# Electric potential energy to kinetic energy

1. Aug 24, 2009

### cerberus9

1. The problem statement, all variables and given/known data
There were two parts to this problem

a) if you do 12 J of work to push 0.001 C of charge from point A to point B in an electric field, what is the voltage difference between A and B?

I already solved this part, I got $$\Delta$$V=12,000 V

b) When the charge is released, what will be its kinetic energy as it flies back past its starting point A? What principle guides your answer?

Here i have absolutely no idea what to do.

2. Relevant equations

$$\Delta$$V=W/q

$$\Delta$$V=$$\Delta$$PEelectric/q

3. The attempt at a solution

I've gotten through part A but i have no idea how to approach part B.

2. Aug 24, 2009

### Villhelm

I think Electric fields are conservative, so if you do work against it, then hold your charge and then release it, then it will gain the same energy going back out ... so would it be 12J at point A?

:/ I try to help

3. Aug 24, 2009

### cerberus9

how did you get 12 J?

4. Aug 24, 2009

### Villhelm

You said 12J above as the work done in moving the charge, which I think is conserved when it comes back out. That's how I understand it, maybe I'm missing something ... it happens

5. Aug 24, 2009

### cerberus9

haha no, don't be upset, you're right. that's the answer in the back of the book. i'm just totally lost as to how to get to it.

6. Aug 24, 2009

### ideasrule

Use the work-energy theorem and the conservation of energy.

7. Aug 24, 2009

### Villhelm

I think it just comes about because electric fields are "conservative" fields ... like in ballistics questions, the ball goes up with V and comes down with -V ... because gravity is conservative too. So, the KE into the ball is spent moving upwards, but it gains the same coming back down.

8. Aug 24, 2009

### cerberus9

thanks a bunch