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Electric Potential: Capacitor

  1. Aug 28, 2012 #1
    1. The problem statement, all variables and given/known data

    A 8.1 mu F capacitor is charged by a 155V battery (see the figure a) and then is disconnected from the battery. When this capacitor C_1 is then connected (see the figure b) to a second (initially uncharged) capacitor, C_2, the final voltage on each capacitor is 20V. What is the value of C_2?

    2. Relevant equations

    Q = CV

    3. The attempt at a solution

    (8.1*10^-6 F)(155 V) = 1.256 * 10^-3 C

    1.256*10^-3 C / 20 V = 6.3*10^-5 F

    I did it this way because I thought charge was conserved.
     
  2. jcsd
  3. Aug 28, 2012 #2

    CWatters

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    I believe you have correctly calculated the combined capacitance (eg C1 and C2 in parallel). That's not quite what the question asked.
     
  4. Aug 30, 2012 #3
    I tried to find the potential energy stored in the capacitors but I don't think I did it correctly.

    I used PE = 1/2 Q*V

    Q = 1.256 * 10^-3 C
    V = 135 V (unsure)

    PE = 0.0847 J

    From there I tried to find the value of C_2 since it said charge is conserved by using:

    C = .5*Q^2/PE

    C = 7.4*10^-3 F
     
  5. Aug 30, 2012 #4

    CWatters

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    If I remember correctly charge will be conserved (nowhere for it to go apart from being shared between the two capacitors).

    So before C2 is connected..

    Q = C1 * Vbefore

    and after C2 is connected..

    Q = (C1+C2) * Vafter

    where

    Vbefore = 155V
    Vafter = 20V

    equate

    C1 * Vbefore = (C1+C2) * Vafter

    Rearrange to give C2...

    C1 * Vbefore = (C1 * Vafter) + (C2 * Vafter)

    C2 * Vafter = C1 * Vbefore - (C1 * Vafter)

    C2 = C1 (Vbefore - Vafter)/Vafter

    Substitute values...

    C2 = 8.1 * 10-6SUP] (155-20)/20

    C2 = 54.7 * 10-6SUP] F

    or 54.7 uF
     
  6. Aug 30, 2012 #5
    Ok, the way you have done it makes much more sense. Thank you.
     
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