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3 years ago

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The voltage across a parallel-plate capacitor with area A = 820 cm2 and separation d = 5 mm varies sinusoidally as V = (14 mV)co

s(170t), where t is in seconds. Find the displacement current between the plates. (Use the following as necessary: t. Do not use other variables, substitute numeric values. Assume that Id is in amperes. Do not include units in your answer.)

1 answer:

shutvik [7]3 years ago

6 0

Answer:

$I_d = -3.454*10^{-10} \ \ (sin ( \ 170 \ t))$

Explanation:

The displacement current $I_d$ is given by the expression;

$I_d = \epsilon _o \frac{d \phi _ E }{dt}$

where

$\phi _ E = A.E \\ \\ \\\phi _E = A. \frac{V}{d} \\ \\ \\\phi _E = \frac{A}{d}(14*10^{-3})(cos \ 170 \ t ) \\ \\ \\\phi _E = \frac{820*10^{-4}}{5*10^{-3}} * (14*10^{-3})(cos \ 170 \ t ) \\ \\ \\\phi _E = 2296*10^{-4} (cos \ 170 \ t)$

$I_d = \epsilon _o \frac{d \phi _ E }{dt}$

$I_d = 8.85*10^{-12}(2296*10^{-4})(-170)(sin(170 \ t))$

$I_d = -3.454*10^{-10} \ \ (sin ( \ 170 \ t))$

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