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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2014
  • Hard
  • MCQ

The gap between the plates of a parallel plate capacitor of area \(A\) and distance between plates \(d\), is filled with a dielectric whose permittivity varies linearly from \({ \varepsilon _1}\) at one plate to \({ \varepsilon _2}\) at the other. The capacitance of capacitor is

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Answer & Solution

Correct Answer

(D) \({ \varepsilon _0}\left( {{ \varepsilon _2} - { \varepsilon _1}} \right)A/\left[ {d\,\ln \left( {{ \varepsilon _2}/{ \varepsilon _1}} \right)} \right]\)

Step-by-step Solution

Detailed explanation

As the permittivity of dielectric varies linearly from \(\varepsilon_{1}\) at one plate to \(\varepsilon_{2}\) at the other, it is governed by equation,

\(k=\left(\frac{\varepsilon_{2}-\varepsilon_{1}}{d}\right) x+\varepsilon_{1}\)

Consider a small element of thickness \(d x\) at a distance \(x\) from plate. Then,

\(d V=\frac{E_{0}}{k} d x \Rightarrow \int_{0}^{V} d V=\int_{0}^{d} \frac{\sigma}{\varepsilon_{0}} \frac{1}{\left(\frac{c_{2}-\varepsilon_{1}}{d}\right) x+\varepsilon_{1}} d x\)

\(V=\frac{d \sigma}{\varepsilon_{0}\left(\varepsilon_{2}-\varepsilon_{1}\right)} \ln \left(\frac{\varepsilon_{2}}{\varepsilon_{1}}\right)\)

\(Q=C V \Rightarrow C=\frac{Q}{V}=\frac{\sigma A}{\frac{d \sigma}{\varepsilon_{0}\left(\varepsilon_{2}-\varepsilon_{1}\right)} \ln \left(\frac{\varepsilon_{2}}{\varepsilon_{1}}\right)}=\frac{\varepsilon_{0}\left(\varepsilon_{2}-\varepsilon_{1}\right) A}{d \ln \left(\frac{\varepsilon_{2}}{\varepsilon_{1}}\right)}\)

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