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 ($\varepsilon_1=2$ and $\varepsilon_2=4$) are introduced between the two plates as shown in the figure, the capacitance becomes $C_2$. The ratio $\frac{C_2}{C_1}$ is

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Capacitance

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} = 2

and

ε_{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}}

Answer: (D)

The capacitor is divided into three regions: two in series (with

ε_{1} = 2

and

ε_{2} = 4

) occupying half the area each, and one region with

ε_{1} = 2

occupying the other half area.

For the series combination (left half):

C = \frac{2 ε _{0} S /2}{d /2} = \frac{2 ε _{0} S}{d}

C^{'} = \frac{4 ε _{0} S /2}{d /2} = \frac{4 ε _{0} S}{d}

Series equivalent:

C_{series} = \frac{C C ^{'}}{C + C ^{'}} = \frac{8 ε _{0} S / d}{6 ε _{0} S / d} = \frac{4 ε _{0} S}{3 d}

For the right half:

C^{''} = \frac{2 ε _{0} S /2}{d} = \frac{ε _{0} S}{d}

Total capacitance:

C_{2} = C_{series} + C^{''} = \frac{4 ε _{0} S}{3 d} + \frac{ε _{0} S}{d} = \frac{7 ε _{0} S}{3 d}

Since

C_{1} = \frac{ε _{0} S}{d}

, we have

\frac{C _{2}}{C _{1}} = \frac{7}{3}

JEE Advanced · 2015 · 24 May 2015 · Paper 2 · One or more correct options