# Figure shows a capacitor made of two circular plates each of radius $12\: cm$, and separated by $5\: cm$. The capacitor is being charged by an external source. The charging current is constant and equal to $0.15\: A$. The rate of change in potential difference between the plates is

$\begin {array} {1 1} (a)\;1.87 \times 10^9 V/s & \quad (b)\;18.7 \times 10^{10} V/s \\ (c)\;18.7 \times 10^9 V/s & \quad (d)\;1.87 \times 10^{10} V/s \end {array}$

Radius of each plate, $r = 12\: cm = 0.12\: m$
Distance between plates, $d = 5\: cm = 0.05\: m$
charging current, $I = 0.15\: A$
$C = \large\frac{ \in_oA}{d}$$=\large\frac{ \in_o \pi r^2 }{d} =\large\frac{ 8.85 \times 10^{-12} \times \pi 0.12^2}{ 0.05} = 80.032\: pF Charge on each plate, q = CV where, V = PD between the plates Differentiation on both sides with respect to time (t) gives: \large\frac{dq}{dt}$$ = \large\frac{CdV}{dt}$
But, $\large\frac{dq}{dt}$ = current (I)
So, $\large\frac{dV}{dt}$$= \large\frac{I}{C} \Rightarrow \large\frac{ 0.15}{(80.032 \times 10^{-12} )}$$ = 1.87 \times 10^9 V/s$
Ans : (a)