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(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$,where $A = \pi r^2$.Given $V = \frac{Q}{C}$,we have $V = \f

Vedclass · Accepted Answer

$1320$

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(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$,where $A = \pi r^2$.Given $V = \frac{Q}{C}$,we have $V = \frac{Q d}{\epsilon_0 A}$.Differentiating with respect to time $t$,we get $\frac{dV}{dt} = \frac{d}{dt} \left( \frac{Q d}{\epsilon_0 A} ight) = \frac{d}{\epsilon_0 A} \frac{dQ}{dt}$.Since $\frac{dQ}{dt} = I$,the rate of change of potential difference is $\frac{dV}{dt} = \frac{I d}{\epsilon_0 A}$.Rearranging for $d$,we get $d = \frac{\epsilon_0 A (dV/dt)}{I} = \frac{\epsilon_0 (\pi r^2) (dV/dt)}{I}$.Substituting the given values: $r = 0.1 \ m$,$I = 0.15 \ A$,$\frac{dV}{dt} = 7 	imes 10^8 \ V/s$,$\epsilon_0 = 9 	imes 10^{-12} \ F/m$,and $\pi = 22/7$.$d = \frac{(9 	imes 10^{-12}) 	imes (22/7) 	imes (0.1)^2 	imes (7 	imes 10^8)}{0.15} \ m$.$d = \frac{9 	imes 10^{-12} 	imes 22 	imes 0.01 	imes 10^8}{0.15} \ m = \frac{9 	imes 22 	imes 10^{-6}}{0.15} \ m = \frac{198 	imes 10^{-6}}{0.15} \ m = 1320 	imes 10^{-6} \ m$.Since $1 \ \mu m = 10^{-6} \ m$,the distance $d = 1320 \ \mu m$.

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