Capacitance Configuration of Four Parallel Metal Sheets with Shorting Combinations

To find the equivalent capacitance for each configuration, we consider the four parallel metal plates $S_1, S_2, S_3, S_4$. The three regions between adjacent plates form three identical parallel plate capacitors:

• Capacitor $C_1$ between $S_1$ and $S_2$, with capacitance $C_1 = C_0 = \frac{\varepsilon_0 a^2}{d}$.
• Capacitor $C_2$ between $S_2$ and $S_3$, with capacitance $C_2 = C_0 = \frac{\varepsilon_0 a^2}{d}$.
• Capacitor $C_3$ between $S_3$ and $S_4$, with capacitance $C_3 = C_0 = \frac{\varepsilon_0 a^2}{d}$.

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Analysis of List-I Items

(P) Capacitance between $S_1$ and $S_4$, with $S_2$ and $S_3$ not connected:

• The intermediate plates $S_2$ and $S_3$ are isolated.
• Consequently, the three capacitors $C_1, C_2,$ and $C_3$ are connected in series between $S_1$ and $S_4$.
• The equivalent capacitance $C_{\text{eq}}$ is: $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{C_0} + \frac{1}{C_0} + \frac{1}{C_0} = \frac{3}{C_0}$ $C_{\text{eq}} = \frac{C_0}{3}$
• Thus, $\text{P} \rightarrow (3)$.

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(Q) Capacitance between $S_1$ and $S_4$, with $S_2$ shorted to $S_3$:

• Since $S_2$ is shorted to $S_3$, $V(S_2) = V(S_3)$. The potential difference across $C_2$ is zero, meaning $C_2$ carries no charge.
• The remaining capacitors $C_1$ (between $S_1$ and $S_2$) and $C_3$ (between $S_3$ and $S_4$) are connected in series through the common shorted node $(S_2 \cup S_3)$.
• The equivalent capacitance $C_{\text{eq}}$ is: $C_{\text{eq}} = \frac{C_1 \cdot C_3}{C_1 + C_3} = \frac{C_0 \cdot C_0}{C_0 + C_0} = \frac{C_0}{2}$
• Thus, $\text{Q} \rightarrow (2)$.

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(R) Capacitance between $S_1$ and $S_3$, with $S_2$ shorted to $S_4$:

• Let $A$ be terminal $S_1$, $B$ be terminal $S_3$, and $X$ be the shorted node $(S_2 \cup S_4)$.
• Analyzing the connections:
  • $C_1$ is connected between terminal $A$ ($S_1$) and node $X$ ($S_2$).
  • $C_2$ is connected between node $X$ ($S_2$) and terminal $B$ ($S_3$).
  • $C_3$ is connected between terminal $B$ ($S_3$) and node $X$ ($S_4$).
• Capacitors $C_2$ and $C_3$ are connected in parallel between node $X$ and terminal $B$: $C_{23} = C_2 + C_3 = C_0 + C_0 = 2C_0$
• This parallel combination $C_{23}$ is in series with $C_1$: $C_{\text{eq}} = \frac{C_1 \cdot C_{23}}{C_1 + C_{23}} = \frac{C_0 \cdot (2C_0)}{C_0 + 2C_0} = \frac{2C_0}{3}$
• Thus, $\text{R} \rightarrow (4)$.

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(S) Capacitance between $S_1$ and $S_2$, with $S_3$ shorted to $S_1$, and $S_2$ shorted to $S_4$:

• Plates $S_1$ and $S_3$ are shorted to form terminal $A = S_1 \cup S_3$.
• Plates $S_2$ and $S_4$ are shorted to form terminal $B = S_2 \cup S_4$.
• Analyzing the connections:
  • $C_1$ is connected between $S_1$ (terminal $A$) and $S_2$ (terminal $B$).
  • $C_2$ is connected between $S_2$ (terminal $B$) and $S_3$ (terminal $A$).
  • $C_3$ is connected between $S_3$ (terminal $A$) and $S_4$ (terminal $B$).
• All three capacitors $C_1, C_2,$ and $C_3$ are connected in parallel between terminals $A$ and $B$: $C_{\text{eq}} = C_1 + C_2 + C_3 = C_0 + C_0 + C_0 = 3C_0$
• Thus, $\text{S} \rightarrow (1)$.

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Conclusion

The correct matching is: $\text{P} \rightarrow 3; \quad \text{Q} \rightarrow 2; \quad \text{R} \rightarrow 4; \quad \text{S} \rightarrow 1$

This corresponds to Option C.