Question

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 $({\epsilon }_1=2$ and ${\epsilon }_2=4)$ 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. 6/5
• B. 5/3
• C. 7/5
• D. 7/3

Answer 1

The correct option is D 7/3





Capacitance $\frac{{ϵ}_{0}A}{d}$
$C_1=\frac{{ϵ}_{0}S}{d}$

From the figure
$C_5=\frac{{ϵ}_1\frac{S}{2}}{d}$

$C_3$ and $C_4$ are in series.

$C_3=\frac{{ϵ}_1\frac{S}{2}}{\frac{d}{2}}$; $C_4=\frac{{ϵ}_2\frac{S}{2}}{\frac{d}{2}}$
$C_{34}\left(\text{series}\right)=\frac{C_3{C}_4}{C_3+C_4}$

$C_{34}\text{and}{C}_5\text{are parllel so}$

$C_2=C_{34}+C_5=\frac{7}{3}\frac{{ϵ}_{0}S}{d}$

$\frac{C_2}{C_1}=\frac{7}{3}$