A capacitor is formed by two square metal plates

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A capacitor is formed by two square metal-plates of edge a, separated by a distance d. Dielectrics of dielectric constant K1 and K2 are filled in the gap as shown in figure (31-E25). Find the capacitance.

Figure

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Capacitance C = ∈0A k1 k2 /d(k1-k2) ln k1/k2


Explanation:

\frac{1}{\mathrm{dcR}}=\frac{1}{\mathrm{dc}_{1}}+\frac{1}{\mathrm{dc}_{2}}=\frac{x \tan \phi}{\varepsilon_{0} \mathrm{k}_{2}(\mathrm{bdx})}+\frac{\mathrm{d}-x \tan \phi}{\varepsilon_{0} \mathrm{k}_{1}(\mathrm{bdx})}

\mathrm{dcR}=\frac{\varepsilon_{0} \mathrm{bd} x}{\frac{x \tan \phi}{\mathrm{k}_{2}}+\frac{(\mathrm{d}-x \tan \phi)}{\mathrm{k}_{1}}}

\text { or } \mathrm{C}_{\mathrm{R}}=\varepsilon_{0} \mathrm{bk}_{1} \mathrm{k}_{2} \int \frac{\mathrm{dx}}{\mathrm{k}_{2} \mathrm{d}+\left(\mathrm{k}_{1}-\mathrm{k}_{2}\right) \times \tan \phi}

=\frac{\varepsilon_{0} \mathrm{bk}_{1} \mathrm{k}_{2}}{\tan \phi\left(\mathrm{k}_{1}-\mathrm{k}_{2}\right)}\left[\log _{\mathrm{e}} \mathrm{k}_{2} \mathrm{d}+\left(\mathrm{k}_{1}-\mathrm{k}_{2}\right) \times \tan \phi\right] \mathrm{a}

\therefore \tan \phi=\frac{d}{a} \text { and } A=a \times a

C_{R}=\frac{\varepsilon_{0} a k_{1} k_{2}}{\frac{d}{a}\left(k_{1}-k_{2}\right)} \quad\left[\log _{e}\left(\frac{k_{1}}{k_{2}}\right)\right]

C_{R}=\frac{\varepsilon_{0} a^{2} k_{1} k_{2}}{d\left(k_{1}-k_{2}\right)} \quad\left[\log _{e}\left(\frac{k_{1}}{k_{2}}\right)\right]

C_{R}=\frac{\varepsilon_{0} a^{2} k_{1} k_{2}}{d\left(k_{1}-k_{2}\right)} \ln \frac{k_{1}}{k_{2}}


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