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A parallel-plate capacitor with circular plates is being discharged. The radius of the circular plate is $10 \mathrm{~cm}$. A circular loop of radius $20 \mathrm{~cm}$ is concentric with the capacitor and located halfway between the plates. If the electric field between the plates is charging at the rate $3.6 \times 10^{12} \mathrm{~V} /(\mathrm{ms})$, then the displacement current through the loop is (Assume $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2$ )
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Verified Answer
The correct answer is:
$1 \mathrm{~A}$
Electric field between plates of a capacitor is
$E=\frac{V}{d}$
But $V=\frac{Q}{C}$ and $C=\frac{\varepsilon_0 A}{d} \Rightarrow Q=A E \varepsilon_0$ $\ldots(\mathrm{i})$
$\therefore \quad E=\frac{Q d}{\varepsilon_0 A d}=\frac{Q}{A \varepsilon_0}$
Displacement currrent is
$I_d=\frac{d Q}{d t}=\frac{d}{d t}\left(E \cdot A \varepsilon_0\right) \quad[\text { from Eq. (i) }]$
$=\varepsilon_0 A\left(\frac{\Delta E}{\Delta t}\right)=\frac{4 \pi \varepsilon_0 A}{4 \pi} \times\left(\frac{\Delta E}{\Delta t}\right)$
$=\frac{4 \pi \varepsilon_0 \cdot \pi r^2}{4 \pi} \cdot\left(\frac{\Delta E}{\Delta t}\right)$
$\Rightarrow \quad I_d=4 \pi \varepsilon_0\left(\frac{r^2}{4}\right)\left(\frac{\Delta E}{\Delta t}\right)$
Substituting values given, we have
$I_d=\frac{1}{9 \times 10^9}\left(\frac{\left(10 \times 10^{-2}\right)^2}{4}\right) \times 3.6 \times 10^{12}=1 \mathrm{~A}$
$\left[\right.$ Since, $\left.\frac{\Delta E}{\Delta t}=3.6 \times 10^{12} \mathrm{~V} / \mathrm{ms}\right]$
$E=\frac{V}{d}$
But $V=\frac{Q}{C}$ and $C=\frac{\varepsilon_0 A}{d} \Rightarrow Q=A E \varepsilon_0$ $\ldots(\mathrm{i})$
$\therefore \quad E=\frac{Q d}{\varepsilon_0 A d}=\frac{Q}{A \varepsilon_0}$
Displacement currrent is
$I_d=\frac{d Q}{d t}=\frac{d}{d t}\left(E \cdot A \varepsilon_0\right) \quad[\text { from Eq. (i) }]$
$=\varepsilon_0 A\left(\frac{\Delta E}{\Delta t}\right)=\frac{4 \pi \varepsilon_0 A}{4 \pi} \times\left(\frac{\Delta E}{\Delta t}\right)$
$=\frac{4 \pi \varepsilon_0 \cdot \pi r^2}{4 \pi} \cdot\left(\frac{\Delta E}{\Delta t}\right)$
$\Rightarrow \quad I_d=4 \pi \varepsilon_0\left(\frac{r^2}{4}\right)\left(\frac{\Delta E}{\Delta t}\right)$
Substituting values given, we have
$I_d=\frac{1}{9 \times 10^9}\left(\frac{\left(10 \times 10^{-2}\right)^2}{4}\right) \times 3.6 \times 10^{12}=1 \mathrm{~A}$
$\left[\right.$ Since, $\left.\frac{\Delta E}{\Delta t}=3.6 \times 10^{12} \mathrm{~V} / \mathrm{ms}\right]$
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