Question

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

Answer 1

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

Explanation:

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

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

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

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

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

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

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

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