Question

A parallel-plate capacitor with circular plates is being discharged. The radius of the circular plate is $10cm$. A circular loop of radius $20cm$ is concentric with the capacitor and located halfway between the plates. If the electric field between the plates is charging at the rate $3.6\times 10^{12}V/(ms)$, then the displacement current through the loop is:
(Assume $\frac{1}{4\pi {\epsilon }_0}=9\times 10^{9}N{m}^2/C^2$ )

• A. $1A$
• B. $2A$
• C. $3A$
• D. $4A$

Answer 1

Answer: (A)

Electric field between plates of a capacitor is
$E=\frac{V}{d}$
But $V=\frac{Q}{C}$ and $C=\frac{{\epsilon }_{0}A}{d}$
$\Rightarrow Q=AE{\epsilon }_0$
$\therefore E=\frac{Qd}{{\epsilon }_{0}Ad}=\frac{Q}{A{\epsilon }_0}$
Displacement currrentis
$I_d=\frac{dQ}{dt}=\frac{d}{dt}\left(E\cdot A{\epsilon }_0\right)$
[from Eq.]
$={\epsilon }_{0}A\left(\frac{\Delta E}{\Delta t}\right)=\frac{4\pi {\epsilon }_{0}A}{4\pi }\times \left(\frac{\Delta E}{\Delta t}\right)$
$=\frac{4\pi {\epsilon }_0\cdot \pi r^2}{4\pi }\cdot \left(\frac{\Delta E}{\Delta t}\right)$
$\Rightarrow I_d=4\pi {\epsilon }_0\left(\frac{r^2}{4}\right)\left(\frac{\Delta E}{\Delta t}\right)$
Substituting values given, we have
$I_d=\frac{1}{9\times 10^9}\left(\frac{{\left(10\times 10^{-2}\right)}^2}{4}\right)\times 36\times 10^{12}=1A$
$\left[\text{ Since, }\frac{\Delta E}{\Delta t}=3.6\times 10^{12}V/ms\right]$