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

A charge of -4.00 x 10

V = k(Q/r)

PE = (QV)/2

KE = (mv

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

Q

Q

v

v

m

r

r

r

V

PE = -KE

KE = (mv

KE = (2.50 x 10

PE = 0.009375

V

r

## 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?## Homework Equations

V = k(Q/r)

PE = (QV)/2

KE = (mv

^{2})/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

Q

_{a}= -4.00 x 10^{-6}CQ

_{c}= -3.00 x 10^{-6}Cv

_{0c}= 15m/sv

_{1c}= 0m/sm

_{c}= 2.50 x 10^{-3}kgr

_{c}= 0.55mr

_{b}= ?r

_{b}= k(Q_{a}/V_{c})V

_{c}= 2PE/Q_{c}PE = -KE

KE = (mv

^{2})/2KE = (2.50 x 10

^{-3}kg x -15m/s)/2 = -0.009375PE = 0.009375

V

_{c}= 2(0.009375)/-3.00 x 10^{-6}C = -6.25 x 10^{3}r

_{b}= 9 x 10^{9}(-4.00 x 10^{-6}C/-6.25 x 10^{3}) = 5.76 m
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