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The parallel-plate capacitor

  1. Sep 10, 2004 #1
    Hi everyone.

    Can someone do this tricky problem?

    I have to find an expression that relates the plate separation x to the potential difference between two parallel-plate capacitor. The charge on the top and bottom plates are +Q and -Q respectively. (Area = A, Voltage = V) The only hint I am given is that it will be a cubic equation.

  2. jcsd
  3. Sep 10, 2004 #2
    well... i know capacitance is the dielectric constant*area/distance of separation, which is also charge contained over voltage... right?

    so wouldnt that work?
  4. Sep 10, 2004 #3
    There is another way to find this if you know the magnitude of the electric field between the plates (you might be able to figure that out somehow, or maybe it's given). Try beginning with some basic principles to derive the formula. A positive test charge is useful when working inside parallel plates. You know that W = Fd*cosΘ, and that work also equals negative the change in electric potential energy of the test charge. Also remember the derived formula which states that the voltage across which a test charge moves equals the change in its electric potential energy divided by its charge. Use these to develop two equations for electric potential energy, equal them, the use F = Eq, where E is field and q is charge of test charge, and you should arrive at something. Just find a way to figure out what the field between the plates is.
  5. Sep 10, 2004 #4

    Doc Al

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    Staff: Mentor

    What exactly is the question? Find an expression for x in terms of V, A, and Q for a parallel plate capacitor? If that's the question, it has nothing to do with cubic equations.

    What have you tried so far? Hint: What's the formula for the capacitance of a parallel plate capacitor?
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