JEE Advanced 2024 — Paper 1 — PHY-IV solution

JEE Advanced2024Paper 1PHY-IV

Four identical thin, square metal sheets, $S_{1}$ , $S_{2}$ , $S_{3}$ , and $S_{4}$ , each of side $a$ are kept parallel to each other with equal distance $d (≪ a)$ between them, as shown in the figure. Let $C_{0} = ε_{0} a^{2} / d$ , where $ε_{0}$ is the permittivity of free space.

Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

List-I	List-II
(P) The capacitance between $S_{1}$ and $S_{4}$ , with $S_{2}$ and $S_{3}$ not connected, is	(1) $3 C_{0}$
(Q) The capacitance between $S_{1}$ and $S_{4}$ , with $S_{2}$ shorted to $S_{3}$ , is	(2) $C_{0} /2$
(R) The capacitance between $S_{1}$ and $S_{3}$ , with $S_{2}$ shorted to $S_{4}$ , is	(3) $C_{0} /3$
(S) The capacitance between $S_{1}$ and $S_{2}$ , with $S_{3}$ shorted to $S_{1}$ , and $S_{2}$ shorted to $S_{4}$ , is	(4) $2 C_{0} /3$
	(5) $2 C_{0}$

A
P $\to$ 3; Q $\to$ 2; R $\to$ 4; S $\to$ 5
B
P $\to$ 2; Q $\to$ 3; R $\to$ 2; S $\to$ 1
C
P $\to$ 3; Q $\to$ 2; R $\to$ 4; S $\to$ 1
D
P $\to$ 3; Q $\to$ 2; R $\to$ 2; S $\to$ 5

Solution

Each pair of adjacent sheets forms a parallel-plate capacitor of value $C_{0} = ε_{0} a^{2} / d$ .

(P) Between $S_{1}$ and $S_{4}$ with $S_{2}, S_{3}$ floating: three gap-capacitors in series, $C_{e q} = C_{0} /3$ . So P $\to$ 3.

(Q) $S_{2}$ and $S_{3}$ are shorted, so the middle gap is short-circuited. The result is two equal capacitors $C_{0}$ in series between $S_{1}$ and the $S_{2}$ - $S_{3}$ node, then through to $S_{4}$ : $C_{e q} = C_{0} /2$ . So Q $\to$ 2.

(R) The equivalent in this case is $2 C_{0} /3$ . So R $\to$ 4.

(S) $S_{3}$ shorted to $S_{1}$ and $S_{2}$ shorted to $S_{4}$ . Between the $S_{1} = S_{3}$ node and the $S_{2} = S_{4}$ node, all three inter-plate gaps connect these two nodes directly, so the three $C_{0}$ capacitors are in parallel: $C_{e q} = 3 C_{0}$ . So S $\to$ 1.

$∴$ Option (C) is correct.

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