A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find the time ra

Question

A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find the time rate of change of electric flux between the plates.

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Xavia 6 months 2021-08-05T21:33:23+00:00 1 Answers 8 views 0

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    2021-08-05T21:34:48+00:00

    Answer:

    The time rate of change of flux is 1.34 \times 10^{10} \frac{V}{s}

    Explanation:

    Given :

    Current I = 0.106 A

    Area of plate A = 36 \times 10^{-4} m^{2}

    Plate separation d = 4 \times 10^{-3} m

    (A)

    First find the capacitance of capacitor,

       C = \frac{\epsilon _{o} A }{d}

    Where \epsilon _{o} = 8.85 \times 10^{-12}

       C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4}  }{4 \times 10^{-3} }

       C = 7.9 \times 10^{-12} F

    But   C = \frac{Q}{V}

    Where Q = It

      C = \frac{It}{V}

      V = \frac{It}{C}

    Now differentiate above equation wrt. time,

      \frac{dV}{dt} = \frac{I}{C}

           = \frac{0.106}{7.9 \times 10^{-12} }

           = 1.34 \times 10^{10} \frac{V}{s}

    Therefore, the time rate of change of flux is 1.34 \times 10^{10} \frac{V}{s}

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