Question

The rate of change of potential difference across capacitor is $5\times {10}^2\text{V/s}$. If capacitance is $2\mu \text{F}$, then displacement current between its plates is

• A. ${10}^{-5}\text{A}$
• B. $2\times {10}^{-3}\text{A}$
• C. ${10}^{-3}\text{A}$
• D. $2.5\times {10}^{-3}\text{A}$

Answer 1

The correct option is C ${10}^{-3}\text{A}$

Let the area of the plates of the capacitor is $A$ and distance between the plate is $d$.

The displacement current flowing through the capacitor is given by

$i_d={\epsilon }_0\frac{d{\varphi }_E}{dt}={\epsilon }_0\frac{d\left(EA\right)}{dt}$

Here, $E$ is the electric field between the plates of the capacitor.

$i_d={\epsilon }_{0}A\frac{dE}{dt}=\frac{{\epsilon }_{0}A}{d}\times d\frac{dE}{dt}$

$i_d=Cd\frac{dE}{dt}[\because \text{capacitive C}=\frac{{\epsilon }_{0}A}{d}]$

If the potential difference between the plates is $V$; then

$i_d=Cd\frac{d}{dt}\left(\frac{V}{d}\right)[\because E=\frac{V}{d}]$

$i_d=Cd\times \frac{1}{d}\times \frac{dv}{dt}$

$i_d=C\frac{dV}{dt}$

Given: $\frac{dV}{dt}=5\times {10}^2\text{V/s}$

$C=2\mu \text{F}=2\times {10}^{-6}\text{F}$

$i_d=2\times {10}^{-6}\times 5\times {10}^2=10\times {10}^{-4}$

$\therefore i_d={10}^{-3}\text{A}$

Hence, option $\left(\text{C}\right)$ is correct.