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I Conceptual question about capacitor plate size

  1. Mar 12, 2016 #1
    You have a capacitor and each plate is of area A, explain qualitatively what happens to the capacitance when you increase the size of only one of the plates and why.

    Edit: adding some conditions:
    1. they are circular discs (for simplicity's sake)
    2. one has diameter d1, one has d2,
    3. make d2 = 2d1, compare to capacitance when d1 = d2.
    4. the separation between the two plates is negligible compared to the diameters.

    this wasn't really a homework problem but it came up in my book and i don't know how to do it.
     
  2. jcsd
  3. Mar 12, 2016 #2

    phinds

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    It isn't looking for a numerical answer. What do you think happens?
     
  4. Mar 12, 2016 #3

    collinsmark

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    As a first step, consider (at least conceptually) calculating the capacitance of single plate (single, finite plane). Suppose this plate is has a uniform charge distribution of [itex] \sigma [/itex] over its surface. Calculate its electric field as a function of distance away from the plane -- remember it's a finite plane, so it's not so simple as using Gauss' Law to get the answer quickly, but it is possible to calculate this electric field, [itex] \vec E [/itex]. Then, determine its electric potential by integrating over the path, starting from infinity, [itex] V = - \int_P \vec E \cdot \vec{d \ell} [/itex]. Finally determine the capacitance, [itex] C = \frac{Q}{V} [/itex].

    Now put that answer away. It was merely a practice exercise.

    Now, to make this more relevant to the given problem, repeat the idea with a planar ring, with inner radius [itex] d_1 [/itex] and outer radius [itex] d_2 [/itex]. Sum that answer with the capacitance of a parallel plate capacitor with both plates having radius of [itex] d_1 [/itex].

    Edit: More to the point about this being a qualitative problem: Is the capacitance of the planar ring even in the same ballpark as the capacitance of the parallel plate capacitor?

    Edit: And perhaps a more important consideration: when the plates are of unequal areas, is it still valid to assume that that charge distribution on larger plate remains uniform? What you would expect, qualitatively speaking, for the charge distribution to be on the larger plate, if both plates had equal and opposite total charge? (If it helps, treat the plates as thin, 3-dimensional conductors. Recall that the electric field inside a conducting material [even a thin material] is always zero. What sort of charge distribution would be necessary, on the larger plate, to facilitate this?)
     
    Last edited: Mar 12, 2016
  5. Mar 12, 2016 #4
    I think it increases by a small amount
    no, its area is too small
    (if large plate is positive and small is negative) there will be a high concentration of positive charges in the large plate in areas that overlaps with the small plate, and a lot less in other areas.
     
  6. Mar 12, 2016 #5

    collinsmark

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    So far so good. :smile:

    But is this increase even significant, or can it be considered negligible for practical purposes?

    I wouldn't concentrate on the area, but rather the potential.

    In the case of the single-plate configuration, when calculating the potential one must integrate the path all the way from infinity. Sure, the electric field decreases as one gets far away from the finite plate, but path that one must integrate over is huge.

    In the parallel plate configuration, the path length is very small -- tiny even. You only need to integrate over the plates' separation (a comparatively tiny distance). Sure, the electric field in-between the plates is double, on account of there being two plates of equal and opposite charge, but the path distance is really tiny.

    So compare the difference in potential between the two configurations, assuming that the total charge |Q| (on a given plate) is the same.

    Now, noting that the potential is in the denominator in [itex] C = \frac{Q}{V} [/itex], how does a larger potential difference affect capacitance for a given charge Q ?

    That sounds right to me! :smile:
     
    Last edited: Mar 12, 2016
  7. Mar 12, 2016 #6
    I just realized that since the potential depends on the number of charges, the amount of charges not on in the area defined by the smaller disk is small enough to ignore. so capacitance is basically the same as two identical plates.
    I'll chew on the rest of what you said at a later time, gotta get back to cramming for physics olympiad semis xD

    thanks for the help! it is very much appreciated!
     
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