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Calculating capacitance

  1. Apr 21, 2006 #1
    Hi, I've attached a diagram of the question. The real question reads differently, but I've done my own interpretation. The interpretation might be wrong/ right, but its left me thinking anyway. So, I'd appreciate any advice/ hints on how I might go about finding the capacitance. I have tried it.
    I reasoned that the voltage across both segments are the same, due to the 'parallel' configuration, and then attempted to solve one of the pairs' capacitance by taking note of the fact that the charge surface density on one is half the other.
    Then I multiplied the result by 2, according to parallel capacitor combinations. I got the wrong answer though.

    Its kinda urgent, cause my exam's coming soon. I'd appreciate any advice at all. Small or big. Thanks a lot.
     

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    Last edited: Apr 21, 2006
  2. jcsd
  3. Apr 21, 2006 #2
    what formula did you use to find the capacitance?
     
  4. Apr 21, 2006 #3
    I used Gauss' law to find the electric field between the plates, then substituted in potential difference E= V/d. After solving for V, I could find C using q = CV.
    But my method is most probably wrong... cause, well, the answer's wrong.

    I didn't use the formula for parallel plate capacitors because in this case, the charges at each plate is different by a 2 factor.
     
  5. Apr 21, 2006 #4

    berkeman

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    Staff: Mentor

    The formula that Hammie is asking for relates the capacitance to epsilon, area A and separation d. It's a handy formula to have memorized.

    You are correct that the total capacitance in this parallel arrangement is twice wha a single side would give, because it is effectively two caps in parallel.
     
  6. Apr 22, 2006 #5
    oh thanks. Yea, it is parallel. I forgot to convert the units so I got my answers wrong.
     
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