JEE Main · PhysicsMedium

$A parallel plate capacitor with circular plates of radius R = 0.1 m is being charged. The displacement current between the plates is I_{d} = 2 A . What is the rate of change of the electric field d E / d t between the plates?$

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Correct answer: A — $7.2 \times 1 0^{12} V/(m s)$

Solution

$The displacement current is I_{d} = ϵ_{0} \frac{d Φ _{E}}{d t} = ϵ_{0} A \frac{d E}{d t} .$
$The area of the plates is A = π R^{2} = π (0.1)^{2} = 0.01 π m^{2} .$
$Rearranging for the rate of change: \frac{d E}{d t} = \frac{I _{d}}{ϵ _{0} A} .$
$Using ϵ_{0} \approx 8.85 \times 1 0^{- 12} F/m, \frac{d E}{d t} = \frac{2}{8.85 \times 1 0 ^{- 12} \times 0.01 π} \approx 7.2 \times 1 0^{12} V/(m s) .$
$Hence the answer is (A).$