Four 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.

Let us consider the four parallel square metal sheets 𝑆₁, 𝑆₂, 𝑆₃, and 𝑆₄ shown in the figure. The area of each sheet is 𝑎² and the separation between consecutive sheets is 𝑑. The capacitance of a single region formed by two adjacent plates separated by distance 𝑑 is given by:

$C_0=\frac{{\epsilon }_0{a}^2}{d}$

The system consists of three individual identical capacitors in series sequence:
- Capacitor 𝐶₁₂ between plates 𝑆₁ and 𝑆₂, with capacitance 𝐶₀.
- Capacitor 𝐶₂₃ between plates 𝑆₂ and 𝑆₃, with capacitance 𝐶₀.
- Capacitor 𝐶₃₄ between plates 𝑆₃ and 𝑆₄, with capacitance 𝐶₀.

Now, let us analyze each case given in List-I step-by-step:

(P) Capacitance between 𝑆₁ and 𝑆₄ with 𝑆₂ and 𝑆₃ not connected:
When 𝑆₂ and 𝑆₃ are isolated (unconnected), the three capacitors formed between adjacent plates (𝑆₁-𝑆₂, 𝑆₂-𝑆₃, and 𝑆₃-𝑆₄) are connected in series between terminal 𝑆₁ and terminal 𝑆₄.
The equivalent capacitance 𝐶_eq is:

$\frac{1}{C_{\text{eq}}}=\frac{1}{C_0}+\frac{1}{C_0}+\frac{1}{C_0}=\frac{3}{C_0}$

$C_{\text{eq}}=\frac{C_0}{3}$

Thus, P → 3.

(Q) Capacitance between 𝑆₁ and 𝑆₄ with 𝑆₂ shorted to 𝑆₃:
When 𝑆₂ and 𝑆₃ are shorted together, they are at the same electric potential. This means there is no potential difference across the middle gap between 𝑆₂ and 𝑆₃, so capacitor 𝐶₂₃ carries no net energy stored across its potential gap and effectively acts as a conducting link between the remaining gaps.
The system reduces to two capacitors in series: 𝐶₁₂ (between 𝑆₁ and 𝑆₂) and 𝐶₃₄ (between 𝑆₃ and 𝑆₄).

$\frac{1}{C_{\text{eq}}}=\frac{1}{C_0}+\frac{1}{C_0}=\frac{2}{C_0}$

$C_{\text{eq}}=\frac{C_0}{2}$

Thus, Q → 2.

(R) Capacitance between 𝑆₁ and 𝑆₃ with 𝑆₂ shorted to 𝑆₄:
Let the potential of terminal 𝑆₁ be 𝑉₁ and terminal 𝑆₃ be 𝑉₃.
Since 𝑆₂ and 𝑆₄ are shorted, let their common potential be 𝑉₂.
- Capacitor 𝐶₁₂ (between 𝑆₁ and 𝑆₂) has potential difference |𝑉₁ - 𝑉₂|.
- Capacitor 𝐶₂₃ (between 𝑆₂ and 𝑆₃) has potential difference |𝑉₃ - 𝑉₂|.
- Capacitor 𝐶₃₄ (between 𝑆₃ and 𝑆₄) has potential difference |𝑉₃ - 𝑉₂|.
Notice that 𝐶₂₃ and 𝐶₃₄ are connected in parallel between potential 𝑉₃ and 𝑉₂, giving a combined capacitance of:

$C_{\text{parallel}}=C_0+C_0=2C_0$

This parallel combination (2𝐶₀) is in series with capacitor 𝐶₁₂ (𝐶₀) across terminals 𝑆₁ and 𝑆₃:

$C_{\text{eq}}=\frac{C_0\times 2C_0}{C_0+2C_0}=\frac{2C_0}{3}$

Thus, R → 4.

(S) Capacitance between 𝑆₁ and 𝑆₂ with 𝑆₃ shorted to 𝑆₁ and 𝑆₂ shorted to 𝑆₄:
Here:
- Terminal A connects 𝑆₁ and 𝑆₃ together.
- Terminal B connects 𝑆₂ and 𝑆₄ together.
Let us analyze the three gaps between terminals A and B:
1. Gap between 𝑆₁ (Terminal A) and 𝑆₂ (Terminal B): forms capacitor 𝐶₁₂ = 𝐶₀.
2. Gap between 𝑆₂ (Terminal B) and 𝑆₃ (Terminal A): forms capacitor 𝐶₂₃ = 𝐶₀.
3. Gap between 𝑆₃ (Terminal A) and 𝑆₄ (Terminal B): forms capacitor 𝐶₃₄ = 𝐶₀.
All three individual capacitors are connected directly in parallel across Terminals A and B.
The net equivalent capacitance is:

$C_{\text{eq}}=C_0+C_0+C_0=3C_0$

Thus, S → 1.

Combining all the matches:
P → 3, Q → 2, R → 4, S → 1.