A parallel plate capacitor of plate separation 2 mm is connected in an electric circuit having source voltage 400 V. If the plate area is 60 cm², then the value of displacement current for 10^-6 sec will be

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1.062 amp

1.062 × 10^-3 amp

1.062 × 10^-4 amp

1.062 × 10^-2 amp

answer is B.

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Detailed Solution

$\begin{array}{rcl} i_{d} & = & ε_{0} \frac{dϕ}{dt} \\ = & ε_{0} \frac{EA}{t} \\ = & ε_{0} \frac{VA}{dt} \\ = & \frac{8.85 x 10^{- 12} x 400 x 60 x 10^{- 4}}{2 x 10^{- 3} x 10^{- 6}} \\ = & 1.062 x 10^{- 2} A \end{array}$

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