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Finding Capacitance of system

  1. May 15, 2015 #1
    1. The problem statement, all variables and given/known data

    Four metallic plates each of surface area S are placed at some distance from each other as shown in figure.
    Then capacitance of the system between P and Q is

    A. ## \frac{ε_0S}{3d} ##

    B. ## \frac{ε_0S}{2d} ##

    C. ## \frac{3ε_0S}{2d} ##

    D. ## \frac{3ε_0S}{d} ##
    2. Relevant equations
    Between 2 parallel plate capacitors without dielectric capacitance is given as
    $$ C = \frac{ε_0A}{d} $$ where A is surface area of plates and d is distance between them.

    3. The attempt at a solution
    I know if two plates are in series then equivalent capacitance is ## \frac{1}{C_{eq}}= \frac{1}{C_1} + \frac{1}{C_2} ##
    If they are in parallel then equivalent C is C1 + C2.
    But I don"t know what the system is in diagram.

    Post Script: How do I disable the red lines coming on words below? I am using first time computer for posting a problem in PF.
  2. jcsd
  3. May 15, 2015 #2


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    The middle plates can be doubled and connected, it is the same set-up. So it is three capacitors, like in the figure, all having the same plate surface area. How are the ones with 2d distance connected?

  4. May 15, 2015 #3
    How did you convert my diagram to that? Is there some method?
  5. May 15, 2015 #4


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    I t
    I told you in the previous post. The middle plates can be thought as two plates, connected together by a wire. The arrngement on the left is equivalent with the one on the right: two capacitors, with plates AB and B'C.
    Originally, the plates A and B are connected. Do the same on the right ones.
  6. May 15, 2015 #5
    Okay, the diagram was the important part.
    Now, 2d plates are in parallel
    So capacitance is adding both 2 capacitances ε0S/d
    Then they are in series with d plate
    So capacitance is 1/2 of ε0S/d which is ε0S/2d . So option B is correct.
    Thanks ehild.
  7. May 15, 2015 #6


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    You are welcome :smile:
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