Q. The charge of a parallel plate capacitor is varying as $q = q_{0}\sin{2\pi ft}$. The plates are very large and close together (area$= A$, separation$= d$). Neglecting edge effects, the displacement current through the capacitor is
A) $\frac{d}{A\varepsilon_{0}}$
B) $\frac{d}{\varepsilon_{0}}\sin{2\pi ft}$
C) $2\pi fq_{0}\cos{2\pi ft}$
D) $\frac{2\pi fq_{0}}{\varepsilon_{0}}\cos{2\pi}ft$
Explanation
(c) $i = \frac{dq}{dt} = \frac{d}{dt}\left( q_{0}\sin{2\pi ft} \right) = q_{0}2\pi f\cos{2\pi ft}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.