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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

 

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

 

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1 Answer

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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)
answered Apr 3, 2014 by balaji.thirumalai
 

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