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14 July, 12:31

If a 500µF can store 0.02J of energy, what must the voltage supplied to the capacitor be

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  1. 14 July, 13:38
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    The energy stored on a capacitor can be expressed in terms of the work done by the battery. Voltage represents energy per unit charge, so the work to move a charge element dq from the negative plate to the positive plate is equal to V dq, where V is the voltage on the capacitor.

    U = 0.5CV^2

    0.02 = 0.5 (500x10^-6) V^2

    Solve for V

    V = 8.94 volts
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