Electric Potential and velocity

[Ryan060]

Homework Statement

A small ball has a mass of mass m 4.0 x 10^-8 kg and a charge of q = -2.0 x 10^-5 C. It enters a box with an initial speed of v0 = 1.0 x 10^4 m/s. The point where the charge leaves the box is at a voltage 350 kV higher than the entry point.

Homework Equations

1/2mv^2(f)-1/2mv^2=qV(f)-qV(i)

The Attempt at a Solution

1/2(4.0 x 10^-8)v^2-1/2(4.0 x 10^-8)(1.0 x 10^4)=q(V(f)-V(i))
1/2(4.0 x 10^-8)v^2-1/2(4.0 x 10^-8)(1.0 x 10^4)^2=-2.0x10^-5(3.5x10^5)
1/2(4.0 x 10^-8)v^2-(2)=-7 <-- there is were it all falls apart as you can see i will end up with a (-) under the sqrt and that is just a fail. I understood everything in physics right up until EP and now the emails i get from my professor are just useless. thanks for your help in advance. i am assuming that the 350kv is the V(f)-V(i) is this wrong?[/B]

 

[vela]

You have a sign error in your original equation. The electric potential energy U is equal to qV, so conservation of energy gives you
$$\frac{1}{2} mv_i^2 + qV_i = \frac 12 mv_f^2 + q V_f.$$

 

[Ryan060]

You have a sign error in your original equation. The electric potential energy U is equal to qV, so conservation of energy gives you
$$\frac{1}{2} mv_i^2 + qV_i = \frac 12 mv_f^2 + q V_f.$$

I rearranged the equation s subtracting qVi and KE(f) in that form i end up with 2 unknowns rewritten out :

2.0+-2.0x10^-5(Vi)=1/2(4.0 x 10^-8)v^2+-2.0x10^-5(Vi+3.5x10^5)

if i had Vi this would be easy should i be solving for that somehow ...i really don't get this. The lecture was a pretty much pointless activity and the book is 2 pages on this section so i feel pretty in the dark at the moment

 

[vela]

You don't need $V_i$. It cancels out.