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Parallel plate capacitor (voltage and charge relative to distance)

  1. Sep 29, 2007 #1
    Hi all.

    I'm struggling to understand something about PP capacitors. Let's say there is no battery connected to a PP capacitor.

    Why does the potential (V) INCREASE when the plates are moved farther apart?

    - we know that: dv = -Ed (in a constant E)
    however as d increases, E decreases with the inverse square of d, falling off exponetially while d falls of linearly, so since E is getting smaller faster than d is getting larger, souldn't dv decrease as d increases?

    - From a non-math perspective I understand it, work must be done to pull the plates apart, thus potential energy is increased, so dv would increase. My math just doesn't agree though.

    Can someone point out my error? Many thanks in advance, this has been driving me nuts for a couple days now!
  2. jcsd
  3. Sep 29, 2007 #2

    Doc Al

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    Staff: Mentor

    As long as the separation distance remains small with respect to the dimensions of the plates, the electric field remains constant. It does not drop off exponentially or as an inverse square with distance (as would happen with a point charge).
  4. Sep 29, 2007 #3
    Thanks Doc Al, I think my problem was that I was using unrealistically high values for 'd' (1 meter or more), it makes sense now.
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