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A parallel plate capacitor has plate area of $1 c m^{2}$ and plate separation 0.5 mm. When the capacitor is being charged, the electric field in the space between the plates is increasing at the rate of $10^{10} V / m - s$ . Then the displacement current at that instant is

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4.85 $μA$

12.15 $μA$

6.25 $μA$

8.85 $μA$

answer is D.

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Detailed Solution

$i_{d} = \in_{0} A \frac{dE}{dt} = 8 .85 \times 10^{- 12} \times 10^{- 4} \times 10^{10} A = 8 .85 μA$

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