Electrostatics 5 Question 2
2. 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 Fig. I, and the second one is filled as shown in Fig. 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)) :
(Main 2019, 12 April I)
(a) $\frac{E_{1}}{E_{2}}=\frac{K_{1} K_{2} K_{3}}{\left(K_{1}+K_{2}+K_{3}\right)\left(K_{2} K_{3}+K_{3} K_{1}+K_{1} K_{2}\right)}$
(b) $\frac{E_{1}}{E_{2}}=\frac{\left(K_{1}+K_{2}+K_{3}\right)\left(K_{2} K_{3}+K_{3} K_{1}+K_{1} K_{2}\right)}{K_{1} K_{2} K_{3}}$
(c) $\frac{E_{1}}{E_{2}}=\frac{9 K_{1} K_{2} K_{3}}{\left(K_{1}+K_{2}+K_{3}\right)\left(K_{2} K_{3}+K_{3} K_{1}+K_{1} K_{2}\right)}$
(d)$\frac{E_1}{E_2}=\frac{\left(K_1+K_2+K_3\right)\left(K_2 K_3+K_3 K_1+K_1 K_2\right)}{9 K_1 K_2 K_3}$
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Answer:
Correct Answer: 2. (d)
Solution:
- Key Idea A capacitor filled with dielectrics can be treated/compared as series/parallel combinations of capacitor having individual dielectric.
e.g.
Case I
Capacitance in the equivalent circuit are
$ \begin{aligned} C_{1}=\frac{\varepsilon_{0} \frac{A}{3}}{d} K_{1} & =\frac{\varepsilon_{0} A}{3 d} K_{1}, C_{2}=\frac{\varepsilon_{0} \frac{A}{3}}{d} K_{2}=\frac{\varepsilon_{0} A}{3 d} K_{2} \\ \text { and } \quad C_{3} & =\frac{\varepsilon_{0} \frac{A}{3}}{d} K_{3}=\frac{\varepsilon_{0} A}{3 d} K_{3} \end{aligned} $
So, equivalent capacitance,
$ \begin{aligned} C_{\mathrm{I}} & =C_{1}+C_{2}+C_{3} \\ & =\frac{\varepsilon_{0} A}{3 d} K_{1}+\frac{\varepsilon_{0} A}{3 d} K_{2}+\frac{\varepsilon_{0} A}{3 d} K_{3} \\ C_{\mathrm{I}} & =\frac{\varepsilon_{0} A}{3 d}\left(K_{1}+K_{2}+K_{3}\right) \end{aligned} $
Case II
Capacitance of equivalent circuit are
$ \begin{aligned} & C_{1}=\frac{\varepsilon_{0} A}{\frac{d}{3}} \cdot K_{1}=\frac{3 \varepsilon_{0} A}{d} K_{1} \\ & C_{2}=\frac{\varepsilon_{0} A}{\frac{d}{3}} K_{2}=\frac{3 \varepsilon_{0} A}{d} K_{2} \end{aligned} $
and
$ C_{3}=\frac{\varepsilon_{0} A}{\frac{d}{3}} K_{3}=\frac{3 \varepsilon_{0} A}{d} K_{3} $
So, equivalent capacitance,
$ \begin{aligned} \frac{1}{C_{\mathrm{II}}} & =\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}} \\ & =\frac{d}{3 \varepsilon_{0} A K_{1}}+\frac{d}{3 \varepsilon_{0} A K_{2}}+\frac{d}{3 \varepsilon_{0} A K_{3}} \\ \Rightarrow \frac{1}{C_{\mathrm{II}}} & =\frac{d}{3 \varepsilon_{0} A} \frac{1}{K_{1}}+\frac{1}{K_{2}}+\frac{1}{K_{3}} \\ & =\frac{d}{3 \varepsilon_{0} A} \frac{K_{2} K_{3}+K_{1} K_{3}+K_{1} K_{2}}{K_{1} K_{2} K_{3}} \\ C_{\mathrm{II}} & =\frac{3 \varepsilon_{0} A}{d} \frac{K_{1} K_{2} K_{3}}{K_{1} K_{2}+K_{2} K_{3}+K_{3} K_{1}} \end{aligned} $
From Eqs. (i) and (ii), we get
$\frac{C_{\mathrm{I}}}{C_{\mathrm{II}}}=\frac{\varepsilon_{0} A}{3 d}\left(K_{1}+K_{2}+K_{3}\right) \times \frac{d\left(K_{1} K_{2}+K_{2} K_{3}+K_{3} K_{1}\right)}{3 \varepsilon_{0} A\left(K_{1} K_{2} K_{3}\right)}$
$ =\frac{\left(K_{1}+K_{2}+K_{3}\right)\left(K_{1} K_{2}+K_{2} K_{3}+K_{3} K_{1}\right)}{9 K_{1} K_{2} K_{3}} $
Now, energy stored in capacitor, $E=\frac{1}{2} C V^{2}$
$ \begin{array}{ll} \Rightarrow & E \propto C \\ \therefore & \frac{E_{\mathrm{I}}}{E_{\mathrm{II}}}=\frac{C_{\mathrm{I}}}{C_{\mathrm{II}}} \end{array} $
