c
The value of diclectric constant is given as

\(\mathrm{K}=\mathrm{K}_{0}+\lambda \mathrm{x}\)

And, \(\mathrm{V}=\int_{0}^{\mathrm{d}} \mathrm{Edr}\)

\(\mathrm{v}=\int_{0}^{\mathrm{d}} \frac{\sigma}{\mathrm{K}} \mathrm{dx}\)

\({=\sigma \int_{0}^{d} \frac{1}{\left(K_{0}+\lambda x\right.} d x}\)

\({=\frac{\sigma}{\lambda}\left[\ln \left(K_{0}+\lambda d-\ln K_{0}\right]\right.}\)

\({=\frac{\sigma}{\lambda} \ln \left(1+\frac{\lambda d}{K_{0}}\right)}\)

Now it is given that capacitance of vacuum \(=1\)

Thus, \(C=\frac{Q}{V}\)

\(=\frac{\sigma . s}{v}\) (Let surface area of plates \(=\) \(s\))

\(=\frac{\sigma}{\lambda} \ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_{0}}\right)\)

\( = \operatorname{s} \,\lambda \,\frac{d}{d}\frac{1}{{\ln \left( {1 + \frac{{\lambda d}}{{{K_0}}}} \right)}}\left( {\because {\text{ in vacuum }}{\varepsilon _0} = } \right.\)

\(c=\frac{\lambda d}{\ln \left(1+\frac{\lambda d}{K_{0}}\right)} \cdot C_{0}\left(\text { here, } C_{0}=\frac{s}{d}\right)\)