A parallel-plate air-filled capacitor has plate area of 10-4 m2 and plate separation of 10-3 m. It is connected to a 0.5V, 4.5 GHz source. The magnitude of the displacement current is (take \(\varepsilon_0=\frac{1}{36 \pi \times 10^9} \mathrm{~F} / \mathrm{m}\) )

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UPSC ESE (Prelims) Electronics and Telecommunication Engineering 19 Feb 2023 Official Paper

Answer 1

The capacitance is,

\(C = \frac{{{\varepsilon _0}A}}{d} = \frac{{10}^{ - 4}}{36 \pi \times 10^9\times{0.001}} = 8.84 \times {10^{ - 13}}\)

The charge on the capacitor is,

\(Q = CV =8.84 \times {10^{ - 13}} \times 0.5 =4.42 \times {10^{ - 13}}\)

Displacement current in one cycle

\(I = \omega Q = 2\pi f Q = 2 \times \pi \times4.5 \times {10^{9}} \times 4.42 \times {10^{-13}} = 12.5\;mA\)