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		List-II
(P)	The capacitance between $$S_1$$ and $$S_4$$, with $$S_2$$ and $$S_3$$ not connected, is	(1)	$$3C_0$$
(Q)	The capacitance between $$S_1$$ and $$S_4$$, with $$S_2$$ shorted to $$S_3$$, is	(2)	$$C_0/2$$
(R)	The capacitance between $$S_1$$ and $$S_3$$, with $$S_2$$ shorted to $$S_4$$, is	(3)	$$C_0/3$$
(S)	The capacitance between $$S_1$$ and $$S_2$$, with $$S_3$$ shorted to $$S_1$$, and $$S_2$$ shorted to $$S_4$$, is	(4)	$$2C_0/3$$
		(5)	$$2C_0$$

The four metal sheets $$S_1,S_2,S_3,S_4$$ are identical squares of area $$A=a^{2}$$ kept successively at equal separations $$d\,(d\ll a)$$.
Between any two adjacent sheets the capacitance is

$$C_0=\frac{\varepsilon_0A}{d}\quad -(1)$$

Thus the stack contains three individual capacitors:
$$S_1\!-\!S_2: C_{12}=C_0,\; S_2\!-\!S_3: C_{23}=C_0,\; S_3\!-\!S_4: C_{34}=C_0$$.

Required: capacitance between $$S_1$$ and $$S_4$$ with $$S_2,S_3$$ left isolated (not connected).
The three capacitors are then in series:

$$\frac{1}{C_{\text{eq}}}= \frac{1}{C_{12}}+\frac{1}{C_{23}}+\frac{1}{C_{34}} =\frac{1}{C_0}+\frac{1}{C_0}+\frac{1}{C_0}= \frac{3}{C_0}$$

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

So (P) corresponds to value $$\mathbf{C_0/3}$$, i.e. List-II (3).

Required: capacitance between $$S_1$$ and $$S_4$$ with $$S_2$$ shorted to $$S_3$$ (equipotential).
Because $$S_2$$ and $$S_3$$ form one common conductor, the region between them is at a single potential and contributes no electric field. The stack now looks like two capacitors in series:

$$S_1\;{\large\|}\;[S_2\!+\!S_3]\;{\large\|}\;S_4$$

Each of the two gaps still has capacitance $$C_0$$, hence

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

Thus (Q) gives $$\mathbf{C_0/2}$$, that is List-II (2).

Required: capacitance between $$S_1$$ and $$S_3$$ when $$S_2$$ is shorted to $$S_4$$.
Define the three electrical nodes:

Node A: $$S_1$$ (one terminal)
Node B: $$S_3$$ (other terminal)
Node C: $$S_2$$ shorted to $$S_4$$ (floating internal conductor)

Network of capacitors:

$$S_1\!\!-\!\!S_2 : C_{AC}=C_0,\quad S_2\!\!-\!\!S_3 : C_{CB}=C_0,\quad S_3\!\!-\!\!S_4 : C_{CB}'=C_0$$

Capacitances between the same pair of nodes add, so between nodes B and C we have

$$C_{BC}=C_{CB}+C_{CB}'=2C_0$$

Now A-C and C-B are in series. The equivalent between A and B is therefore

$$C_{\text{eq}}=\frac{C_{AC}\,C_{BC}}{C_{AC}+C_{BC}} =\frac{C_0\,(2C_0)}{C_0+2C_0} =\frac{2C_0}{3}$$

Hence (R) gives $$\mathbf{2C_0/3}$$, i.e. List-II (4).

Required: capacitance between $$S_1$$ and $$S_2$$ when $$S_3$$ is shorted to $$S_1$$ and $$S_2$$ is shorted to $$S_4$$.
Thus we have two final nodes:

Node A: $$S_1$$ and $$S_3$$ (same potential)
Node B: $$S_2$$ and $$S_4$$ (same potential)

All three original capacitors now connect directly between these two nodes:

$$C_{12}=C_0,\; C_{23}=C_0,\; C_{34}=C_0$$

They are in parallel, so

$$C_{\text{eq}}=C_{12}+C_{23}+C_{34}=3C_0$$

Therefore (S) corresponds to $$\mathbf{3C_0}$$, which is List-II (1).

Collecting the four matches:

P → 3,  Q → 2,  R → 4,  S → 1

The option that lists this combination is Option C.