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Discharging of Capacitors

Two identical...

Question

Two identical parallel plate capacitors, of capacitance \(C\) each, have plates of area \(A\), separated by a distance \(d\). The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants \(K_{1},\)\(K_{2}\) and \(K_{3}\). The first capacitor is filled as shown in Figure \(I\), and the second one is filled as shown in Figure \(II\). If these two modified capacitors are charged by the same potential \(V\), the ratio of the energy stored in the two, would be:
\((E_{1}\) refers to capacitor \((I)\) and \(E_{2}\) to capacitor \((II))\):

(I) (II)

A

\frac{E_{1}}{E_{2}} = \frac{9 K_{1} K_{2} K_{3}}{(K_{1} + K_{2} + K_{3}) (K_{2} K_{3} + K_{3} K_{1} + K_{1} K_{2})}

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B

\frac{E_{1}}{E_{2}} = \frac{K_{1} K_{2} K_{3}}{(K_{1} + K_{2} + K_{3}) (K_{2} K_{3} + K_{3} K_{1} + K_{1} K_{2})}

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C

\frac{E_{1}}{E_{2}} = \frac{(K_{1} + K_{2} + K_{3}) (K_{2} K_{3} + K_{3} K_{1} + K_{1} K_{2})}{K_{1} K_{2} K_{3}}

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D

\frac{E_{1}}{E_{2}} = \frac{(K_{1} + K_{2} + K_{3}) (K_{2} K_{3} + K_{3} K_{1} + K_{1} K_{2})}{9 K_{1} K_{2} K_{3}}

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Solution

The correct option is A $\frac{E_{1}}{E_{2}} = \frac{9 K_{1} K_{2} K_{3}}{(K_{1} + K_{2} + K_{3}) (K_{2} K_{3} + K_{3} K_{1} + K_{1} K_{2})}$
In figure $I$ , all the three dielectrics are in series,

$\Rightarrow \frac{1}{C_{1}} = \frac{(\frac{d}{3})}{K_{1} ε_{0} A} + \frac{(\frac{d}{3})}{K_{2} ε_{0} A} + \frac{(\frac{d}{3})}{K_{3} ε_{0} A}$

$\Rightarrow C_{1} = \frac{3 K_{1} K_{2} K_{3} ε_{0} A}{d (K_{1} K_{2} + K_{2} K_{3} + K_{3} K_{1})}$

In figure $(I I)$ , all the three dielectrics are in parallel,

$\Rightarrow C_{2} = \frac{K_{1} ε_{0} (\frac{A}{3})}{d} + \frac{K_{2} ε_{0} (\frac{A}{3})}{d} + \frac{K_{3} ε_{0} (\frac{A}{3})}{d}$

$\Rightarrow C_{2} = \frac{(K_{1} + K_{2} + K_{3}) ε_{0} A}{3 d}$

The energy stored in a capacitor is given by,

$E = \frac{1}{2} C V^{2}$

As both the capacitors are connected to same $V$

$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{\frac{1}{2} C_{1} V^{2}}{\frac{1}{2} C_{2} V^{2}}$

$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{9 K_{1} K_{2} K_{3}}{(K_{1} + K_{2} + K_{3}) (K_{2} K_{3} + K_{3} K_{1} + K_{1} K_{2})}$

Hence, option $(C)$ is correct.

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