# Help on some electric potential energy questions

1. Jan 9, 2005

### thursdaytbs

Hey - just asking for some help on some electric potential energy questions.
First I used the equation 1/2mv^2 = q(Va-Vb)
so v = sqrt (((2)(25000)(q)) / m )

and Va-Vb = W / q, so q = W/(Va-Vb)
but then i'm stuck and dont know where to go from there since the mass and the charge (q) isnt' given.

So I said that mgh(of a) = 1/2mv^2 (of b)
so (9.8)h = 1/2(42^2), h = 90. That means that the potential difference between A and B is 90?, and point A is at a higher potential since the charge/mass flows TO B FROM A?

This one I'm completely lost. Leme think, so Va-Vb = 320V = W/q (and q = 1300).

So 320V = W / 1300, W = 416,000J? and that just needs to be converted into horsepower?

2. Jan 9, 2005

### vincentchan

1. you need the mass and charge of an electron, check it online or your textbook...
2. it didn't say the particle is accelarate under the effect of gravity, mgh is not working in this case, instead you need E=qV=1/2mv^2 where V is your unknown
3. yes, how many joles is a horsepower?

3. Jan 9, 2005

### thursdaytbs

For number 2, I thought it was two situations, one with a charge and another with a mass?

When you say qV = 1/2mv^2, thats for the charge - what would the mass equal to?

I'm a bit confused, could you explain a little bit more please?

(The other two problems I figured out with your hints, thanks for the help.)

4. Jan 10, 2005

### thursdaytbs

Can anyone help me with number two please? Any other suggestions?

Potential difference between the block's A and B position is 90 correct?
since 1/2mv^2 = mgh
becomes 1/2v^2 = gh
v = 42, g = 9.8, that solves for h = 90. so potential difference = 90?

Can anyone tell me if that's correct and how the charge differs?

5. Jan 10, 2005

### vincentchan

in number 2, the question is asking you a charged particle q with mass m under the influence of ELECTRIC POTENTIAL, not GRAVITITIONAL POTENTIAL.

the formulas of electric potential is qV, and the formulas of gravititional potential is mgh, so replace your mgh by qV, and you will get the answer

6. Jan 10, 2005

### learningphysics

It's the same particle for which mass and charge are given. Just plug them in and solve.

The question says nothing about the particle falling from a height. As vincentchan said, the acceleration is only due to electric field... no gravitational field so you don't need mgh.

Last edited: Jan 10, 2005