AC voltage V(t) = 20 sin$\omega$t of frequency 50 Hz is applied to a parallel plate capacitor. The separation between the plates is 2 mm and the area is 1 m². The amplitude of the oscillating displacement current for the applied AC voltage is _________. [Take $\varepsilon$₀ = 8.85 $\times$ 10^$-$12 F/m]

Given,<br/><br/>AC voltage, V(t) = 20 sin $\omega$t volt.<br/><br/>Frequency, f = 50Hz<br/><br/>Separation between the plates, d = 2 mm = 2 $\times$ 10^$-$3 m<br/><br/>Area, A = 1 m²<br/><br/>As, $C = {{{\varepsilon _0}A} \over d}$<br/><br/>where, ${{\varepsilon _0}}$ = absolute electrical permittivity of free space = 8.854 $\times$ 10^$-$12 N^$-$1 kg²m^$-$2<br/><br/>$C = {{{\varepsilon _0} \times 1} \over {2 \times {{10}^{ - 3}}}}$ .... (i)<br/><br/>Capacitive reactance $({X_C}) = {1 \over {\omega C}}$ .... (ii)<br/><br/>From Eqs. (i) and (ii), we get<br/><br/>$${X_C} = {{2 \times {{10}^{ - 3}}} \over {2 \times 50\pi \times {\varepsilon _0}}}$$ ($\because$ $\omega$ = 2$\pi$f)<br/><br/>$= {{2 \times {{10}^{ - 3}}} \over {25 \times 4\pi {\varepsilon _0}}}$<br/><br/>$\Rightarrow {X_C} = {{2 \times {{10}^{ - 3}}} \over {25}} \times 9 \times {10^9}$<br/><br/>$\Rightarrow {X_C} = {{18} \over {25}} \times {10^6}\,\Omega$<br/><br/>By using Ohm's law,<br/><br/>As, $${I_0} = {{{V_0}} \over {{X_C}}} = {{20 \times 25} \over {18}} \times {10^{ - 6}} = 27.78 \times {10^{ - 6}}$$<br/><br/>$\Rightarrow$ I₀ = 27.78$\mu$A<br/><br/>$\therefore$ The amplitude of the oscillating displacement current for applied AC voltage will be approximately 27.79 $\mu$A.