Question

A parallel plate capacitor is charged to $60\mu C$. Due to a radioactive source, the plate loses charge at the rate of $1.8\times {10}^{-8}C/s$. The magnitude of displacement current is

• A. $1.8\times {10}^{-8}C/s$
• B. $3.6\times {10}^{-8}C/s$
• C. $4.1\times {10}^{-11}C/s$
• D. $5.7\times {10}^{-12}C/s$

Answer 1

The correct option is A

$1.8\times {10}^{-8}C/s$

Step 1: Given Data

Charge, $q=60\mu C$

Rate of charge loss, $\frac{dq}{dt}=1.8\times {10}^{-8}C/s$

Step 2: Find displacement current

Displacement current is given by

$$\begin{gathered} I=\frac{dq}{dt} \\ I=1.8\times {10}^{-8}C/s \end{gathered}$$

Hence, option A is correct.