Question

A +50?C point charge is placed 32cmfrom an identical +50?Ccharge.

How much work would be required for an external force to move a +0.50?C test charge from a point midway between them to a point 13cm closer to either of the charges?

Express your answer using two significant figures.

Answer 1

Answer 2

a similar question is solved below, but with different values. hope this helps you.

A 50?C point charge is placed 35 from an identical50 ?C charge.
How much work would be required to move a0.30 ?C test charge from a point midway between themto a point 13 cm closer to either of the charges?

answer

First off I am assuming the 50 ?C charges are placed 35cm away from each other.

Let's define variables first in the standard units

Q = 50 *10^-6 C
q = 0.30 * 10^-6 C

r = 0.35 m

r1 = 0.35 + 0.13 = 0.48 m

r2 = 0.35 - 0.13 = 0.22 m

The key equation in this problem is W = q( V(b) - V(a) ) If weare going from point a to point b, W is the work done moving acharge q.

V(a) = kQ/r + kQ/r

V(a) = 2*kQ/r

V(b) = kQ/r1 + kQ/r2

W = q ( V(b) - V(a) )

W = q ( kQ/r1 + kQ/r2 - 2*kQ/r )

Simplified

W = kQq( 1/r1 + 1/r2 -2/r )

We convert units and evaluate to find

W = -2.23795 Joules   

Answer 3

Answer 4

work done equal to change in energy of the particle after its displacement

when placed at 13 cm from either of those point charges

energy of the system = K * q1*q2/r + q1*q2/(32-r)

= K*50*0.5(1/0.13 + 1/0.19)*10^(-12)

K = 9*10^9

initial energy = 2.914J

final energy = K * q1*q2/r + q1*q2/(32-r)

=K*50*0.5(1/0.16 + 1/0.16)*10^(-12)

final energy =2.8125J

work done = final - initial = 0.101 J