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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
${ \varepsilon _0}\left( {{ \varepsilon _1} + { \varepsilon _2}} \right)A/d$
${ \varepsilon _0}\left( {{ \varepsilon _2} + { \varepsilon _1}} \right)A/2d$
${ \varepsilon _0}\,A/\left[ {d\,\ln \left( {{ \varepsilon _2}/{ \varepsilon _1}} \right)} \right]$
${ \varepsilon _0}\left( {{ \varepsilon _2} - { \varepsilon _1}} \right)A/\left[ {d\,\ln \left( {{ \varepsilon _2}/{ \varepsilon _1}} \right)} \right]$
Solution
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,
$d V=\frac{E_{0}}{k} d x \Rightarrow \int_{0}^{V} d V=\int_{0}^{d} \frac{\sigma}{\varepsilon_{0}} \frac{1}{\left(\frac{c_{2}-\varepsilon_{1}}{d}\right) x+\varepsilon_{1}} d x$
$V=\frac{d \sigma}{\varepsilon_{0}\left(\varepsilon_{2}-\varepsilon_{1}\right)} \ln \left(\frac{\varepsilon_{2}}{\varepsilon_{1}}\right)$
$Q=C V \Rightarrow C=\frac{Q}{V}=\frac{\sigma A}{\frac{d \sigma}{\varepsilon_{0}\left(\varepsilon_{2}-\varepsilon_{1}\right)} \ln \left(\frac{\varepsilon_{2}}{\varepsilon_{1}}\right)}=\frac{\varepsilon_{0}\left(\varepsilon_{2}-\varepsilon_{1}\right) A}{d \ln \left(\frac{\varepsilon_{2}}{\varepsilon_{1}}\right)}$
