Class 12 NEET Physics Electromagnetic Waves Questions

Given, $V = V_{0}\sin\omega t$ ......(1)The displacement current flowing between the plates of the capacitor is$I_{d} = \varepsilon_{0}\dfrac{d\Phi_{E}}{dt}$For a capacitor this equals the conduction (charging) current, so$I_{d} = C\dfrac{dV}{dt}$$= C\dfrac{d}{dt}\left(V_{0}\sin\omega t\right)$ (using equation 1)$= C\,V_{0}\,\omega\cos\omega t$$I_{d} = V_{0}\,\omega C\cos\omega t$Hence the displacement current leads the applied voltage by $\pi/2$ and has amplitude $V_{0}\omega C$, i.e. $V_{0}/X_{C}$ with $X_{C}=\dfrac{1}{\omega C}$.So the option $I_{d}=V_{0}\,\omega C\cos\omega t$ (the symbol printed as $\varepsilon$ in the option is the angular frequency $\omega$, a typographical artifact) is the correct choice, i.e. option A.Options of the form $I_{d}=\dfrac{V_{0}}{\omega C}\cos\omega t$ or $I_{d}=\dfrac{V_{0}}{\omega C}\sin\omega t$ are dimensionally and physically wrong, and $I_{d}=V_{0}\,\omega C\sin\omega t$ is wrong because the current is proportional to $\dfrac{dV}{dt}$, which is a cosine, not a sine.