Question

A parallel plate capacitor of plate separation $2mm$ is connected in an electric circuit having source voltage $400V$. If the plate area is $60c{m}^3$ then the value of displacement current for time period of ${10}^{-6}sec$ will be

• A. $1.062amp$
• B. $1.062\times {10}^{-2}amp$
• C. $1.062\times {10}^{-1}amp$
• D. $1.602\times {10}^{-2}amp$

Answer 1

The correct option is D $1.602\times {10}^{-2}amp$

We know that

$I_D={\epsilon }_0\frac{d{\varphi }_E}{dt}={\epsilon }_0\frac{EA}{t}={\epsilon }_0\left(\frac{V}{d}\right)\frac{A}{t}$

$=\frac{8.85\times {10}^{-12}\times 400\times 60\times {10}^{-4}}{{10}^{-3}\times {10}^{-6}}=1.602\times {10}^{-2}amp$