\(AC\) 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 \(1 \,{m}^{2}\). The amplitude of the oscillating displacement current for the applied \(AC\) voltage is ...... \(\mu {A}\). [Take \(\left.\varepsilon_{0}=8.85 \times 10^{-12} \,{F} / {m}\right]\)

\({C}=\frac{\epsilon_{0} {A}}{{d}}={k} \frac{\epsilon_{0} \times 1}{2 \times 10^{-3}} {F}\)

\(\therefore {X}_{{c}}=\frac{1}{\omega {C}}=\frac{2 \times 10^{-3}}{2 \times 50 \pi {\epsilon}_{0}}=\frac{2 \times 10^{-3}}{25 \times 4 \pi \epsilon_{0}} \Omega\)

\(\therefore {X}_{{c}}=\frac{2 \times 10^{-3}}{25} \times 9 \times 10^{9}=\frac{18}{25} \times 10^{6} \Omega\)

\(\therefore {i}_{0}=\frac{{V}_{0}}{{X}_{{c}}}=\frac{20 \times 25}{18} \times 10^{-6} {A}=27.47 \mu {A}\)

The value of amplitude of displacement current will be same as value of amplitude of conventional current.