99 ×109×3.0×6.0

F2= 13.485 ×109N

Step 3: Find the net force on the upper corner charge. Since the forces

between the upper corner charge and the two lower corner charges are along the

same line but in opposite directions, the net force will be the difference of these

two forces.

Net Force =F2−F1

Net Force = 13.485 ×109N−13.485 ×109N

Net Force = 0

Question 14

Question

Three point charges are arranged along the x-axis as follows: q1=−2.0µC at

x= 0.0m, q2= 4.0µC at x= 4.0m, and q3=−6.0µC at x= 8.0m. Calculate

the magnitude and direction of the net electrostatic force on q2.

Solution

Step 1: Calculate the force between q2and q1. The force between two charges

is given by Coulomb’s Law:

F=k·|q1·q2|

where Fis the magnitude of the force, kis Coulomb’s constant (8.99 ×109N·

m2/C2), q1, q2are the point charges, ris the distance between the charges.

The distance between q1and q2is r= 4.0m. Plugging in the values, we get:

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

(4.0)2= 3.59 N

Step 2: Calculate the force between q2and q3. The distance between q2and

q3is r= 4.0m. Plugging in the values, we get:

F23 = 8.99 ×109·|4.0×10−6·6.0×10−6|

(4.0)2= 6.74 N

Step 3: Calculate the net force on q2. The net force on q2is the vector sum

of F12 and F23. Since F12 is to the left (negative x-direction) and F23 is to the

right (positive x-direction), their magnitudes will be subtracted to find the net

force.

Fnet =|F23 −F12|=|6.74 −3.59|= 3.15 N

The direction of the net force will be towards the positive x-axis.