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Finding electric potential difference

  • Thread starter wendo
  • Start date
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 :(
 

berkeman

Mentor
55,708
5,791
Well, one way to do it would be to assign an arbitrary area to the plates, and solve for the voltage. Then change the area of the plates, and notice that the voltage..... does what?

Try this: set the parallel plates to a size of 1m X 1m, so the charge on each plate is 33.3nC. Now you know that Q=CV, and you have an equation that you can use for the capacitance of a parallel plate capacitor in terms of d, A, and epsilon, correct? What voltage do you calculate?

Now try the same thing with the plates twice as big (4x area)....what voltage do you get?
 

Doc Al

Mentor
44,803
1,061
1st:

Electric field=n/epsilon knot= 3.33x10-9/8.85x10-12
=376.3

V=Ed
=752.6(0.142m)
=106.8C/m
Looks like you doubled the field from one formula to the next, for some reason.

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
Those formulas are for point charges--nothing to do with this problem.
 
15
0
Using an arbitrary area worked!!! THANK YOU SOOOOOO MUCH!!!
(i'm gonna warn you now that i might have a lot of questions today...)

So the voltage will be the same regardless of what the area is, but in order to keep it the same the charge on the plates will change. I see!!!

I was just really confused at how to be able to find out what the charge was from a charge density without knowing the dimensions, so now I know just to assume a 1m^2 area!

Thank youuu!!!!!
 

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