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(B) The capacitance $C$ of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$,where $\varepsilon_0$ is the permittivity of free sp

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$1/d^2$

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(B) The capacitance $C$ of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$,where $\varepsilon_0$ is the permittivity of free space,$A$ is the area of the plates,and $d$ is the separation distance.To find the rate of change of capacitance with respect to time $t$,we differentiate $C$ with respect to $t$:$\frac{dC}{dt} = \frac{d}{dt} \left( \frac{\varepsilon_0 A}{d} \right) = \varepsilon_0 A \frac{d}{dt} (d^{-1})$.Using the chain rule,$\frac{dC}{dt} = \varepsilon_0 A (-d^{-2}) \frac{dd}{dt}$.Since the plates are pulled apart with velocity $v$,the rate of change of separation is $\frac{dd}{dt} = v$.Substituting this,we get $\frac{dC}{dt} = -\frac{\varepsilon_0 A}{d^2} v$.The magnitude of the rate of change is $\left| \frac{dC}{dt} \right| = \frac{\varepsilon_0 A v}{d^2}$.Thus,$\left| \frac{dC}{dt} \right| \propto \frac{1}{d^2}$.

The plates of a parallel plate capacitor are pulled apart with a velocity $v$. If at any instant their mutual distance of separation is $d$,then the magnitude of the time rate of change of capacity depends on $d$ as follows:

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