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A parallel-plate capacitor of area , plate separation and capacitance is filled with four dielectric materials having dielectric constants, , and as shown in the figure below. If a single dielectric material is to be used to have the same capacitance in this capacitor, then its dielectric constant is given by
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Here the capacitance, \(C=\frac{A k \epsilon_0}{d}\)
\(C_{K_1}, C_{K_2}, C_{K_3}\) are in parallel combination and they are connected in series with \(\mathrm{C}_{\mathrm{K}_4}\)
Here, \(\mathrm{C}_{\mathrm{K}_1}=\frac{(\mathrm{A} / 3) \mathrm{K}_1 \epsilon_0}{\mathrm{~d} / 2}=\frac{2 \mathrm{~K}_1}{3} \mathrm{C}\),
\(\mathrm{C}_{\mathrm{k}_2}=\frac{(\mathrm{A} / 3) \mathrm{k}_2 \mathrm{\epsilon}_0}{\mathrm{~d} / 2}=\frac{2 \mathrm{k}_2}{3} \mathrm{C}\)
\(C_{K_3}=\frac{(\mathrm{A} / 3) \mathrm{k}_3 \epsilon_0}{\mathrm{~d} / 2} \frac{2 \mathrm{k}_3 \mathrm{C}}{3} \mathrm{C}\)
\(\mathrm{C}_{\mathrm{K}_4}=\frac{(\mathrm{A}) \mathrm{k}_4 \epsilon_0}{\mathrm{~d} / 2}=2 \mathrm{k}_4 \mathrm{C}\)
Now the equivalent capacitance for the combination of four capacitors is
\(\begin{aligned}
& \frac{1}{C_{c q}}=\frac{1}{\left(C_{k_1}+C_{k_2}+C_{k_3}\right)}+\frac{1}{C_{k_4}} \\
& C_{e q}=k C
\end{aligned}\)
\(\frac{1}{k C}=\frac{3}{2 C}\left[\frac{1}{k_1+k_2+k_3}\right]+\frac{1}{2 k_4 C}\)
or \(\frac{1}{k}=\frac{3}{2\left(k_1+k_2+k_3\right)}+\frac{1}{2\left(k_4\right)}\)
or \(k=\frac{2\left(k_1+k_2+k_3\right)}{3}+2\left(k_4\right)\)
\(C_{K_1}, C_{K_2}, C_{K_3}\) are in parallel combination and they are connected in series with \(\mathrm{C}_{\mathrm{K}_4}\)
Here, \(\mathrm{C}_{\mathrm{K}_1}=\frac{(\mathrm{A} / 3) \mathrm{K}_1 \epsilon_0}{\mathrm{~d} / 2}=\frac{2 \mathrm{~K}_1}{3} \mathrm{C}\),
\(\mathrm{C}_{\mathrm{k}_2}=\frac{(\mathrm{A} / 3) \mathrm{k}_2 \mathrm{\epsilon}_0}{\mathrm{~d} / 2}=\frac{2 \mathrm{k}_2}{3} \mathrm{C}\)
\(C_{K_3}=\frac{(\mathrm{A} / 3) \mathrm{k}_3 \epsilon_0}{\mathrm{~d} / 2} \frac{2 \mathrm{k}_3 \mathrm{C}}{3} \mathrm{C}\)
\(\mathrm{C}_{\mathrm{K}_4}=\frac{(\mathrm{A}) \mathrm{k}_4 \epsilon_0}{\mathrm{~d} / 2}=2 \mathrm{k}_4 \mathrm{C}\)
Now the equivalent capacitance for the combination of four capacitors is
\(\begin{aligned}
& \frac{1}{C_{c q}}=\frac{1}{\left(C_{k_1}+C_{k_2}+C_{k_3}\right)}+\frac{1}{C_{k_4}} \\
& C_{e q}=k C
\end{aligned}\)
\(\frac{1}{k C}=\frac{3}{2 C}\left[\frac{1}{k_1+k_2+k_3}\right]+\frac{1}{2 k_4 C}\)
or \(\frac{1}{k}=\frac{3}{2\left(k_1+k_2+k_3\right)}+\frac{1}{2\left(k_4\right)}\)
or \(k=\frac{2\left(k_1+k_2+k_3\right)}{3}+2\left(k_4\right)\)
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