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A parallel-plate capacitor of area $A,$ plate separation $d$ and capacitance $C$ is filled with four dielectric materials having dielectric constants $K_1,K_2,K_3$ and $K_4$ as shown in the figure. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor, then its dielectric constant $K$ is given by
$\frac{2}{K} = \frac{3}{{{K_1} + {K_2} + {K_3}}} + \frac{1}{{{K_4}}}\;\;\;\;$
$\;\frac{1}{K} = \frac{1}{{{K_1}}} + \frac{1}{{{K_2}}} + \frac{1}{{{K_3}}} + \frac{3}{{2{K_4}}}$
$K=K_1+K_2+K_3+3K_4$
$K=$ $\frac{2}{3}\left[ {{K_1} + {K_2} + {K_3}} \right] + 2{K_4}$
Solution
Here, $C_{1}=\frac{2 \varepsilon_{0} k_{1} A}{3 d}, C_{2}=\frac{2 \varepsilon_{0} k_{2} A}{3 d}$
$C_{3}=\frac{2 \varepsilon_{0} k_{3} A}{3 d}, C_{4}=\frac{2 \varepsilon_{0} k_{4} A}{d}$
Given system of $C_{1}, C_{2}, C_{3}$ and $C_{4}$ can be simplified as $\therefore \frac{1}{C_{A B}}=\frac{1}{C_{1}+C_{2}+C_{3}}+\frac{1}{C_{4}}$
Suppose, $C_{A B}=\frac{k \varepsilon_{0} A}{d}$
$\quad \frac{1}{k\left(\frac{\varepsilon_{0} A}{d}\right)}=\frac{1}{\frac{2}{3} \frac{\varepsilon_{0} A}{d}\left(k_{1}+k_{2}+k_{3}\right)}+\frac{1}{\frac{2 \varepsilon_{0} A}{d} k_{4}}$
$\Rightarrow \frac{1}{k}=\frac{3}{2\left(k_{1}+k_{2}+k_{3}\right)}+\frac{1}{2 k_{4}} \therefore \frac{2}{k}=\frac{3}{k_{1}+k_{2}+k_{3}}+\frac{1}{k_{4}}$
