# Electric Fields , Finding final velocity. NEED HELP

1. Jun 5, 2012

### skye204

1. The problem statement, all variables and given/known data
Mass= 15g
Q1= 8.0uC
Q2= 25 uC (stationary)

Q1 is released from a point 12cm from Q2. How fast will Q1 be moving when the separation is 20cm.

2. Relevant equations

F=ma
F=qe
F=kQq/r2
E=kQ/r2
Ke=mv2/2
W=FD

3. The attempt at a solution
F=(9x109)(25x10^-6)(8x10^-6)/(.12^2)
F=125N

F=MA
125/0.015= A
A= 8333m/s2

vf2=0+2(8333)(.20-.12)
vf=36.5m/s

Thanks for help!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 5, 2012

### Infinitum

Hi skye204!! Welcome to PF

The formula that you've used i.e "vf2=vi2+2ad" is applicable only when the acceleration is constant. In this case acceleration changes, so this is not applicable. I would suggest you apply the energy conservation principle instead

3. Jun 5, 2012

### pcm

vf2=0+2(8333)(.20-.12)
vf=36.5m/s

you can not use the above formula as the acceleration is not constant.
you should try using energy method.

4. Jun 5, 2012

### skye204

I get it

W=FD
W=125N x.12
W= 15

F=(9x109)(25x10^-6)(8x10^-6)/(.2^2)
F=45N

W=45 x .2
W= 9

pe=pe' + ke'
15= 9 + 0.015(v^2)/2
v= 28.28

Thanks!