A parallel-plate capacitor of area A , plate separation d and capacitance C is filled with four dielectric…

A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials having dielectric constants k1, k2, k3 and k4 as shown in the figure below. 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

  1. k=k1+k2+k3+3k4
  2. k=23k1+k2+k3+2k4
  3. 2k=3k1+k2+k3+1k4
  4. k=2k43k1k1+k4+k2k2+k4+k3k3+k4

Solution

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
$\begin{aligned} & \frac{1}{C_{cq}}=\frac{1}{\left(C_{k_1}+C_{k_2}+C_{k_3}\right)}+\frac{1}{C_{k_4}} \\ & C_{eq}=k C \end{aligned}$ \(\frac{1}{k C}=\frac{3}{2 C}\left[\frac{1}{k_1+k_2+k_3}\right]+\frac{1}{2 k_4 C}\)
or \(\frac{1}{k}=\frac{3}{2\left(k_1+k_2+k_3\right)}+\frac{1}{2\left(k_4\right)}\)
or \(k=\frac{2\left(k_1+k_2+k_3\right)}{3}+2\left(k_4\right)\)

Asked in: NEET 2016 (Phase 2)

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