- 15

- 0

**1. Homework Statement**

Two parallel plates having charges of equal magnitude but opposite sign are separated by 14.2 cm. Each plate has a surface charge density of 33.3 nC/ m2. A proton is released from rest at the positive plate. Determine the potential difference between the plates.

**2. Homework Equations**

E=surface charge density/epsilon knot (8.85X10-12)

V=Ed

**3. The Attempt at a Solution**

I tried to relate the electric field equation to voltage difference by V=Ed but this doesn't seem to work. Is there something that I'm missing?

Here are my two approraches:

1st:

Electric field=n/epsilon knot= 3.33x10-9/8.85x10-12

=376.3

V=Ed

=752.6(0.142m)

=106.8C/m

This was wrong so then i tried to find q by solving it from E=kq/r^2

and then using the q to find what potential difference would be through the equation electric potential=kq/r

so from there I got electric potential is 53.43 and since we need to find the difference of voltage i did:

delta v= 53.43- (-53.43)

=106.8 the same thing as before!!!!

What am I missing and what am I doing wrong?? I only have 2 tries left :(