Question

In this problem, consider electrostatic forces only. In the figure, there is a particle of charge +Q at x = 0, and there is a particle of charge +16Q at x = +4a. You are going to bring in a third particle, with a charge of ?4Q, and place it at an appropriate spot on the x-axis. Use Q = 60.0 ? 10^?6C, and a = 40.0 cm. Also, use k = 9.00 ? 10⁹ N · m²/C².

(a)First, place the particle of charge ?4Q at the correct location so that the +Q charge experiences no net force because of the other two charged particles. In that situation, calculate the magnitude of the force that just one of the other charged particles exerts on the +Q charge.

(b)Instead, at what location on the x-axis would you place the particle of charge ?4Q so that the +16Q charge experiences no net force because of the other two charged particles? Note that we're looking for an answer in units of centimeters.

(c)Finally, place the particle of charge ?4Q at the correct location so that the ?4Q charge experiences no net force because of the other two charged particles. In that situation, calculate the magnitude of the force that just one of the other charged particles exerts on the ?4Q charge.

Answer 1

Given

Q = 60.0 x 10^-6 C

a = 40.0 x 10^-2 m

k = 9.00 x 10⁹ Nm²/C²

Solution

A - No force on Q

The force acting on +Q by +16Q, in the absence of -4Q

F₁ = kQ(16Q)/(4a)²

F₁ = 9 x 10⁹ x 60.0 x 10^-6 x 16 x 60.0 x 10^-6 / ( 4 x 40.0 x 10^-2)²

F₁ = 202.5 N

For Q to feel no force -4Q must provide a force equal to this in magnitude and opposite in direction

Let’s say the charge -4Q is placed at a distance d from Q.

F₂ = kQ(4Q)/(d)²

kQ(16Q)/(4a)²= kQ(4Q)/(d)²

4/16a²=1/d²

d = 2a = 80.0 cm

Magnitude of the exerted by one of the other charges on Q is 202.5 N

B - No force on +16Q

The force acting on +16Q , in the absence of -4Q

F₁ = k(16Q)Q/(4a)²

Let’s say the charge -4Q is placed at a distance d from +16Q.

F₂ = k(16Q)(4Q)/(d)²

k(16Q)Q/(4a)² = k(16Q)(4Q)/(d)²

1/(4a)² = 4/(d)²

1/16a² = 4/d²

d = 8a = 8 x 40 = 320 cm

d is the distance of -4Q from +16Q

The position of d on the x axis is

x = 4a - d

x = 160 - 360

x = -160 cm

C - No force on -4Q

Let’s say the charge -4Q is placed at a distance d from Q.

F₁= kQ(4Q)/(d)²

F₂ = k(16Q)(4Q)/(4a-d)²

k(16Q)(4Q)/(4a-d)² = kQ(4Q)/(d)²

16/(4a-d)² = 1/d²

Taking root on both sides

4/4a-d = 1/d

4d = 4a-d

5d=4a

d = 4a/5

d = 32 cm = 0.32 m

F₁= kQ(4Q)/(d)²

F₁ = 9 x 10⁹ x 60.0 x 10^-6 x 4 x 60.0 x 10^-6 / ( 0.32)²

F₁ = 1265.625 N

F₁ = 1.26 x 10³ N

Magnitude of force exerted on -4Q by one of the other charges is 1.26 kN