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Question

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

Vedclass · Accepted Answer

$\frac{2}{K} = \frac{3}{{{K_1} + {K_2} + {K_3}}} + \frac{1}{{{K_4}}}\;\;\;\;$

Vedclass · Answer

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 $	herefore \frac{1}{C_{A B}}=\frac{1}{C_{1}+C_{2}+C_{3}}+\frac{1}{C_{4}}$

Suppose, $C_{A B}=\frac{k \varepsilon_{0} A}{d}$

$\quad \frac{1}{k\left(\frac{\varepsilon_{0} A}{d}ight)}=\frac{1}{\frac{2}{3} \frac{\varepsilon_{0} A}{d}\left(k_{1}+k_{2}+k_{3}ight)}+\frac{1}{\frac{2 \varepsilon_{0} A}{d} k_{4}}$

$\Rightarrow \frac{1}{k}=\frac{3}{2\left(k_{1}+k_{2}+k_{3}ight)}+\frac{1}{2 k_{4}} 	herefore \frac{2}{k}=\frac{3}{k_{1}+k_{2}+k_{3}}+\frac{1}{k_{4}}$

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