Question 113 of 180

The gap between the plates of a parallel plate capacitor of area $A$ and distance between plates $d$, is filled with a dielectric whose relative permittivity varies linearly from ${ϵ}_{1}$ at one plate to ${ϵ}_{2}$ at the other. The capacitance of the capacitor is

A$\dfrac{{ϵ}_{0}\left({\in }_{2}-{\in }_{1}\right)A}{\left[d\ln \left({\in }_{2}/{\in }_{1}\right)\right]}$B$\dfrac{{ϵ}_{0}\left({\in }_{2}+{\in }_{1}\right)A}{2d}$C$\dfrac{{ϵ}_{0}\left({\in }_{1}+{\in }_{2}\right)A}{d}$D$\dfrac{{ϵ}_{0}A}{\left[dln\left({\in }_{2}/{\in }_{1}\right)\right]}$

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Answer: (A) $\dfrac{{ϵ}_{0}\left({\in }_{2}-{\in }_{1}\right)A}{\left[d\ln \left({\in }_{2}/{\in }_{1}\right)\right]}$

$\dfrac{1}{dC}=\dfrac{dx}{{ϵ}_{0}ϵA}$
$=\dfrac{dx}{{ϵ}_{0}\left[{ϵ}_{1}+\dfrac{\left({ϵ}_{2}-{ϵ}_{1}\right)}{d}x\right]A}$
$\dfrac{1}{C}=\int \dfrac{1}{dC}=\dfrac{1}{{\in }_{0}A}\int _{0}^{x}\dfrac{dx}{{\in }_{1}+\left(\dfrac{{\in }_{2}-{\in }_{1}}{d}\right)}$
$=\dfrac{\ln \left({\in }_{2}/{\in }_{1}\right)d}{{\in }_{0}A\left({\in }_{2}-{\in }_{1}\right)}$
$C=\dfrac{{\in }_{0}A\left({\in }_{2}-{\in }_{1}\right)}{\ln \left({\in }_{2}/{\in }_{1}\right)\cdot d}$