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A parallel palate capacitor with square plates is filled with four dielectrics of dielectric constants $K_1, K_2, K_3, K_4$ arranged as shown in the figure. The effective dielectric constant $K$ will be
$K = \frac{{({K_1} + {K_3})({K_2} + {K_4})}}{{{K_1} + {K_2} + {K_3} + {K_4}}}$
$K = \frac{{({K_1} + {K_2})({K_3} + {K_4})}}{{2({K_1} + {K_2} + {K_3} + {K_4})}}$
$K = \frac{{({K_1} + {K_2})({K_3} + {K_4})}}{{{K_1} + {K_2} + {K_3} + {K_4}}}$
$K = \frac{{({K_1} + {K_4})({K_2} + {K_3})}}{{2({K_1} + {K_2} + {K_3} + {K_4})}}$
Solution
$C_{1}=\frac{\varepsilon_{0} K_{1} \frac{L^{2}}{2}}{\frac{d}{2}}+\frac{\varepsilon_{0} K_{3} \frac{L^{2}}{2}}{\left(\frac{d}{2}\right)}=\frac{\varepsilon_{0} L^{2}}{d}\left(K_{1}+K_{3}\right)$
$C_{2}=\frac{\varepsilon_{0} K_{2} \frac{L^{2}}{2}}{\frac{d}{2}}+\frac{\varepsilon_{0} K_{4} \frac{L^{2}}{2}}{\frac{d}{2}}=\frac{\varepsilon_{0} L^{2}}{d}\left(K_{2}+K_{4}\right)$
$\therefore \quad \frac{1}{c}=\frac{1}{c_{1}}+\frac{1}{c_{2}}$
$\Rightarrow \quad \frac{\mathrm{d}}{\varepsilon_{0} \mathrm{KL}^{2}}=\frac{\mathrm{d}}{\varepsilon_{0} \mathrm{L}^{2}\left(\mathrm{K}_{1}+\mathrm{K}_{3}\right)}+\frac{\mathrm{d}}{\varepsilon_{0} \mathrm{L}^{2}\left(\mathrm{K}_{2}+\mathrm{K}_{4}\right)}$
