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Effect of Fringing on Capacitance

  1. Nov 14, 2005 #1

    I've been reading electromagnetism of late...in particular about parallel plate capacitors. So here are 2 questions about them.

    We generally consider the field between the two capacitor plates to be approximately equal to that generated by a pair of infinite conducting planes with equal and opposite charges on them. In doing so, we neglect fringing of the fields just outside the region of the capacitance. The analysis yields the relation

    [tex]C = \frac{\epsilon_{0}A}{d}[/tex]

    between capacitance C and geometric parameters (A = area of cross-section of plates and d = distance between them).

    Question 1: Now, if we were to take fringing into consideration, would the capacitance increase or decrease? I think it should increase --as it will increase with area A even if the linear relationship above does not hold--because if we include the region just outside the capacitor in our new 'effective' capacitance we will have a new capacitance with increased area of cross-section.

    Question 2: If the two plates forming a parallel plate capacitor are of two different cross-sectional areas then how is the above expression modified. My reasoning is that we ought to use the minimum of the two areas in the expression as the smaller sheet would govern how much it area of the larger one it sees. But if this is so, how does the charge "adjust" on the larger plate so that for all practical purposes, we have identical charge distributions facing each other (in terms of magnitude of charge only).

    Thanks for your help...

    Last edited: Nov 14, 2005
  2. jcsd
  3. Nov 19, 2005 #2
    Just a reminder...
  4. Nov 19, 2005 #3

    Physics Monkey

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    Regarding question 1, the capacitance is slightly larger than the ideal result. If you keep the plate area fixed and let d become large, the capacitance remains above the ideal result and limits to a d independent value for large d (related to the self capacitance of a single metal plate).

    Regarding question 2, if the larger plate is "infinite" then you have the case of a finite conductor above an infinite conducting plane. The solution will be some charge density on the finite plate, and it is easy to see that that effect of the infinite plane should correspond to the equal and opposite charge distribution behind the infinite plane (from images). Thus in some sense the system is equivalent to a parallel plate capacitor system (made of the smaller area). I leave it to you to fill in the details and decide to what extent the analogy is relevant.
    Last edited: Nov 19, 2005
  5. Nov 21, 2005 #4
    Thanks :-)

    Oh I was thinking of keeping d (distance between the plates) fixed and allow A to increase "effectively".
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