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Effects of capacitors on potential difference

  1. Jan 30, 2017 #1
    1. The problem statement, all variables and given/known data
    Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the quantity of charge you place on capacitor 1 is six times the quantity you place on capacitor 2.

    How do the potential differences across each of the two capacitors compare to each other? (V1/V2 = ?)

    2. Relevant equations

    I tried two different approaches:

    V(A->B) = -W(A->B)/qtest = -(integral of E*dl from A to B)

    V = q/C
    C = εA/d

    3. The attempt at a solution

    q1 = 6q2
    r1 = 2q2

    V1/V2 = (kq1/r1^2)*r1 / (kq2/r2^2)*r^2 = (6q2/2r2)/(q2/r2) = 3

    This is not correct.

    C1/C2 = d1/d2 = 2

    V = q/C = 6/2 = 3

    I must be misunderstanding potential difference but I'm not sure where I'm going wrong.
     
  2. jcsd
  3. Jan 30, 2017 #2
    ??? Why do you think 3 is wrong ???
     
  4. Jan 31, 2017 #3
    Is it not wrong? When I tried the answer on MasteringPhysics it was marked as incorrect.
     
  5. Jan 31, 2017 #4
    Well it's pretty straight forward. V is proportional to q and inversely proportional to d, and that is all that changed. Hard to get a different answer.
     
  6. Jan 31, 2017 #5

    gneill

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    Staff: Mentor

    Capacitance is inversely proportional to d, so potential is proportional to d. That is:
    $$V = \frac{q}{C} = \frac{q}{ε \frac{A}{d}} = \frac{q \cdot d}{ε \cdot A}$$
    $$V \propto q \cdot d$$
     
  7. Jan 31, 2017 #6
    Oh, I suck! I didn't even look at the equations. It just seemed so obvious and natural that the potential difference MUST get bigger as the plates get closer. You know, it seemed so obvious that I'm having trouble getting my head back on straight.
     
  8. Jan 31, 2017 #7

    gneill

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    Staff: Mentor

    Sorry about that :smile:
     
  9. Jan 31, 2017 #8
    Wow, I can't believe I messed that up. Thank you!
     
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