# Algebra Clarification Requested - The Fate of an Electron in a Uniform Electric Field

1. Jan 20, 2005

### evilempire

The electron, having been held at height h0, is now released from rest. Calculate its speed v when it reaches the top plate.

I found the initial energy to be:

-q_e*E*h_0

and the final energy to be:

1/2*m*v^2+-q_e*E*h_1

And I need to solve for v, so I set the two equal to each other as per the conservation of energy thereom and solve for v, correct?

-q_e*E*h_0 = 1/2*m*v^2+-q_e*E*h_1

My problem is isolating v and cancelling/clearing up variables. My algebra is lagging a bit, admittedly. Thanks for any help.

2. Jan 20, 2005

### dextercioby

Pay attention with the signs...The skeleton of your equation is
$$A=B+v^{2}C$$

Isolate v^{2} and then take root of second order...

Daniel.

3. Jan 20, 2005

### evilempire

Thanks. I only get one shot at an answer (it's an online-based assignment), so I would appreciate any response saying if this is correct or not, and what I did wrong if it is incorrect:

$$sqrt((2(h_0-h_1))/(m))=v$$

4. Jan 20, 2005

### evilempire

Can anyone else help me out please?

EDIT: Nevermind, I got it.

Last edited: Jan 20, 2005
5. Jan 21, 2005

### dextercioby

Excellent,that was the idea,to get it,without too much help... Anyways,i was asleep... :tongue2:

Daniel.

P.S.I didn't dream of your exercise... :tongue2: