JEE PYQ: Capacitance Question 31
Question 31 - 2019 (12 Apr Shift 1)
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)):
(1) $\frac{E_1}{E_2} = \frac{K_1 K_2 K_3}{(K_1 + K_2 + K_3)(K_2 K_3 + K_3 K_1 + K_1 K_2)}$
(2) $\frac{E_1}{E_2} = \frac{(K_1 + K_2 + K_3)(K_2 K_3 + K_3 K_1 + K_1 K_2)}{K_1 K_2 K_3}$
(3) $\frac{E_1}{E_2} = \frac{9K_1 K_2 K_3}{(K_1 + K_2 + K_3)(K_2 K_3 + K_3 K_1 + K_1 K_2)}$
(4) $\frac{E_1}{E_2} = \frac{(K_1 + K_2 + K_3)(K_2 K_3 + K_3 K_1 + K_1 K_2)}{9K_1 K_2 K_3}$
Show Answer
Answer: (3)
Solution
For Fig. I (series): $C_1 = \frac{3K_1 K_2 K_3 \epsilon_0 A}{d(K_1 K_2 + K_2 K_3 + K_3 K_1)}$. For Fig. II (parallel): $C_2 = \frac{(K_1 + K_2 + K_3)\epsilon_0 A}{3d}$. $\frac{E_1}{E_2} = \frac{C_1}{C_2} = \frac{9K_1 K_2 K_3}{(K_1 + K_2 + K_3)(K_2 K_3 + K_3 K_1 + K_1 K_2)}$.
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