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Simple Capacitor and Conservation of Charge Question.

  1. Jul 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Capacitors C1 = 6.18 µF and C2 = 1.69 µF are charged as a parallel combination across a 250 V battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Calculate the resulting charge on capacitor C1.

    Q1= charge on C1 when in parallel
    Q2= charge on C2 when in parallel.
    q1= charge on C1 when connected in series.
    q2= charge on C2 when connected in series.
    Qt= total charge

    2. Relevant equations

    C=Q/ΔV



    3. The attempt at a solution
    V1= q1/C1, V2= q2/C2, V1= V2, q2= q1*(C2/C1)

    q1+q2= Qtotal sub in q2 from above

    q1+q1*(C2/C1) = Qtotal

    Isolate q1

    q1= Qt*(C1/(C1+C2))

    Now I know how to get the answer, the problem is my understanding.

    In order to get the correct answer you must take Qt = Q1 - Q2,

    I can't understand why this is. I thought the total would be Q1+Q2 = Qt just like q1+q2=Qt
    If I can't understand it I can't move on.
     
    Last edited: Jul 26, 2012
  2. jcsd
  3. Jul 26, 2012 #2

    tiny-tim

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    Welcome to PF!

    Hi Sabastien! Welcome to PF! :smile:
    Are you trying to use formulas? :redface:

    Just draw the circuit …

    suppose the capacitors were originally 5 and -5, and 4 and -4

    then the question says the 5 is connected to the -4 and the -5 is connected to the 4 …

    so how much charge is there on each side? :wink:
     
  4. Jul 27, 2012 #3
    No I created the the Q1+Q2=q1+ q2 based on my assumption of conservation of charge.

    My drawings looks something like this:
    (x) is just empty space ignore them.

    Original set-up (1):
    x--+C1--x
    |xxxxxxx|
    x--+C2--x
    |xxxxxxx|
    x--+dV--x

    Afterwards (2):

    x--+C1--+C2--x
    |xxxxxxxxxxxxx|
    x-------------x

    What I did to find the total charge is that I know that V1 = V2 = dV so based on that capacitor they both hold a specific charge. Q1 and Q2 respectively. Now I believed that the total charge in the system would obviously be Q1 + Q2. From there I said that the total in the system couldn't change, so when placed in the alternate formation with no voltage. The charges would rearrange but the total charge would remain the same. So q1 + q2 must also be equal to Qtotal.

    Where I seem to go wrong, is the belief that Qtotal = V1*C1 + V2*C2 . I always take those values to be positive because they have both positive sides splitting and both negative side combining. When rearranged they seem to lose charge, which shouldn't happen if charge is conserved. If the total charge is reduced when placed positive to negative how does that conserve charge? Also wouldn't you also be able to say that Qt = Q2-Q1 ,then you have a negative Qt, but is Qt just a magnitude? Is that why it's always positive? I feel like I am rambling in my own confusion. The only way that I can justify why the total charge would change would be that when the positive is connected to the negative some charges are "cancelled" in the exchange and that the total HAS to be a magnitude. Is this correct?

    Side question: If I had a third arrangement that just takes away the dV and keeps the positive to positive and negative to negative, would that hold the original charge in equilibrium and give me the Q1+Q2=Qt that I expected, or would the capacitors start to discharge until they made the second arrangement and hold that in equilibrium?

    Arrangement 3:

    x--+C1--x
    |xxxxxxx|
    x--+C2--x
     
  5. Jul 29, 2012 #4

    rude man

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    My advice, which you may not like, is to connect C1 and the flipped C2 via a resistor, then write the equations to solve for V1, the new voltage on C1.

    I say that because philosophically one violates energy conservation if one seeks final quiescent (dc) capacitor voltages and does not assume a finite resistance in reconnecting the capacitors.

    I imagine a charge conservation argument can also be invoked but I hesitate to rush in where angels fear to tread .... :biggrin:
     
  6. Jul 29, 2012 #5

    rude man

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    OK, you can look at it in terms of charge conservation too, but then you have to understand that, when C2 is flipped before re-connecting, you now have the situation where Q2 is negative. Since C1 > C2, q1 and q2 will both be > 0.

    Since you obviously really want to understand the situation it would still behoove you to contemplate the apparent lack of energy conservation: Q1*V1/2 + Q2*C2/2 > q1*C1/2 + q2*C2/2 where Q1 and Q2 are the original charges, i.e C1*V + C2*V, V = 250V. Don't be misled by those who claim the lost energy is radiated out! Which explains my first post.
     
  7. Jul 30, 2012 #6
    Thank you, I feel like I understand it now.
     
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