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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
$6 / 5$
$5 / 3$
$7 / 5$
$7 / 3$
Solution
$C _{10}=\frac{4 \varepsilon_0 \frac{ S }{2}}{ d / 2}=\frac{4 \varepsilon_0 S }{ d }$
$C _{20}=\frac{2 \varepsilon_0 S }{ d }, C _{30}=\frac{\varepsilon_0 S }{ d }$
$\frac{1}{ C _{10}^{\prime}}=\frac{1}{ C _{10}}+\frac{1}{ C _{10}}=\frac{ d }{2 \varepsilon_0 S }\left[1+\frac{1}{2}\right]$
$\Rightarrow C _{10}^{\prime}=\frac{4 \varepsilon_0 S }{3 d }$
$C _2= C _{30}+ C _{10}^{\prime}=\frac{7 \varepsilon_0 S }{3 d }$
$\frac{ C _2}{ C _1}=\frac{7}{3}$
