Q.24 A parallel plate capacitor is maintained at potential of 400 V. If area of plates is 40 cm2 and separation between plates is 40 mm, calculate displacement current in 1 µs.
Best answer
Answer: \(I_d \approx 3.54\times10^{-4}\) A
The paper gives V = 400 V, A = 40 cm2 \(= 40\times10^{-4}\) m2, d = 40 mm \(= 40\times10^{-3}\) m and t = 1 µs \(= 10^{-6}\) s. It gives no separate rate of change of voltage, so the question is read as: the plates acquire the potential of 400 V (and the corresponding charge) in 1 µs. The displacement current then equals the rate of change of charge on the plates, \(I_d = \varepsilon_0\dfrac{d\Phi_E}{dt} = \dfrac{dq}{dt} = \dfrac{CV}{t}\).
Capacitance:
\[C = \frac{\varepsilon_0A}{d} = \frac{8.85\times10^{-12}\times40\times10^{-4}}{40\times10^{-3}} = 8.85\times10^{-13}\ \text{F}\]Charge: \(q = CV = 8.85\times10^{-13}\times400 = 3.54\times10^{-10}\) C.
\[I_d = \frac{q}{t} = \frac{3.54\times10^{-10}}{1\times10^{-6}} = 3.54\times10^{-4}\ \text{A}\]Note: once the potential is held steady at 400 V, the electric flux between the plates no longer changes and the displacement current is zero. The value above is the average displacement current while the charge builds up in 1 µs, which is what the question asks for.
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