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# A $30\: pF$ capacitor has circular plates of area $100\: cm^2$. It is charged by a $70\: V$ battery through a $2\Omega$ resistor. At the instant, the battery is disconnected; the electric field between the plates is changing most rapidly. At this instant, find the rate of change of electric field between the plates.

$\begin {array} {1 1} (a)\;39.5 \times 10^{12} \: V/ms & \quad (b)\;3.95 \times 10^{12} \: V/ms \\ (c)\;39.5 \times 10^{14} \: V/ms & \quad (d)\;3.95 \times 10^{14} \: V/ms \end {array}$

$E = 70\: V, \: C = 30\: pF = 30 \times 10^{-12} pF, R = 2\Omega,\: A = 100 cm^2 = 10^{-2}m^2$
For $RC$ circuit, $q = CE(1-e^{\large\frac{-1}{RC}} )$
$\large\frac{dq}{dt}$$= \large\frac{CE}{RC}$$ \times e^{\large\frac{-1}{RC}} = \large\frac{E}{R}$$\times e^{\large\frac{-1}{RC}} At, t=0, \large\frac{dq}{dt}$$ = \large\frac{E}{R}$$= \large\frac{70}{2}$$ A = 35A$
i.e, $I_D = 35A$
$I_D = \in_oA \times \large\frac{dE}{dt}$
$\Rightarrow \large\frac{dE}{dt} $$= \large\frac{I_D}{\in_oA} \Rightarrow \large\frac{dE}{dt}$$ \large\frac{35}{(8.85 \times 10^{-12} \times 10^{-2} )} V/ms$
$= 3.95 \times 10^{14} \: V/ms$
Ans : (d)