Electrostatics 5 Question 19
21. A parallel plate capacitor having plates of area $S$ and plate separation $d$, has capacitance $C _1$ in air. When two dielectrics of different relative permittivities
$\left(\varepsilon _1=2\right.$ and $\left.\varepsilon _2=4\right)$ are introduced between the two plates as shown in the figure, the capacitance becomes $C _2$. The ratio $\frac{C _2}{C _1}$ is
(a) $\frac{6}{5}$
(b) $\frac{5}{3}$
(c) $\frac{7}{5}$
(d) $\frac{7}{3}$
(2015 Adv.)
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Solution:
$$ C _1=\frac{\varepsilon _0 s}{d}, C=\frac{2 \varepsilon _0 \frac{s}{2}}{\frac{d}{2}}=\frac{2 \varepsilon _0 s}{d} \Rightarrow C^{\prime}=\frac{4 \varepsilon _0 \frac{s}{2}}{\frac{d}{2}}=\frac{4 \varepsilon _0 s}{d} $$
$$ \begin{aligned} & \text { and } \quad C^{\prime \prime}=\frac{2 \varepsilon _0 \frac{s}{2}}{d}=\frac{\varepsilon _0 s}{d} \\ & C _2=\frac{C C^{\prime}}{C+C^{\prime}}+C^{\prime \prime}=\frac{4}{3} \frac{\varepsilon _0 s}{d}+\frac{\varepsilon _0 s}{d}=\frac{7}{3} \frac{\varepsilon _0 s}{d} \frac{C _2}{C _1}=\frac{7}{3} \end{aligned} $$
