સમાંતર પ્લેટ કેપેસીટરમાં A સપાટીનું ક્ષેત્રફળ | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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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