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Capacitors and switches problem

  1. Mar 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Two capacitors, one with C1=2pF and V1=30V, and the other with C2=3pF and V2=20V are connected to each other through a switch (positive end of one connected to the negative end of the other). The other end of each capacitor is kept open. What happens to V1 and V2when the switch is closed?

    2. Relevant equations

    3. The attempt at a solution

    Both the capacitors have the same charge, i.e. 60pC. When the switch is closed there will be no charge flowing between the two capacitors as two capacitors connected in series hold the same charge. Moreover, since one end of each capacitor is open, there can be no current flow in any case. Hence there will be no change in V1 and V2 when the switch is closed.

    The answer gven in the book mentions that when the switch is closed the potential difference across C1 and C2 will become equal to the average of V1 and V2, i.e. = 25 V. I am not able to reconcile with this logic. Hence would appreciate some views of the Forum members.
  2. jcsd
  3. Mar 21, 2010 #2


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    Homework Helper
    Gold Member

    Re: Electrostatics

    You are right, the voltage remains the same across both capacitors.

    The situation is the same as if you had two uncharged capacitors, 2 pF and 3 pF, connected in series, and then charged to 60 pC. The resultant capacitance is 1.2 pF, so the overall potential difference is V=q/C=50 V. V is shared between the two capacitors: V=V1+V2. V1=q/C1=30 V and V2=20 V. The book states nonsense.

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