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Square and Circle Capacitor

  1. Jul 31, 2006 #1
    I need to know how the capacitance of a capacitor built from a
    square plate and a large rectangular plate compares to that of a
    capacitor built from a circle and a large rectangular plate, if the
    diameter of the circle equals the length of the side of the square.
  2. jcsd
  3. Jul 31, 2006 #2

    Andrew Mason

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    I think you have to apply Gauss' law to the smaller plate on the theory that the field above the smaller plate and the charge distribution on it will be uniform. The charge distribution will not be uniform on the larger plate so the field will not be uniform. (Let's say the smaller plate is on the bottom). Applying Gauss' law to the smaller plate: [itex]EA = Q/\epsilon[/itex] or [itex]E = \sigma/\epsilon[/itex]

    Since E = V/d over the smaller plate does not depend on surface area, [itex]\sigma[/itex] does not depend on surface area. Therefore, the amount of charge on the smaller plate varies in proportion to the plate area. Since [itex]C = Q/V[/itex], the capacitance would vary in proportion to the area of the smaller plate.

    Apply that principle to your problem.

  4. Aug 1, 2006 #3

    When the plates are close together, the simple law C = epso S/d applies. Therefore, in this limit case, the ratio of the capcacitance will be the ratio of the surface, which is pi/4 = 0.785398, simply.

    When the separation increases, the fringing fields may increase the capacitance considerably as compared to the simplified formula above. This is because of charge accumulation in the edges and in the corners (for rectangles).

    For moderate separations, there is a Kirchhoff formula to evaluate, but I can't find it back right now. This may allow you to estimate your ratio for moderate separations. Unfortunately, I am not sure it can be adapted for rectangular or square plates.

    Last edited: Aug 1, 2006
  5. Aug 5, 2006 #4
    Although Andrew Mason and lalbatros way would work, I'd prefer to solve Laplace equation in both cartesian and spherical coordinates, it'll be a lot easier since your only changing one coordinate.
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