A capacitor is charged until its stored energy is 4.00 J. A second capacitor is then connected to it in parallel. (a) If the charge distribu

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A capacitor is charged until its stored energy is 4.00 J. A second capacitor is then connected to it in parallel. (a) If the charge distributes equally, what is the total energy stored in the electric fields

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Verity 4 years 2021-08-28T23:46:45+00:00 1 Answers 27 views 0

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  1. kimcuc
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    2021-08-28T23:48:26+00:00
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    Explanation:

    (a)   It is known that energy present in electric field is the sum of energies of the capacitors. Since, there are only two capacitors so net energy of the capacitor will be as follows.

            U = U_{1} + U_{2}

                = \frac{q^{2}_{1}}{2C_{1}} + \frac{q^{2}_{2}}{2C_{2}}

    As the charge is equally distributed, so q_{1} = q_{2} = \frac{q_{o}}{2}

    Also, potential difference of both of them is equal.

    So,       V_{1} = V_{2}

    or,       \frac{q_{1}}{C_{1}} = \frac{q_{2}}{C_{2}}

             \frac{q_{o}}{2C_{1}} = \frac{q_{o}}{2C_{2}}

    as      C_{1} = C_{2} = C

                  U = \frac{q^{2}_{o}}{8C} + \frac{q^{2}_{o}}{8C}

                     = 2 J

    Thus, we can conclude that if the charge distributes equally, then total energy stored in the electric fields is 2 J.

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