A parallel plate capacitor with circular plat | vedclass

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Question

(A) The total displacement current $I_d$ between the plates is equal to the conduction current $i$ in the wires,so $I_d = i = \varepsilon_0 A \frac{dE

Vedclass · Accepted Answer

$\frac{3}{4} i$

Vedclass · Answer

(A) The total displacement current $I_d$ between the plates is equal to the conduction current $i$ in the wires,so $I_d = i = \varepsilon_0 A \frac{dE}{dt} = \varepsilon_0 (\pi R^2) \frac{dE}{dt}$.The displacement current density $J_d$ is uniform across the plates,given by $J_d = \frac{I_d}{A} = \frac{i}{\pi R^2}$.The displacement current $I_d'$ in the annular region between $r = \frac{R}{2}$ and $r = R$ is the integral of the current density over that area:$I_d' = J_d 	imes A' = \frac{i}{\pi R^2} 	imes [\pi R^2 - \pi (\frac{R}{2})^2]$$I_d' = \frac{i}{\pi R^2} 	imes [\pi R^2 - \frac{\pi R^2}{4}]$$I_d' = \frac{i}{\pi R^2} 	imes [\frac{3}{4} \pi R^2] = \frac{3}{4} i$.

$A$ parallel plate capacitor with circular plates of radius $R$ is being charged as shown. At the instant shown,the displacement current in the region between the plates enclosed between $\frac{R}{2}$ and $R$ is given by

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