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Simple Parallel Plate Capacitor Question

  1. Jan 4, 2010 #1
    Assuming a Parallel Plate Capacitor of area A, separation distance d, plate charges [tex]\pm Q[/tex]: and plate charge densities [tex]\pm\sigma[/tex]:

    In my textbook and in Wikipedia,

    Electric Field of a charged plane with large or infinite area: [tex]\frac{\sigma}{\epsilon}[/tex]
    Electric Field between plates of parallel plate capacitor: [tex]\frac{\sigma}{\epsilon}[/tex]
    Voltage difference between plates of parallel plate capacitor = [tex]\frac{Q \cdot d}{\epsilon \cdot A}[/tex]

    My question is since a parallel plate capacitor contains two plates of equal and opposite charge, shouldn't there be an additional factor of two? In other words,

    shouldn't the net electric field between the two plates be: [tex]\frac{2 \cdot \sigma}{\epsilon}[/tex]
    and shouldn't the voltage difference between the two plates be: [tex]\frac{2 \cdot Q \cdot d}{\epsilon \cdot A}[/tex]
  2. jcsd
  3. Jan 4, 2010 #2


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    Ah... this is not the first time I've come across this particular conundrum.

    That formula applies to the case of a conductor with one plane surface, where the entire charge density [itex]\sigma[/itex] lies on that one surface. But each plate in a capacitor has two surfaces. If you define [itex]\sigma[/itex] as the charge density on each of those surfaces individually, the total charge on each plate is
    [tex]Q = 2\sigma A[/tex]
    so the charge density on each surface is
    [tex]\sigma = \frac{Q}{2A}[/tex]
    In this sense, yes, the electric field at the center is
    but that still works out to
    [tex]V = \frac{Qd}{\epsilon A}[/tex]
    which you could also calculate from Gauss's law.

    It is also possible to define [itex]\sigma[/itex] as the total charge density on each plate, i.e. as the sum of the charge densities on both sides of the plate. In that case, the factor of 2 disappears, so that
    [tex]Q = \sigma A[/tex]
    But with this definition of [itex]\sigma[/itex], the electric field between the plates is just [itex]\sigma/\epsilon[/itex], so it still works out to
    [tex]V = \frac{Qd}{\epsilon A}[/tex]
  4. Jan 4, 2010 #3
    Ah... two-sided plates. Makes perfect sense. I wish they explained that in my textbook or in Wikipedia...
  5. Jan 4, 2010 #4


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    Me too - that particular ambiguity has caused a lot of confusion.

    Which Wikipedia page did you look at? Maybe it needs to be clarified.
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