Question

A parallel-plate capacitor with circular plates is being discharged. The radius of the circular plate is $10 \,cm$. A circular loop of radius $20 \,cm$ 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 $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \,Nm ^{2} / C ^{2}$ )

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

Answer 1

Answer: (A)

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