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Question

The value of diclectric constant is given as

$\mathrm{K}=\mathrm{K}_{0}+\lambda \mathrm{x}$

And, $\mathrm{V}=\int_{0}^{\mathrm{d}} \mathrm{Edr}$

Vedclass · Accepted Answer

$C\, = \,\frac{{\lambda d}}{{\ln \,(1 + \lambda d/{K_0})}}{C_0}$

Vedclass · Answer

The value of diclectric constant is given as

$\mathrm{K}=\mathrm{K}_{0}+\lambda \mathrm{x}$

And, $\mathrm{V}=\int_{0}^{\mathrm{d}} \mathrm{Edr}$

$\mathrm{v}=\int_{0}^{\mathrm{d}} \frac{\sigma}{\mathrm{K}} \mathrm{dx}$

${=\sigma \int_{0}^{d} \frac{1}{\left(K_{0}+\lambda xight.} d x}$

${=\frac{\sigma}{\lambda}\left[\ln \left(K_{0}+\lambda d-\ln K_{0}ight]ight.}$

${=\frac{\sigma}{\lambda} \ln \left(1+\frac{\lambda d}{K_{0}}ight)}$

Now it is given that capacitance of vacuum $=1$

Thus, $C=\frac{Q}{V}$

$=\frac{\sigma . s}{v}$ (Let surface area of plates $=$ $s$)

$=\frac{\sigma}{\lambda} \ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_{0}}ight)$

$ = \operatorname{s} \,\lambda \,\frac{d}{d}\frac{1}{{\ln \left( {1 + \frac{{\lambda d}}{{{K_0}}}} ight)}}\left( {\because {	ext{ in vacuum }}{\varepsilon _0} = } ight.$

$c=\frac{\lambda d}{\ln \left(1+\frac{\lambda d}{K_{0}}ight)} \cdot C_{0}\left(	ext { here, } C_{0}=\frac{s}{d}ight)$

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