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## 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?

## Homework Equations

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!