Question 15

(P) Since $S_2$ and $S_3$ are not connected (isolated), the system forms three capacitors in series. $1/C_{eq} = 1/C_0 + 1/C_0 + 1/C_0 = 3/C_0 \Rightarrow C_{eq} = C_0/3$. Thus, P matches 3.

(Q) When $S_2$ is shorted to $S_3$, they are at the same potential. The system reduces to two capacitors in series (gap 1-2 and gap 3-4). $1/C_{eq} = 1/C_0 + 1/C_0 = 2/C_0 \Rightarrow C_{eq} = C_0/2$. Thus, Q matches 2 . (R) Let $S_1$ be at potential $V_A$ and $S_3$ be at $V_B$. $S_2$ and $S_4$ are shorted to $V_C$. Capacitor $C_{12}$ is between $V_A$ and $V_C$. Capacitors $C_{23}$ and $C_{34}$ are in parallel between $V_B$ and $V_C$. Total capacitance $C = \frac{C_{12}(C_{23} + C_{34})}{C_{12} + C_{23} + C_{34}} = \frac{C_0(2C_0)}{3C_0} = 2C_0/3$. Thus, R matches 4.

(S) $S_1, S_3$ are at $V_A$ and $S_2, S_4$ are at $V_B$. Gaps 1-2, 2-3, and 3-4 are all between $V_A$ and $V_B$. They are in parallel.

$C_{eq} = C_0 + C_0 + C_0 = 3C_0$. Thus, S matches 1.

Correct Option: C