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Potential difference in capacitor help

  1. Apr 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Initially, the switch in the figure is in position A and capacitors C_2 and C_3 are uncharged. Then the switch is flipped to position B. (Figure attached)
    a) Afterward, what is the charge on C_1 capacitor?
    b) Afterward, what is the potential difference across C_1 capacitor?
    c) Afterward, what is the charge on C_2 capacitor?
    d) Afterward, what is the potential difference across C_2 capacitor?
    e) Afterward, what is the charge on C_3 capacitor?
    f) Afterward, what is the potential difference across C_3 capacitor?

    2. Relevant equations

    V= IR

    3. The attempt at a solution

    How do I start this problem?
     
  2. jcsd
  3. Apr 12, 2009 #2

    LowlyPion

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    Re: Capacitor

    This would be a start.
     
  4. Apr 12, 2009 #3
    Re: Capacitor

    and here is the figure...
     

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  5. Apr 12, 2009 #4

    LowlyPion

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    Re: Capacitor

    So what do you think is going to happen?

    What is the equivalent capacitance of the additional capacitors that are not initially connected?
     
  6. Apr 12, 2009 #5
    Re: Capacitor

    we know C = Q/V
    V = 100 V
    C1 = 15*10^-6 F
    so Q = C1V = 100 V(15*10^-6 F) = 1.5*10^-3 C

    C2 and C3 is in series. so 1/C = (1/C2 + 1/C3) = (1/30*10^-6)+(1/30*10^-6) = 66666.67
    so C = 1.5*10^-5

    would the Q be same after switching to point b? If yes, then we can figure out V using above equation.
     
  7. Apr 12, 2009 #6

    LowlyPion

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    Re: Capacitor

    When the connection is broken then there is a fixed charge on the C1. When the switch connects to C2 and C3 then the charge will be shared according to what the equivalent capacitance for all 3 of them.

    C2 and C3 = 2*C1

    C2 and C3 equivalent C = C1

    When you put them in || you have then C1 + C1 = 2 C1

    Q = V*C

    So with the new C = 2*old C and charge the same, then Vnew = 1/2*Vold
     
  8. Apr 12, 2009 #7
    Re: Capacitor

    there are like 6 parts of this problem.
    so for part a)
    C = Q/V
    V = 100 V
    C1 = 15*10^-6 F
    so Q = C1V = 100 V(15*10^-6 F) = 1.5*10^-3 C
    and how do I get the potential difference?
     
  9. Apr 13, 2009 #8

    rl.bhat

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    Re: Capacitor

    When uncharged capacitors are connected to the charged capacitors the common potential difference is given by V = (C1V1 + o)/(C1 + C2)
     
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