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21 July, 20:27

An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 6.0 μC flows to the positive plate. The 4.0 V battery is then disconnected and replaced with a 7.0 V battery, with the positive and negative terminals connected in the same manner as before.

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  1. 21 July, 20:36
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    1 μC extra charge will be flow here

    Explanation:

    Given data

    battery V1 = 4.0 V

    flows Q1 = 6.0 μC

    replace battery V2 = 7.0 V

    to find out

    what happen if we replace battery

    solution

    we apply here principal of capacitor

    that is Q directly proportional voltage

    so we say Q2/Q1 = V2 / V1

    put all value here

    Q2/Q1 = V2 / V1

    Q2/6 = 7 / 6

    Q2 = 7

    so we see here 7 μC will be flow

    and Q = Q2 - Q1 = 7 - 6 = 1 μC

    so we also say that 1 μC extra charge will be flow here
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