# Point Charge Location Question

Sorry for the poor title - I'm not entirely sure how to describe the type of problem!

## Homework Statement

A charge of -4.00 x 10-6 C is fixed in place. From a horizontal distance of 55.0cm a particle of mass 2.50 x 10-3 kg and charge -3.00 x 10-6 C is fired with an initial speed of 15.0m/s directly towards the fixed charge. How far does the particle travel before it stops and begins to return back?

V = k(Q/r)

PE = (QV)/2

KE = (mv2)/2

## The Attempt at a Solution

I've tried this a few times and haven't found anything that works so here is my latest attempt:

A__________________________B_______________________C
A = fixed charge
B = unknown distance
C = Starting point for moving charge

Qa = -4.00 x 10-6 C
Qc = -3.00 x 10-6 C
v0c = 15m/s
v1c = 0m/s
mc = 2.50 x 10-3 kg
rc = 0.55m
rb = ?

rb = k(Qa/Vc)
Vc = 2PE/Qc
PE = -KE
KE = (mv2)/2

KE = (2.50 x 10-3 kg x -15m/s)/2 = -0.009375
PE = 0.009375

Vc = 2(0.009375)/-3.00 x 10-6 C = -6.25 x 103

rb = 9 x 109 (-4.00 x 10-6 C/-6.25 x 103) = 5.76 m

Last edited:

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rl.bhat
Homework Helper
KE = 1/2*m*V^2
PE = k*Qa*Qc/rb
In this problem ra is not needed.

Doc Al
Mentor

## Homework Equations

V = k(Q/r)

PE = (QV)/2
Don't confuse the PE between two point charges with the energy stored in a capacitor. Look up the correct formula for PE.

KE = (mv2)/2
Good.

Hint: Compare the PE at point C with the PE at point B.

Thanks - I knew I was doing something wrong and just couldn't seem to sort out the problem correctly!