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Capacitor problem

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  1. Jan 30, 2016 #1
    1. The problem statement, all variables and given/known data
    A capacitor of capacitance 5µF is charged to 24 V and another capacitor of capacitance 6µF is charged to 12 V. The positive plate of first capacitor is now connected to negative plate of second and vice versa.Fid ew charges on capacitor.

    2. Relevant equations
    Q=CV

    E= QV/2 = CV^2 / 2

    3. The attempt at a solution
    After the capacitors are connected , finally potential difference between both the capacitors would be same. Hence final charges would be in the ratio of their capacitances. Hence it must be 960/11 µC and 1152/11 µC.
    But the answer is given as 21.8µC and 26.2µC
     
  2. jcsd
  3. Jan 30, 2016 #2

    gneill

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    Staff: Mentor

    How did you determine the charge that remains after the capacitors are connected together? Did the plates that were connected have the same sign of charge?
     
  4. Jan 30, 2016 #3

    cnh1995

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    Homework Helper

    Right. This will give you one equation in terms of Q1 and Q2. Find another equation in Q1 and Q2 and solve them simultaneously.
     
  5. Jan 30, 2016 #4
    No
     
  6. Jan 30, 2016 #5

    cnh1995

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    Homework Helper

    Whatever charge leaves the +ve plate of 5μF is trapped on the -ve plate of 6μF capacitor(assuming conventional current). Can you write the second equation from this?
     
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