The space between the plates of a parallel plate capacitor is...

The value of dielectric constant is given as, \begin{aligned} &\mathrm{K}=\mathrm{K}_0+\lambda \mathrm{x} \\ &\text{ And, } \mathrm{V}=\int_0^{\mathrm{d}} \mathrm{Edr} \\ &\mathrm{V}=\int_0^{\mathrm{d}} \frac{\sigma}{\mathrm{K}} \mathrm{dx} \end{aligned}\begin{aligned} &=\sigma \int_0^{\mathrm{d}} \frac{1}{\left(\mathrm{~K}_0+\lambda \mathrm{x}\right.} \mathrm{dx} \\ &=\frac{\sigma}{\lambda}\left[\ln \left(\mathrm{K}_0+\lambda \mathrm{d}-\ln \mathrm{K}_0\right]\right. \\ &=\frac{\sigma}{\lambda} \ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right) \end{aligned} Now it is given that capacitance of vacuum $=\mathrm{C}_0$. Thus, $\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}$ \begin{aligned} &=\frac{\sigma . \mathrm{s}}{\mathrm{v}}(\text{ Let surface area of plates }=\mathrm{s}) \\ &=\frac{\sigma . \mathrm{s}}{\frac{\sigma}{\lambda} \ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)} \end{aligned} $=\mathrm{s} \lambda \cdot \frac{\mathrm{d}}{\mathrm{d}} \frac{1}{\ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)}\left(\because\right.$ in vacuum $\left.\varepsilon_0=1\right)$\mathrm{c}=\frac{\lambda \mathrm{d}}{\ln \left(1+\frac{\lambda \mathrm{d}}{\mathrm{K}_0}\right)} \cdot \mathrm{C}_0 \quad\left(\right.$here, \left.\mathrm{C}_0=\frac{\mathrm{s}}{\mathrm{d}}\right)$