A positive charge of 1 µC is taken from points A & B such that VA-VB = 100 V. Then the energy of charge increases by 10-3 J

Question

A positive charge of 1 µC is taken from points A & B such that VA-VB = 100 V. Then the
energy of charge increases by 10-3 J
energy of charge decreases by 10-3 J
energy of charge remains unchanged
energy of charge decreases by 103

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Maris 2 weeks 2021-09-02T05:53:52+00:00 1 Answers 0 views 0

Answers ( )

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    2021-09-02T05:55:33+00:00

    Explanation:

    It is given that,

    Let Charge of 1\ \mu C is taken from points A & B such that V_A-V_B=1000\ V.

    We need to find the energy of charge. Electric potential is defined as the work done per unit of electric charge. So,

    W=(V_B-V_A)q\\\\W=-1000\times 10^{-6}\\\\W=E=-10^{-3}\ J

    So, the energy of charge decreases by 10^{-3}\ J. Hence, the correct option  is (a).

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