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) $\frac{E _1}{E _2}=\frac{9 K _1 K _2 K _3}{\left(K _1+K _2+K _3\right)\left(K _2 K _3+K _3 K _1+K _1 K _2\right)}$
True/False
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Answer:
Correct Answer: 2. (b)
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,
$$ \begin{aligned} C _I & =C _1+C _2+C _3 \\ & =\frac{\varepsilon _0 A}{3 d} K _1+\frac{\varepsilon _0 A}{3 d} K _2+\frac{\varepsilon _0 A}{3 d} K _3 \\ C _I & =\frac{\varepsilon _0 A}{3 d}\left(K _1+K _2+K _3\right) \end{aligned} $$
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,
$$ \begin{aligned} \frac{1}{C _{II}} & =\frac{1}{C _1}+\frac{1}{C _2}+\frac{1}{C _3} \\ & =\frac{d}{3 \varepsilon _0 A K _1}+\frac{d}{3 \varepsilon _0 A K _2}+\frac{d}{3 \varepsilon _0 A K _3} \\ \Rightarrow \frac{1}{C _{II}} & =\frac{d}{3 \varepsilon _0 A} \frac{1}{K _1}+\frac{1}{K _2}+\frac{1}{K _3} \\ & =\frac{d}{3 \varepsilon _0 A} \frac{K _2 K _3+K _1 K _3+K _1 K _2}{K _1 K _2 K _3} \\ C _{II} & =\frac{3 \varepsilon _0 A}{d} \frac{K _1 K _2 K _3}{K _1 K _2+K _2 K _3+K _3 K _1} \end{aligned} $$
From Eqs. (i) and (ii), we get
$\frac{C _I}{C _{II}}=\frac{\varepsilon _0 A}{3 d}\left(K _1+K _2+K _3\right) \times \frac{d\left(K _1 K _2+K _2 K _3+K _3 K _1\right)}{3 \varepsilon _0 A\left(K _1 K _2 K _3\right)}$
$$ =\frac{\left(K _1+K _2+K _3\right)\left(K _1 K _2+K _2 K _3+K _3 K _1\right)}{9 K _1 K _2 K _3} $$
Now, energy stored in capacitor, $E=\frac{1}{2} C V^{2}$
$$ \begin{array}{ll} \Rightarrow & E \propto C \\ \therefore & \frac{E _I}{E _{II}}=\frac{C _I}{C _{II}} \end{array} $$
