Question

A parallel plate capacitor having plate area $A$ and plate separation $d$ is joined to a battery of emf $V$ and internal resistance $R$ at $t=0$. Consider a plane surface of area $A/2$, parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time. The charge on the capacitor at time $t$ is given by $q=CV(1-e^{-t/\tau }),$ where $\tau =CR$.

• A. $\frac{V}{R}{e}^{-\frac{td}{{ϵ}_{0}AR}}$
• B. $\frac{V}{2R}{e}^{-\frac{td}{{ϵ}_{0}AR}}$
• C. $\frac{V}{R}{e}^{\frac{td}{{ϵ}_{0}AR}}$
• D. $\frac{V}{2R}{e}^{\frac{td}{{ϵ}_{0}AR}}$

Answer 1

The correct option is B $\frac{V}{2R}{e}^{-\frac{td}{{ϵ}_{0}AR}}$

Given:

$q=CV(1-e^{-t/\tau }),$ where $\tau =CR$

$\therefore$ Surface charge density, $\sigma =\frac{q}{A}=\frac{CV}{A}(1-e^{-t/\tau })$

Electric field between the plates of capacitor,

$E=\frac{\sigma }{{ϵ}_0}=\frac{CV}{{ϵ}_{0}A}(1-e^{-t/\tau })$

Electric flux from the given area,

${\varphi }_E=\frac{EA}{2}=\frac{CV}{2{ϵ}_0}(1-e^{-t/\tau })$

Displacement current, $i_d={ϵ}_0\frac{d{\varphi }_E}{dt}$

$\Rightarrow i_d={ϵ}_0\frac{d}{dt}\left[\frac{CV}{2{ϵ}_0}\right(1-e^{-t/\tau }\left)\right]=\frac{CV}{2\tau }{e}^{-t/\tau }$

Substituting, $\tau =CR$

$\Rightarrow i_d=\frac{V}{2R}{e}^{-t/CR}$

Again substituting, $C=\frac{{ϵ}_{0}A}{d}$

$i_d=\frac{V}{2R}{e}^{-\frac{td}{{ϵ}_{0}AR}}$

Hence, option $\left(B\right)$ is the correct answer.