# Homework Help: Electric Field Problem

1. Sep 14, 2006

### avb203796

A proton is placed between two oppositely charged plates, with a potential difference of 20.0 volts. if the distnace between the plates is 0.20 cm, calculate the following:

a. The ratio of the acceleration of the proton to an electron between the plates.

The electric field between the plates would have no charge because the charges of the plates would cancel each other out, right? So then it by itself would not produce any force correct? The proton which would have a positive charge would obviously be attracted to the negatively charged plate but how do I go about determining what force it produces?

b. The amount of work needed to move the proton halfway back across the potential to the negative plate.

Not even sure where to begin with this part of the problem

c. What potential difference would be needed to accelarate a proton from rest to a velocity of 0.30 the speed of light over a distance of 1.0 cm

I know the speed of light is = 3.0 x 10^8 so the velocity we want to attain would then be 9.0 x 10^7 but where do I go from there?

2. Sep 14, 2006

### Hootenanny

Staff Emeritus
You may wish to rethink this statement, an electric field cannot have a charge. Also, because the plates are oppositely charged there will be no point of zero potential between the plates. Think about it like this; there are two plates, one positively charged and one negatively charged. Now place a proton at any point between the plates. The negative charge is going to produce an attractive force on the proton towards the negative plate. Now, the positive charge is going to produces a repulsive force on the proton, this force will also act towards the negative plate. Therefore, there is no way the two electric fields from the two charges could possibly 'cancel' as you suggest. Does that make sense?
HINT: Voltage is work done per unit charge.
Refer to above hint