JEE PYQ: Capacitance Question 34
Question 34 - 2019 (09 Jan Shift 2)
A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants $K_1$, $K_2$, $K_3$, $K_4$ arranged as shown in the figure. The effective dielectric constant K will be:
(1) $K = \frac{(K_1 + K_3)(K_2 + K_4)}{K_1 + K_2 + K_3 + K_4}$
(2) $K = \frac{(K_1 + K_2)(K_3 + K_4)}{2(K_1 + K_2 + K_3 + K_4)}$
(3) $K = \frac{(K_1 + K_2)(K_3 + K_4)}{K_1 + K_2 + K_3 + K_4}$
(4) None of these
Show Answer
Answer: (4)
Solution
$C_{12} = \frac{K_1 K_2}{K_1 + K_2} \cdot \frac{\epsilon_0 L^2}{d}$ and $C_{34} = \frac{K_3 K_4}{K_3 + K_4} \cdot \frac{\epsilon_0 L^2}{d}$. $C_{eq} = C_{12} + C_{34}$. Comparing with $K\frac{\epsilon_0 L^2}{d}$: $K_{eq} = \frac{K_1 K_2(K_3 + K_4) + K_3 K_4(K_1 + K_2)}{(K_1 + K_2)(K_3 + K_4)}$. This doesn’t match any given option.
BETA
