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Question

As $\mathrm{K}$ is variable we take a plate element of Area $A$ and thickness $dx$ at distance $\mathrm{x}$

Capacitance of element

$\mathrm{d} \

Vedclass · Accepted Answer

$\frac{\mathrm{AK} \varepsilon_{0}}{\mathrm{d}}\left(1+\frac{\alpha \mathrm{d}}{2}\right)$

Vedclass · Answer

As $\mathrm{K}$ is variable we take a plate element of Area $A$ and thickness $dx$ at distance $\mathrm{x}$

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

$\frac{1}{\mathrm{C}}=\int \frac{1}{\mathrm{d} \mathrm{C}}=\int_{0}^{\mathrm{d}} \frac{\mathrm{dx}}{\mathrm{AK} \varepsilon_{0}(1+\alpha \mathrm{x})}$

$\frac{1}{\mathrm{C}}=\frac{1}{\alpha \mathrm{AK} \varepsilon_{0}} \ln \left(\frac{1+\alpha \mathrm{d}}{1}\right)$

$=\frac{1}{\mathrm{C}}=\frac{1}{\alpha \mathrm{AK} \varepsilon_{0}}\left(\alpha \mathrm{d}-\frac{(\alpha \mathrm{d})^{2}}{2}+\frac{(\alpha \mathrm{d})^{3}}{3}+\ldots .\right)$

$\Rightarrow \frac{1}{\mathrm{C}}=\frac{\alpha \mathrm{d}}{\alpha \mathrm{AK} \varepsilon_{0}}\left(1-\frac{\alpha \mathrm{d}}{2}+\frac{(\alpha \mathrm{d})^{2}}{3}+\ldots .\right)$

$\frac{1}{\mathrm{C}}=\frac{\mathrm{d}}{\mathrm{AK} \varepsilon_{0}}\left(1-\frac{\alpha \mathrm{d}}{2}\right)$

$C=\frac{A K \varepsilon_{0}}{d}\left(1+\frac{\alpha d}{2}\right)$

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