$AC$ voltage $V\left(t\right)=20 \sin \omega t$ of frequency $50 \mathrm{Hz}$ is applied to a parallel plate capacitor. The separation between the plates is $2 \mathrm{mm}$ and the area is $1{ \mathrm{m}}^2$. The amplitude of the oscillating displacement current for the applied $AC$ voltage is $\left[\text{Take }{\mathrm{\epsilon }}_0=8.85\times {10}^{-12} \mathrm{F} {\mathrm{m}}^{-1}\right]$

• A

  $21.14 \text{μA}$
• B

  $83.37 \text{μA}$
• Correct answer
  C

  $27.79 \text{μA}$
• D

  $55.58 \text{μA}$

Hence, the correct answer is option: 'C'.