# Electric Potential Energy Problem

## Homework Statement

A charge of –4.00 µC is fixed in place. From a horizontal distance of
55.0 cm, a particle of mass 2.50 × 10–3 kg and charge –3.00 µC is fired with an
initial speed of 15.0 m/s directly toward the fixed charge. How far does the
particle travel before it stops and begins to return back?

KE = 1/2 mv^2
Pba = -W = -qEd
F = kQ1Q2/r^2

## The Attempt at a Solution

1) Found the Kinetic energy of the moving particle :
KE = 1/2mv^2
= 1/2 (2.5x10^-3)(15)^2
= 0.281 J

2) Set the value I found for KE to PE and used the Potential Energy eqn:
PE = -qEd

Since E = kQ/d^2

PE = -qd(kQ/d^2)

Therefore: d = -qkQ/PE

d = 0.38 m

I'm not sure if I did that right. But the answer I came up with looks like it could work. Any help would be appreciated!

Related Introductory Physics Homework Help News on Phys.org
You should not be finding $$\Delta$$PE that way.
$$\Delta$$PE = -qEd only if E is constant. But E is not constant here.

Write down V at the initial point and the final point.
How is PE related to V?
Then subtract the two to find $$\Delta$$PE.

:)