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 (${ϵ}_1=1$ and ${ϵ}_2=4$) are introduced between the two plates as shown in the figure, the capacitance become $C_2$. The ratio $\frac{C_2}{C_1}$ is

• A. $6/5$
• B. $5/3$
• C. $7/5$
• D. $7/3$

Answer 1

Answer: (D)

$7/3$

$C=2{\epsilon }_0\frac{s}{d};$ $C^{\prime }=4{\epsilon }_0\frac{s}{d}$

They are in series, their equivalent $=\frac{4{\epsilon }_0\frac{s}{d}\times 2{\epsilon }_0\frac{s}{d}}{4{\epsilon }_0\frac{s}{d}+2{\epsilon }_0\frac{s}{d}}$ $=\frac{4}{3}{\epsilon }_0\frac{s}{d}$

$C_3=\frac{2{\epsilon }_{0}s}{2d}$ $=\frac{{\epsilon }_{0}s}{d}$

$C_2=\frac{4{\epsilon }_{0}s}{3d}$ $+\frac{{\epsilon }_{0}s}{d}$ $=\frac{7}{3}{\epsilon }_0\frac{s}{d}$

And, $C_1={\epsilon }_0\frac{s}{d}$

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