F our identical thin, square metal sheets, 𝑆₁, 𝑆₂, 𝑆₃ and 𝑆₄, each of side 𝑎 are kept parallel to each other with equal distance 𝑑 (≪ 𝑎) between them, as shown in the figure. Let 𝐶₀ = 𝜀₀𝑎²/𝑑, where 𝜀₀ 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 𝑆1 and 𝑆4, with 𝑆2 and 𝑆3 not connected, is	(1) 3𝐶0
(Q) The capacitance between 𝑆1 and 𝑆4, with 𝑆2 shorted to 𝑆3, is	(2) 𝐶0/2
(R) The capacitance between 𝑆1 and 𝑆3, with 𝑆2 shorted to 𝑆4, is	(3) 𝐶0/3
(S) The capacitance between 𝑆1 and 𝑆2, with 𝑆3 shorted to 𝑆1, and 𝑆2 shorted to 𝑆4, is	(4) 2𝐶0/3
	(5) 2𝐶0

The correct option is P → 3; Q → 2; R → 4; S → 1.

Explanation:

The four parallel metal plates S₁, S₂, S₃, and S₄ shown in the figure form three individual parallel-plate capacitors:
1. Capacitor C₁₂ between plates S₁ and S₂
2. Capacitor C₂₃ between plates S₂ and S₃
3. Capacitor C₃₄ between plates S₃ and S₄
Since each sheet is of side a and the equal distance between adjacent sheets is d, each of these capacitors has an individual capacitance:
$C_{12}=C_{23}=C_{34}=C_0=\frac{{\epsilon }_0{a}^2}{d}$

(P) The capacitance between S₁ and S₄ with S₂ and S₃ not connected:
When plates S₂ and S₃ are isolated (not connected to any external potential or to each other), the three capacitors C₁₂, C₂₃, and C₃₄ are connected in series between plates S₁ and S₄.
The equivalent capacitance C_eq is given by:
$\frac{1}{C_{\mathrm{eq}}}=\frac{1}{C_0}+\frac{1}{C_0}+\frac{1}{C_0}=\frac{3}{C_0}$
$C_{\mathrm{eq}}=\frac{C_0}{3}$
Therefore, P → 3.

(Q) The capacitance between S₁ and S₄ with S₂ shorted to S₃:
When S₂ and S₃ are connected by a wire (shorted), they are at the same electric potential. This shorts out the middle capacitor C₂₃, meaning no charge is stored on it.
The remaining two capacitors, C₁₂ (between S₁ and S₂) and C₃₄ (between S₃ and S₄), are connected in series.
The equivalent capacitance C_eq is:
$\frac{1}{C_{\mathrm{eq}}}=\frac{1}{C_0}+\frac{1}{C_0}=\frac{2}{C_0}$
$C_{\mathrm{eq}}=\frac{C_0}{2}$
Therefore, Q → 2.

(R) The capacitance between S₁ and S₃ with S₂ shorted to S₄:
Let the terminal connected to S₁ be A and the terminal connected to S₃ be B. Let plates S₂ and S₄ be shorted to a common node Y.
The capacitors are connected as follows:
- C₁₂ is connected between S₁ (terminal A) and S₂ (node Y).
- C₂₃ is connected between S₃ (terminal B) and S₂ (node Y).
- C₃₄ is connected between S₃ (terminal B) and S₄ (node Y).
Thus, C₂₃ and C₃₄ are in parallel between terminal B and node Y, giving an equivalent capacitance of:
$C_{\mathrm{BY}}=C_{23}+C_{34}=C_0+C_0=2C_0$
This combination C_BY is connected in series with C₁₂ (which is between terminal A and node Y).
The overall equivalent capacitance C_eq between A and B is:
$C_{\mathrm{eq}}=\frac{C_{12}\cdot C_{\mathrm{BY}}}{C_{12}+C_{\mathrm{BY}}}=\frac{C_0\cdot 2C_0}{C_0+2C_0}=\frac{2}{3}{C}_0$
Therefore, R → 4.

(S) The capacitance between S₁ and S₂ with S₃ shorted to S₁, and S₂ shorted to S₄:
Let plates S₁ and S₃ be connected together to form terminal A, and plates S₂ and S₄ be connected together to form terminal B.
The three capacitors are connected between these terminals as:
- C₁₂ between S₁ (terminal A) and S₂ (terminal B).
- C₂₃ between S₃ (terminal A) and S₂ (terminal B).
- C₃₄ between S₃ (terminal A) and S₄ (terminal B).
All three capacitors are connected in parallel between terminals A and B.
The equivalent capacitance C_eq is:
$C_{\mathrm{eq}}=C_{12}+C_{23}+C_{34}=C_0+C_0+C_0=3C_0$
Therefore, S → 1.