If the voltage across the plates is increasing at the rate of 220 V/s, what is the displacement current in the capacitor

Question

If the voltage across the plates is increasing at the rate of 220 V/s, what is the displacement current in the capacitor

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Kiệt Gia 6 months 2021-07-31T12:41:57+00:00 1 Answers 19 views 0

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    2021-07-31T12:43:46+00:00

    Complete Question

    A parallel-plate capacitor has square plates 20 cm on a side and 0.50 cm apart. If the voltage across the plates is increasing at the rate of 220 V/s, what is the displacement current in the capacitor?

    Answer:

    The value is  I  = 1.416*10^{-8} \  A

    Explanation:

    From the question we are told that

           The  length and breath of the square plate is  l =  b  =  20 \ cm  = 0.2 \ m

            The  distance of separation between each plate is k = 0.50 \ cm  =  0.005 \  m

           The rate of voltage  increase is  \frac{dV}{dt}  =  200 \  V/s

    Generally the charge on the plate is mathematically represented as

      Q =  CV

    Now C is the capacitance of the capacitor which is mathematically represented as

       C = \frac{\epsilon_o *  A}{k}

    Here A is the cross-sectional area which is mathematically represented as

        A = l^2

    =>   A =  0.2^2

    =>   A =  0.04 \  m^2

    So  

         C = \frac{8.85 *10^{-12} *  0.04}{0.005}

              C = 7.08*10^{-11} \  F

    Now the change of the charge flowing through the plates with time is mathematically represented as

            \frac{d Q}{dt}  =  C  \frac{dV}{dt}

    So  

           \frac{d Q}{dt}  =  7.08 *10^{-11} *  200

           \frac{d Q}{dt}  = 1.416*10^{-8}

    Generally   \frac{d Q}{dt}  = \  current \ i.e  \  I

    So

         I  = 1.416*10^{-8} \  A

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