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A C1= 2.50uF capacitor is charged to 862 V and a C2= 6.80uF capacitor is charged to 660 V . These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other.
What will be the potential difference across each? [Hint: charge is conserved.]
What will be the charge on each?

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Steve

  1. C = Q/V (That is the meaning of C)

    how much charge on 2.5*10^-6 at 862
    Q1 = C V = 2.5*10^-6 * 862 = 2155*10^-6 coulombs

    how much charge on 6.8*10^-6 at 660 ?
    Q2 = 6.8*10^-6 * 660 = 4488*10^-6

    Total charge = Q1+Q2 = 6643*10^-6
    Total Capacitance = (2.5+6.8)10^-6 = 9.30*10^-6 Farads

    V = same on each now = 6643/9.30 = 714.3 volts

    now just do
    Q1 = C1 (714)
    Q2 = C2 (714
    for the charge on each

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    Damon

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