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Capacitor prob

  1. Oct 14, 2008 #1
    Hi,

    If I have two capacitors [itex]C_1=1 \mu F[/itex] ..... [itex]C_2=2 \mu F[/itex],and I start by taking each and connecting it to 5V battery in series, so the first cap has [itex]Q_1=C_1 V=5\mu C[/itex] then seperatley I connect the other cap to another 5V source so it gains charge [itex]Q_2=C_2 V=10 \mu C[/itex].

    Now I remove the batteries and connect the two caps in series with each other, so that the opposing charge plates face each other. How would I calculate the final voltage across the 2 microfarard cap?

    My reasoning was if you have two caps in series and nothing else, it's basically two seperate wires. The " two wires" have 10microC at one end and -5microC at the other, so current initially flows until 7.5microC have travelled, leaving 2.5microC at both ends, similarly for the "other wire". So this reasoning would make me think on the 2microF cap the final voltage would be V=q/C_2=2.5/2=1.25V

    This doesnt seem to be the correct answer however, can anyone please shed some light?
     
  2. jcsd
  3. Oct 14, 2008 #2
    charge cannot accumulate in a loop so both capacitors should have equal and opposite charge Q. Q = CV so the voltage on each capacitor will be V = Q/C; The 1 uF cap. has twice the voltage as the 2 uF cap. following the above relation. This can only be 3.33 V for the 1 uF cap. and 1.67 V for the 2 uF cap.
    Also notice that V1 + V2 = V total since the caps. are in series.
     
  4. Oct 15, 2008 #3
    But this means that each capacitor has [itex] Q_1=Q_2=3.33 \mu C [/itex], I don't know how you get to this value of charge on each capacitor, because by my reasoning above each should end up value being 2.5microC (7.5 flows off ten leaving it with 2.5, and this 7.5 goes onto the -5 taking it also to plus 2.5, otherwise why would current stop flowing and give the final value?).

    Also I don't see why the relation V1+V2=5V should hold, as the caps aren't in series with the battery now just each other. So by Kirchoff's loop you would expect V1+V2=0
     
  5. Oct 15, 2008 #4
    I've attached a picture of my reasoning, I know it's wrong but I don't know why it's wrong, or what is the correct way to do this problem.
     

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  6. Oct 15, 2008 #5

    atyy

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    ???????????

    Let the initial charge on C1 be Q1.
    Let the initial charge on C2 be Q2.

    The opposing charge plates face each other.
    C1 final inner plate charge is Q1+dq
    C2 final inner plate charge is -Q2-dq
    C2 final outer plate charge is Q2+dq
    C1 final outer plate charge is -Q1-dq

    The connected plates are equipotential.
    C1 and C2 outer plate potential is 0
    C1 and C2 inner plate potential is V

    By definition of capacitance:
    Q1+dq=C1.V
    Q2+dq=C2.V

    Solving:
    dq=(C2.Q1-C1.Q2)/(C1-C2)
    V=(Q1-Q2)/(C1-C2)

    ???????????
     
  7. Oct 15, 2008 #6

    Doc Al

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    You know the total charge on the capacitors, since charge is conserved. And you also know that the charges will rearrange until the voltage across both capacitors is equal, since they are in parallel. Set up those equations and solve.

    You are assuming that the charge will be equal on each capacitor, but that's not true: the voltage will be equal, not the charge.
     
  8. Oct 15, 2008 #7
    OK I can see by Kirchoff's loop that [itex]V_1+V_2=0 [/itex] which leads to [itex] \frac{q_1}{C_1}+\frac{q_2}{C_2}=0 [/itex], and this gives [itex] q_1=-\frac{C_1}{C_2}q_2 [/itex].

    Then If I use charge conservation. We know the initial charge on both caps(i.e. will denote initial charges by capital Q's, and final charges by lowercase as above) [itex] Q_1=10 [/itex] and [itex]Q_2=5 [/itex]. Whatever we take off one capacitor dq must end up on the other capacitor or vice versa. So this leads me to:
    [itex] Q_1-dq=q_1 [/itex] and [itex] Q_2+dq=q_2 [/itex], adding these you get [itex] Q_1+Q_2=q_1+q_2 [/itex]. Adding initial charges [itex] Q_1+Q_2=10+5=15 [/itex]. Combining this with the first relation from Kirchoff:

    [itex] q_2(1-\frac{C_1}{C_2})=15 [/itex] which gives q_2=30, and a voltage of 15V accross the second cap which is clearly wrong.??
     
  9. Oct 15, 2008 #8
    Maybe I should be looking at the opposite plates of each capacitor since theyre the ones charge is actually moving between giving [itex] Q_1+Q_2=10-5=5 [/itex] but even then is comes out wrong
     
  10. Oct 15, 2008 #9
    You're not using Kirchoff's law correctly.
    If the caps. are in series as you say, it should be V1+V2+V = 0 so that V1+V2 = -V. You now understand?
    Also use charge conservation laws: charge doesn't accumulate in a loop.
     
  11. Oct 15, 2008 #10

    met4t you are wrong, the formula V1+V2+V=0 would be correct IF the battery was still present in series with the 2 caps, but it isnt, it's just the two caps connected in a loop together, the battery is gone.

    From Kirchoffs loop you would get [itex] \frac{q_1}{C_1}+\frac{q_2}{C_2}=0 [/itex] leading to [itex] q_1=- \frac{C_1}{C_2} q_2 [/itex]. So like Doc Al commented the voltage accross each cap has equal magnitude.
    See post number 7. I already used charge cons to deduce [itex] Q_1+Q_2=q_1+q_2 [/itex]
     
  12. Oct 15, 2008 #11
    Oops, then u should have 7.5 uC on each capacitor since charge doesn't accumulate in a loop.
    Divide by capacitances to get voltages on each capacitor. That would be 7.5 V and 3.75 V for 1 uF and 2 uF respectively.
     
  13. Oct 15, 2008 #12

    Doc Al

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    Here's how it works.

    You have the original charges on the capacitors: q1 = 5 μC and q2 = 10 μC.

    When you reconnect them, you'll be putting + and - charges together, so the total charge on the reconnected capacitors will be 10 - 5 = 5 μC. (One side is positive, the other negative, of course.)

    Thus the final charges must satisfy:
    Q1 + Q2 = 5 μC

    When the capacitors are reconnected, they will have the same voltage, so:
    V = Q1/C1 = Q2/C2

    Combine those two equations to solve for the charges and the voltage.
     
  14. Oct 15, 2008 #13

    Doc Al

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    This is where you're getting yourself mixed up.

    q1 + q2 = 10 μC + 5 μC = 15 μC

    But:

    Q1 + Q2 = 10 μC - 5 μC = 5 μC
     
  15. Oct 15, 2008 #14
    What you have posted is definitely correct, and gives the correct answers I have here (this is GRE problem btw)

    What I don't understand is why Kirchoffs loop isn't working however. If I do the Kirchoff procedure, and assume a direction for the current then go round the loop in this direction, I will go from - to + on cap 1, so add V_1, then(since theyre connected with opp ends together) I'll go from - to + on cap 2 aswell, so add V_2. This is implies V_1+V_2=0 .....and it gives me V_1=-V_2.

    So why is Kirchoff telling me the voltages are equal and opp instead of just equal?
     
  16. Oct 15, 2008 #15

    atyy

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    Would you mind posting the numerical answer? Thanks!
     
  17. Oct 15, 2008 #16
    The correct answer comes if you use q_1+q_2=5, and assume voltage is equal so that q_1/C_1=q_2/C_2 which implies q_1=C_1/C_2*q_2. Subbing the latter expression into the former: q_2(1+C_1/C_2)=5 ----->q_2(3/2)=5----->q_2=3.33, and the voltage accross is at 1.66V. Charge on cap 1 is then 1/2 charge on cap 2, and one finds it also has voltage 1.66V.

    I can understand this solution now by different means, thinking of the capacitors as splitting the circuit into two sep wires, each wire must be an equipotential line (since they are connected) (see my original diagram). Similarly for the other wire, this obviously leads to the conclusion that the voltage accross each cap must be eqaul. Also the overall charge on each segment of wire is 10-5=+5, this charge has nowhere to go so no matter how charge distribution occurs the sum of the final charges q1+q2=5. These two facts allow correct solution, that I'm happy with.

    I just don't understand why the way I initially was trying to do it wasn't working, doesn't Kirchoff apply here, why wont his loop rule tell me the caps have equal voltages, instead of equal and opp voltages?
     
  18. Oct 15, 2008 #17

    Doc Al

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    Because you are looking at the voltage from the opposite direction. Let's label the ends of the capacitors so we are clear.

    C1: A(+) and B(-) [q1 = 5 μC]
    C2: C(+) and D(-) [q2 = 10 μC]

    Note: I'm showing the signs they have before they are connected; once they are connected, the charges will rearrange so that B&C have positive charges and A&D have negative charges.

    Around the loop we'll have: V(AB) + V(CD) = 0
    So: V(AB) = - V(CD)

    This should make sense since AB = (-)(+) while CD = (+)(-)

    But that's the same as saying V(AB) = V(DC), which is how I look at it. (Realize that when you have two things connected, you have a choice of viewing them as in series or in parallel. :wink:)

    Make sense?
     
  19. Oct 15, 2008 #18
    Yep makes sense,thanks, I think the key is initially it's -+-+ but finally it's -+ +- so when you go around the loop from left to right there's a voltage increase then a voltage drop over cap 2. So with correct signs V_1-V_2=0.

    Although without actually using Kirchoffs law implicitley its hard to actually justify the second cap flipping signs. Since all we have is charge cons q1+q5=5 in the middle so without knowing already that the voltage must be equal on each plate, the charge could in principle redistribute any old way. So I guess although Kirchoff does apply at every instance, you need something extra to Kirchoff to actually be able to correctly apply his rule at the final time. The only way I can think of doing this is by the argument about the two sep wires being equipotentials which leads naturally to the voltage across each cap being equal. Then you know this and charge con, you can see the cap flips because of this equal voltages rule, and then it allows you to apply Kirchoff with the correct signs.

    Kirchoff is like an afterthought to this problem, rather than a tool I think
     
  20. Oct 15, 2008 #19

    atyy

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    h0dgey84bc and Doc Al, thanks for the answers!

    In my attempted solution (#5), I used Kirchoff's current law by making the change in charge on adjacent plates equal and opposite. Then I used the equilibrium condition by making the "outer" plates have potential 0, and the "inner" plates have potential V, so the potential drop across the capacitors is equal and opposite. Equilibrium is immediate without a resistor between the capacitors.

    I think my mistake in the above equations was that both LHS are the charge on the "positive" plate, but RHS V is in one case the potential difference going from the negative to positive plate, and in the other case from the positive to negative plate. So if I fudge:

    By definition of capacitance:
    Q1+dq=C1.V
    Q2+dq=-C2.V

    V=(Q1-Q2)/(C1+C2)=(5-10)/(1+2)=-5/3=-1.66
    dq=(-C1.Q2-C2.Q1)/(C1+C2)=(-1.10-2.5)/3=-20/3
    Charge on C1 inner plates is 5-20/3=-1.66
    Charge on C2 inner plates is -10+20/3=-3.33

    So the answers are numerically right, but I am not sure if the fudge is justified.
     
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