Question

A parallel-plate capacitor having plate area $A$ and plate separation $d$ is joined to a battery of emf $E$ and internal resistance $R$ at $t=0$. Consider a plane surface of area $A/2$, parallel to the plates and situated symmetrically between them. If the displacement current through this surface as a function of time is given as:
$i_d=\frac{E}{nR}{e}^{-\frac{td}{{\epsilon }_{0}AR}}$
Where, ${\epsilon }_0$ is the permittivity of free space.

The value of $n$ is :

Answer 1

Let the charge on the capacitor at time $t$ is $Q$.

Then, the electric field between the plates of the capacitor is

$E=\frac{Q}{{\epsilon }_{0}A}$

The flux through the area considered is

${\varphi }_E=\frac{Q}{{\epsilon }_{0}A}\times \frac{A}{2}=\frac{Q}{2{\epsilon }_0}$

Therefore, the displacement current flowing through area $\frac{A}{2}$ is

$i_d={\epsilon }_0\frac{d{\varphi }_E}{dt}$

$i_d={\epsilon }_0\times \left(\frac{1}{2{\epsilon }_0}\right)\frac{dQ}{dt}=\frac{1}{2}\frac{dQ}{dt}$

We know that when a capacitor is connected to a $\text{DC}$ voltage, the charge on the plates of the capacitor at any instant is given by

$Q=EC[1-e^{-t/RC}]$

Here, $C=$ Capacitance of the capacitor

Therefore,
$i_d=\frac{1}{2}\frac{d\left[EC\right(1-e^{-t/RC}\left)\right]}{dt}$

$i_d=\frac{1}{2}\times EC\times -\left(\frac{-1}{RC}\right)\times e^{-t/RC}$

$\therefore i_d=\frac{E}{2R}{e}^{-t/RC}$

Here, capacitance is given by

$C=\frac{{\epsilon }_{0}A}{d}$

$\therefore i_d=\frac{E}{2R}{e}^{-\frac{td}{{\epsilon }_{0}AR}}$

Hence, $n=2$.