Q. A capacitor having a capacity of 2 pF. Electric field across the capacitor is changing with a value of $10^{12}\ Vs^{- 1}$. The displacement current is
A) 2 A
B) 4 A
C) 6 A
D) 10 A
Explanation
(a) $i = \frac{dQ}{dt} = \frac{d}{dt}(CV)$
$= C\frac{dV}{dt} = 2 \times 10^{- 12} \times 10^{12} = 2\ A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.