Physics - Capacitor Question with Solution | TestHub

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