JEE Advanced 2024 Paper 1 · Q15 · Capacitors

Question

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 $d\,(\ll a)$ between them, as shown in the figure. Let $C_0 = \varepsilon_0 a^2/d$, where $\varepsilon_0$ 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:
(P) The capacitance between $S_1$ and $S_4$, with $S_2$ and $S_3$ not connected, is
(Q) The capacitance between $S_1$ and $S_4$, with $S_2$ shorted to $S_3$, is
(R) The capacitance between $S_1$ and $S_3$, with $S_2$ shorted to $S_4$, is
(S) The capacitance between $S_1$ and $S_2$, with $S_3$ shorted to $S_1$, and $S_2$ shorted to $S_4$, is

List-II: (1) $3C_0$; (2) $C_0/2$; (3) $C_0/3$; (4) $2C_0/3$; (5) $2C_0$.

[Figure: four parallel sheets $S_1, S_2, S_3, S_4$ separated by equal gap $d$ each.]

SolveKar AI · Accepted Answer

Each adjacent pair of sheets forms a capacitor of $C_0$ (gap $d$, area $a^2$). The four sheets give three series gaps $C_0$—$C_0$—$C_0$.

(P) $S_2, S_3$ floating ⇒ three $C_0$'s in series: $C = C_0/3$. → (3).

(Q) $S_2$–$S_3$ shorted ⇒ the middle gap is shorted out; two $C_0$'s ($S_1$–$S_2$ and $S_3$–$S_4$) in series: $C = C_0/2$. → (2).

(R) $S_2$ shorted to $S_4$, measure $S_1$–$S_3$. Between nodes $S_1$ and $S_3$: $S_1$–$S_2$ ($C_0$) in series with [$S_2$–$S_3$ direct $C_0$, parallel with $S_2(=S_4)$–$S_3$ another $C_0$] $= C_0$ in series with $2C_0 = 2C_0/3$. → (4).

(S) $S_3$ shorted to $S_1$ and $S_2$ shorted to $S_4$. All three gaps connect a node in $\{S_1,S_3\}$ to a node in $\{S_2,S_4\}$, so the three $C_0$'s are in parallel: $C = 3C_0$. → (1). Mapping P→3, Q→2, R→4, S→1 = option (C).

JEE Advanced 2024 Paper 1 · Q15 · Capacitors

Solution