Q. A parallel plate capacitor of plate separation $2\ mm$ is connected in an electric circuit having source voltage $400\ V$. If the plate area $60cm^{2}$, then the value of displacement current for $10^{- 6}s$ will be
A) $1.062\ amp$
B) $1.062 \times 10^{- 2}\ amp$
C) $1.062 \times 10^{- 3}\ amp$
D) $1.062 \times 10^{- 4}\ amp$
Explanation
(b) $I_{D} = \varepsilon_{0}\frac{d\phi_{E}}{dt} = \varepsilon_{0}\frac{EA}{t} = \varepsilon_{0}\left( \frac{V}{d} \right).\frac{A}{t}$
$= \frac{8.85 \times 10^{- 12} \times 400 \times 60 \times 10^{- 4}}{2 \times 10^{- 3} \times 10^{- 6}} = 1.602 \times 10^{- 2}amp$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.