A parallel plate capacitor has plates of area \(A\) separated by distance \(d\) between them. It is filled with a dielectric which has a dielectric constant that varies as \(\mathrm{k}(\mathrm{x})=\mathrm{K}(1+\alpha \mathrm{x})\) where \(\mathrm{x}\) is the distance measured from one of the plates. If \((\alpha \text {d)}<<1,\) the total capacitance of the system is best given by the expression 

Capacitance of element

\(\mathrm{d} \mathrm{C}=\frac{(\mathrm{A}) \mathrm{K}(1+\alpha \mathrm{x}) \varepsilon_{0}}{\mathrm{dx}}\)

Now all such elements are is series so

equivalent capacitance

\(\frac{1}{\mathrm{C}}=\int \frac{1}{\mathrm{d} \mathrm{C}}=\int_{0}^{\mathrm{d}} \frac{\mathrm{dx}}{\mathrm{AK} \varepsilon_{0}(1+\alpha \mathrm{x})}\)

\(\frac{1}{\mathrm{C}}=\frac{1}{\alpha \mathrm{AK} \varepsilon_{0}} \ln \left(\frac{1+\alpha \mathrm{d}}{1}\right)\)

\(=\frac{1}{\mathrm{C}}=\frac{1}{\alpha \mathrm{AK} \varepsilon_{0}}\left(\alpha \mathrm{d}-\frac{(\alpha \mathrm{d})^{2}}{2}+\frac{(\alpha \mathrm{d})^{3}}{3}+\ldots .\right)\)

\(\Rightarrow \frac{1}{\mathrm{C}}=\frac{\alpha \mathrm{d}}{\alpha \mathrm{AK} \varepsilon_{0}}\left(1-\frac{\alpha \mathrm{d}}{2}+\frac{(\alpha \mathrm{d})^{2}}{3}+\ldots .\right)\)

\(\frac{1}{\mathrm{C}}=\frac{\mathrm{d}}{\mathrm{AK} \varepsilon_{0}}\left(1-\frac{\alpha \mathrm{d}}{2}\right)\)

\(C=\frac{A K \varepsilon_{0}}{d}\left(1+\frac{\alpha d}{2}\right)\)