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Question

Physics Question on Capacitors and Capacitance

The gap between the plates of a parallel plate capacitor of area A and distance between plates dd, is filled with a dielectric whose permittivity varies linearly from ϵ1\epsilon_{1} at one plate to ϵ2\epsilon_{2} at the other. The capacitance of capacitor is :

A

ϵ0(ϵ1+ϵ2)A/d\epsilon_{0} \left(\epsilon_{1}+\epsilon_{2}\right)A/d

B

ϵ0(ϵ1+ϵ2)A/2d\epsilon_{0} \left(\epsilon_{1}+\epsilon_{2}\right)A/2d

C

ϵ0A/[dln(ϵ2/ϵ1)]\epsilon_{0}A/\left[d ln\left(\epsilon_{2}/ \epsilon_{1}\right)\right]

D

ϵ0(ϵ2ϵ1)A/[dn(2/ϵ1)]\epsilon_{0}\left(\epsilon_{2}-\epsilon_{1}\right) A /\left[ d \ell n \left(\in_{2} / \epsilon_{1}\right)\right]

Answer

ϵ0(ϵ2ϵ1)A/[dn(2/ϵ1)]\epsilon_{0}\left(\epsilon_{2}-\epsilon_{1}\right) A /\left[ d \ell n \left(\in_{2} / \epsilon_{1}\right)\right]

Explanation

Solution

ε=(ε2ε1d)x+ε1\varepsilon=\left(\frac{\varepsilon_{2}-\varepsilon_{1}}{d}\right) x+\varepsilon_{1} dC=ε0εAdxd C=\frac{\varepsilon_{0} \varepsilon A}{d x} Ceq =dcC_{\text {eq }}=\int dc