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Q.

A parallel plate capacitor is charged to 60 $μ$ C. Due to a radioactive source, the plate loses charge at the rate of 1.8 X 10^-8 Cs^-1. The magnitude of displacement current is

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a

$3.6 \times 10^{- 8} {Cs}^{- 1}$

b

$1.8 \times 10^{- 8} {Cs}^{- 1}$

c

$4.1 \times 10^{- 11} {Cs}^{- 1}$

d

$5.7 \times 10^{- 12} {Cs}^{- 1}$

answer is A.

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Detailed Solution

Displacement current is given by

$I_{d} = \frac{d q}{d t} = 1.8 \times 10^{- 8} {Cs}^{- 1}$

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