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Charging capacitors in series with another capacitor

  1. May 18, 2015 #1
    This question was taken from the public examination in my country/region.
    The answer from the marking scheme is, only statement 1 is correct and the others are wrong. Can anyone explain why the other statements are wrong?
    1. The problem statement, all variables and given/known data

    The following components are connected in series in a circuit in the following order:
    a charged capacitor C, two uncharged capacitors C1 and C2, an open switch S.
    ( I couldn't draw it on the computer...)
    It is given that C1 > C2 and the resistances of the connecting wires are negligible. Which of the following is/are correct after closing the switch S?
    Statement 1: p.d across C1 < p.d. across C2
    Statement 2: The sum of magnitudes of charges on all three capacitors = the original charge (before closing switch) on C
    Statement 3: Total energy stored in three capacitors = energy stored in C before closing switch
    2. Relevant equations
    Energy stored in capacitor= ½CΔV2 = ½QΔV = ½Q2/C
    Definition of capacitance: C=Q/ΔV
    3. The attempt at a solution
    (1) was proven to be true with definition of capacitance. Tried to state out heat is released for (3) but it is not true due to R(wire)=0Ω → P=0W as P=I2R.
  2. jcsd
  3. May 18, 2015 #2


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    Hello LWfromHK, welcome to PF :smile: !

    From your wording, I take it you acccept that (1) is a correct answer and are in doubt about (3).

    In case you want to check the answers, do you know how much energy is stored in a capacitor that is charged with a charge Q ?
    If so, you should be able to calculate the energy stored after the new equalibrium is established with the energy stored before.

    (2) isn't so much a calculation as a matter of understanding. Where can the inital charge on C go and what are the consequences ?
  4. May 18, 2015 #3
    Thanks for your reply! It helped a lot in both my thinking, not just only on this question.
    For (2), the initial charges go into the capacitors C1 and C2, by charge sharing. The charge at C1 is equal to that at C2. For the initial charge on C transferred to C2, the same amount will be transferred on C1. Let this charge transferred to be q. Then the charge loss at C is q, charge gain at C1 is q, and that at C2 is also q. Plates that are connected by the wire from C1 to C2 will be charged by the same amount but opposite in sign, through electrostatic induction. Hence, there is a net charge gain of -q (at C) +q (at C1) +q (at C2) = +q ≠ 0 for the system
    ∴ (2) is incorrect

    For (3), I think I would do the calculation using the result from (2),
    Knowing that initial energy stored in C = Ei = ½Q2/C for initial charge Q and p.d. ΔV
    Final total energy stored in all three capacitors = ½ [(Q-q)2/C + q2/C1 + q2/C2
    = ½ {(Q2 -2Qq+q2)/C+q2/C1+q2/C2}
    =Ei + q2/2Ceq - Qq/C, where Ceq is the equivalent capacitance of all three capacitors.
    Take net change in energy stored in the capacitors (ΔE) = q2/2Ceq - Qq/C
    Then, final total energy stored in all three capacitors = Ei + ΔE
    Since q, Q, C, C1, C2 are arbitrary values, it is uncertain whether a net change in energy stored in the capacitors (ΔE) is registered. Hence, statement (3) is wrong, as a combination of any values may lead to ΔE≠ 0. We cannot consider it correct if there exists a case which ΔE≠0.
    Did I get it correctly? Please notify me if I didn't. Thanks!
  5. May 18, 2015 #4


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    Almost. The last line is correct, but not for the reason you state in the preceding line :frown:.

    Once the switch is closed, you have three capacitors in series. Whatever charge there is, it can't get out of the circuit!
    If a charge q is transferred from C to C1, then a charge -q is transferred from the bottom plate of C to the bottom plate of C2. So the net charge gain is zero. That reason (2) is incorrect nevertheless is because the sentence "sum of magnitudes of charges on all three capacitors" nastily wants you to add |Q-q| on C, |q| on C1 and |q| on C2.

    If this is established, we can move on to (3) comfortably, using your equation. You are well underway, but there is more to be said here. If you want a simple case: take C as in this problem and replace C1 and C2 by one uncharged capacitor with capacity C as well. What can you say about the energies before and after in that case ?

    Again (3) is indeed wrong, but I suspect it is always wrong.

    (Afterwards, there will be the complicating question: where did the energy go ?! :rolleyes: intriguing, but fortunately not within the scope of this exercise. More later...)

  6. May 19, 2015 #5
    I see, the magnitudes of charges cannot be added through summation.

    The charge on both capacitors will be equal, that is, half of the original of the charge carried by C. and thus the energy stored in each capacitor will be (½)2=¼ times the original C. Hence, the total energy will be 2×¼=½ times the original.

    I suggest the energy difference is the work done to push the charges from one capacitor to another. Otherwise, no charges can be accumulated at the other uncharged capacitor.
  7. May 19, 2015 #6


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    Good work. So (2) is done with, right ?

    For (3), you have shown that for two equal capacities the energy is reduced to one half the original energy. You could work it out further with unequal capacities (good exercise in dealing with inequalities!)

    To explain it is probably beyond the scope of the exercise: in the case of ideal capacitors, ideal zero resistance connections etc., the current when the switch closes would jump to infinity, which is rather impossible. And once this big current is flowing, it wants to keep going (by self-inductance): the discharge goes on beyond equilibrium and the current reverses. An oscillating circuit that radiates off electromagnetic waves that carry away energy. All very far fetched: a really small resistance already limits the current and dissipates electric energy into heat (or a flash of light, for example).

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