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Finding electric potential difference

  1. Jul 13, 2007 #1
    1. The problem statement, all variables and given/known data

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


    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:


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


    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 :(
  2. jcsd
  3. Jul 13, 2007 #2


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    Staff: Mentor

    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?
  4. Jul 13, 2007 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Looks like you doubled the field from one formula to the next, for some reason.

    Those formulas are for point charges--nothing to do with this problem.
  5. Jul 13, 2007 #4
    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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