Physics

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The drawing shows two fully charged capacitors (C1 = 7 µF, q1 = 4 µC; C2 = 16 µF, q2 = 4 µC). The switch is closed, and charge flows until equilibrium is reestablished (i.e., until both capacitors have the same voltage across their plates). Find the resulting voltage across either capacitor

  • Physics -

    you have 8microC total charge.

    You have 23microF capacitance.

    Voltage= Qtotal/C=8/23 volts.

    on C1, q=C1*8/23=7micro*8/23=56/23 micro F
    on C2, q=16micro*8/23=128/23 microC

    adding q1+q2, 184/23 microF=8microF

  • Physics -

    Depending on which way they are wired:assume they are wired positive plate to negative plate of each other.

    Then Qeff=16*4-7*4=36uC.(If wired positive plate to positive plate then charges will add up)

    This charge will get redistributed.

    Applying conservation of charge:

    Qeff=(C1+C2)Vfinal

    Vfinal=1.56v

  • Physics -

    The first answer is correct...final voltage will be: Qtotal/(C1+C2)=0.347

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