# Kinetic energy of a charge

1. Jul 14, 2007

### Aerosion

1. The problem statement, all variables and given/known data

I don't want to sound like I'm asking anyone to do my homework for me, but I really don't know how to find this.

Let me give an example problem: A uniform electric field of 2kN/C is in the x direction. A point charge Q=3 microcoloumbs initially at reast at the origin is released. What is the kinetic energy when it is at x=4m and what is the change in potential energy of the charge from x=0 to x=4m?

2. Relevant equations

3. The attempt at a solution

I know that the potential energy difference is the negative of the kinetic energy of the charge, and I know how to get the potential energy (integrate the electric field with regard to the increase in x and multiply by the negative of the charge) but I'd rather know exactly how to find the kinetic energy of a charge.

2. Jul 14, 2007

### Staff: Mentor

Sounds to me like you just described how to find the KE of a charge as it is accelerated in an electric field. So I'm unclear what the problem is. Why don't you show your work for that sample problem and tell us where you get stuck?

3. Jul 14, 2007

### Aerosion

Well, that's the thing: I don't know how to find the kinetic energy of a charge; as in I'm unclear as to what the formula is. That's why I said that I don't want anyone to think that they're doing homework for me: because I don't know how to get started.

Like, I know that KE=1/2mv^2, but I don't think that's any use in this particular problem. If the formula for KE of a charge is similiar to the potential energy difference of a charge, and if I've already found it out (unknowingly), then could someone tell me?

4. Jul 14, 2007

### Aerosion

Oh oh...and the work I did for PE is...

I used the change in potential energy equation -q=(integrate)E*ds, and subsituated 3microcoloumbs for q, 2kn/C for E, and 0 and 4 for the definite integral, such that it looked like

-3mC*(integrate fr. 0 to 4)2 ds.

5. Jul 14, 2007

### Staff: Mentor

Realize that the field is constant here, so no real integration is needed. The work done on the charge by the field is just F*d = Eqd. The work done will equal the change in KE of the charge. (Note that the work done by the field is just the negative of the change in PE.)