Electric Fields , Finding final velocity. NEED HELP

[skye204]

Homework Statement

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.

Homework Equations

F=ma
F=qe
F=kQq/r²
E=kQ/r²
Ke=mv²/2
W=FD
vf²=vi²+2ad

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=vi²+2ad
vf²=0+2(8333)(.20-.12)
vf=36.5m/s

ACTUAL ANSWER IS : 28.3 according to the answer key

Thanks for help!

 

[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

 

[pcm]

vf2=vi2+2ad
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.

 

[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!