# Electric Potential related to velocity

1. Dec 16, 2011

### jeandempsey

1. The problem statement, all variables and given/known data
A +3μC charge with a mass of 2 x 10 ^-4 kg is moving with a speed of 5 m/s at a position where the electric potential is 1000 V. How fast will it be moving when it gets to a position where the electric potential is 200 V? Assume energy is conserved.

2. Relevant equations
V=electric potential energy / q
D=Vot+1/2at2
KE=1/2mv^2

3. The attempt at a solution
I got the change in electric potential energy to be .0024. And then I set that equal to the kinetic energy equation. I got my answer to be 24m/s, but the answer is 7 m/s. I think I might be doing this totally wrong.

2. Dec 16, 2011

### Staff: Mentor

In what way did you "set that equal to the kinetic energy equation"? Did you account for the kinetic energy due to the initial velocity? Perhaps you should show the calculation that you performed.

3. Dec 16, 2011

### jeandempsey

I did this: my electric potential energy was -2.4x10^-3
So- -2.4x10^-3 = 1/2(2x10^-4)v^2

4. Dec 17, 2011

### Staff: Mentor

There was an initial velocity, so an initial kinetic energy. The change in potential results in a change in the kinetic energy, not the total.

5. Dec 17, 2011

### jeandempsey

Thank you, but I'm still kind of confused as to how you would get the velocity from that

6. Dec 17, 2011

### Staff: Mentor

What is the relationship between kinetic energy and speed? What, then, is the initial kinetic energy of the charged particle to begin with? Then, when it has moved across the 1000V to
200V potential difference its kinetic energy will have CHANGED from that initial value.

If you want to look at the situation as a single formula, then using the law of conservation of energy you can write that the sum of the kinetic energy and potential energy for the system is constant. So that

$KE_1 + PE_1 = KE_2 + PE_2$