Question

To establish an instantaneous displacement current of magnitude $i\text{A}$ in between the plates of a parallel plate capacitor of capacitance $\frac{1}{2}\text{F}$, the value of $\frac{dV}{dt}$ is

• A. $2i$
• B. $\frac{i}{2}$
• C. $\frac{1}{2i}$
• D. $i$

Answer 1

The correct option is A $2i$

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

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

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

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

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

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

Here, $i_d=i,C=\frac{1}{2}\text{F}$

$\Rightarrow i=\frac{1}{2}\frac{dV}{dt}$

$\therefore \frac{dV}{dt}=2i$

Hence, option $\left(A\right)$ is correct.