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Two capacitors in parallel: potential difference/energy

  1. Sep 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Two capacitors, one that has a capacitance of 4 μF and one that has a capacitance of 12 μF are connected in parallel. The parallel combination is then connected across the terminals of a 12 V battery. Next, they are carefully disconnected so that they are not discharged. They are then reconnected to each other - the positive plate of each capacitor connected to the negative plate of the other.

    (a) Find the potential difference across each capacitor after they are reconnected.

    (b) Find the energy stored in the capacitors before they are disconnected from the battery, and find the energy stored after they are reconnected.


    2. Relevant equations

    (In parallel) Ceq = C1 + C2 + ... + Cn
    C = Q/V
    E (energy) = (1/2) C V2
    3. The attempt at a solution

    I know that in parallel, the voltage across all of the capacitors is the same (which is 12 V in this case). I also know that Ceq = C1 + C2 + ... + Cn, which in this case should turn into (4 * 10-6 ) + (12 * 10-6) = 1.6 * 10-5 F

    If I plug this into the formula E (energy) = (1/2) C V2 then I get:

    E = (1/2)(1.6 * 10-5)(122) = 1.152 * 10-3 J

    With this, it seems that I have solved the first half of (b), which was the question about how much energy is stored in the capacitors before they disconnect from the battery. I don't know if it is right however, since this is written homework and not webassign. Furthermore, I do not know how to do (a) or the 2nd half of (b). Can someone guide me in the right direction on how to do (a) and the 2nd half of (b), and also verify whether or not my first half of (b) is correct? Thank you.
     
  2. jcsd
  3. Sep 24, 2016 #2

    BvU

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    Hi.
    It might be that the composer of the exercise wants two answers (energy in each of the capacitors), but that's somewhat ambiguous.

    Now, for part a, what happens when the two capacitors are reconnected ? What will remain constant and what will even out ?
     
  4. Sep 24, 2016 #3
    I don't know for sure what remains constant and what evens out, but my educated hypothesis leads me to believe that the total charge on the capacitors will balance out so that the positive and negative plates have equal, but opposite charges. Is this correct?

    As for what remains constant, does the equivalent capacitance remain constant?

    If my assertions are correct, then since the total charge on the capacitors to begin with was 1.92 * 10-4 C (I calculated this), and since the equivalent capacitance is 1.6 * 10-5 F, I believe that if I divide my charge by 2 (so that both capacitors have the same amount of charge on them) and use V = Q/C, then I can do (9.6 * 10-5) / (1.6 * 10-5) = 6 V (Note that the 9.6 * 10-5 came from dividing my total charge by 2)

    Is 6 V the correct answer for (a) by my process?

    Also, are you saying that I got the right answer for (b), and that the answer for the first half of (b) is the same as that of the second?
     
  5. Sep 25, 2016 #4

    BvU

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    Let's go over them. I agree with the 1.92 10-4C. The picture below divides the sequence in four steps. Maybe that makes it easier.

    In step 1 both capacitors have the same voltage. What about the charge ?
    Step2: disconnecting does not change either of these two.
    In step 3 one of the two is flipped. No change except for the signs. You may want to add the signs to the picture.

    Perhaps now I can repeat my question: what will remain constant and what will even out ? (compare 4 and 1)

    Your belief
    does not hold up: imagine the smaller capacitor has almost zero capacity: it can never hold half the charge.



    upload_2016-9-25_11-20-53.png
     
  6. Sep 25, 2016 #5
    Well I notice that in #4, the 4 μF capacitor's plates are reversed as compared to how they were in #1. Since E = V/d (where E is the electric field between the two plates) , and since the sign of the E field is determined by the direction that a positive test charge would move, I suppose this reverse in position of the plates would cause the voltage on the capacitor to change signs while keeping the same magnitude. In terms of what balances out, I suppose this means that the capacitor that reverses would have -12 V while the other capacitor would maintain a voltage of 12 V. This would make the whole system balance out to 0 V which would make sense since they are no longer connected to the battery.

    Is my understanding here correct?
     
  7. Sep 26, 2016 #6

    BvU

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    That is correct
    Correct as well
    Not correct. The driving force is the Voltage difference between the two capacitors. When the difference is zero, charge has no reason to move any more.

    As you can see from the picture in #4, there is nowhere the total charge on the top plates can go. But if the charge on one of the top plates is negative and on the other is positive, something is surely bound to happen...
     
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