Q. A parallel plate capacitor is charged to$60\mu C$. Due to a radioactive source, the plate loss charge at the rate of $1.8\ \times \ 10^{- 8}\ {Cs}^{- 1}.$ The magnitude of displacement current is
A) $1.8 \times {10}^{- 8}\ Cs^{- 1}$
B) $3.6 \times 10^{- 8}{Cs}^{- 1}$
C) $4.1 \times 10^{- 11}{Cs}^{- 1}$
D) $5.7 \times 10^{- 12}\ {Cs}^{- 1}$
Explanation
(a) Displacement current is given by
$I_{d} = \frac{dq}{dt}\ = \ 1.8\ \times \ 10^{- 8}\ {Cs}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.