Physics

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A 11-V battery is connected to three capacitors in series. The capacitors have the following capacitances: 5.3 µF, 10.7 µF, and 31.8 µF. Find the voltage across the 31.8-µF capacitor. Please Help!

  • Physics -

    A charge Q comes out of the battery.
    That charge shows up on the top of the first C
    It draws -Q from top of second, leaving + Q on second
    That draws -Q from third, leaving +Q on top of #3
    so
    each capacitor has charge Q
    C = Q/V
    so
    V1 = Q/C1
    V2 = Q/C2
    V3 = Q/C3
    and
    11 = V1+V2+V3
    11 = Q (1/C1 + 1/C2 + 1/C3)
    solve for Q
    then V3 = Q/(31.8*01^-6)

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