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27 January, 17:56

It takes 101 J of work to move 2 C of charge from the negative plate to the positive plate of a parallel plate capacitor. What voltage difference exists between the plates

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  1. 27 January, 18:37
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    10.05volts

    Explanation:

    Work done to move a charge from the positive side of a plate to the negative side is represented mathematically as 1/2CV² which is also the energy stored by the capacitor.

    Work done = 1/2CV² where

    C is the charge

    V is voltage difference that exists between the plates.

    Given Work done = 101Joules

    C = 2Coulombs

    Substituting the values in the formula to get V, we will have;

    101 = 1/2 (2) V²

    101 = V²

    V = √101

    V = 10.05volts

    Therefore the voltage difference between the plates is 10.05colt's
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