Question

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 \(C_0 = \epsilon_0\frac{a^2}{d}, \)where \(\epsilon_0\) 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 S1 and S4, with S2 and S3 not connected, is	I	\(3C_0\)
Q	The capacitance between S1 and S4, with S2 shorted to S3, is	II	\(\frac{C_0}{2}\)
R	The capacitance between S1 and S3, with S2 shorted to S4, is	III	\(\frac{C_0}{3}\)
S	The capacitance between S1 and S2, with S3 shorted to S1, and S2 shorted to S4, is	IV	\(2\frac{C_0}{3}\)
			\[2C_0\]

• A. P → 3; Q → 2; R → 4; S → 5
• B. P → 2; Q → 3; R → 2; S → 1
• C. P → 3; Q → 2; R → 4; S → 1
• D. P → 3; Q → 2; R → 2; S → 5

Answer 1

The Correct Option is C

To solve the problem of matching List-I to List-II, we need to analyze the arrangement of metal sheets and the resulting capacitance based on known principles. We're considering four identical sheets S1, S2, S3, S4 with spacing \(d\) where \(d \ll a\). Given \(C_0 = \epsilon_0 \frac{a^2}{d}\), let's determine each scenario:

P. The capacitance between S1 and S4, with S2 and S3 not connected:

Effectively, this is the series combination of 3 capacitances, each \(\frac{C_0}{2}\):
\(\frac{1}{C} = \frac{1}{(\frac{C_0}{2})} + \frac{1}{(\frac{C_0}{2})} + \frac{1}{(\frac{C_0}{2})}\)
Thus, the total capacitance \(C = \frac{C_0}{3/2} = \frac{2C_0}{3}\).

Q. The capacitance between S1 and S4, with S2 shorted to S3:

By shorting S2 to S3, the central plates form a single capacitor leaving two capacitors in series:
\(\frac{1}{C} = \frac{1}{C_0} + \frac{1}{C_0}\), therefore, the capacitance becomes \(C = \frac{C_0}{2}\).

R. The capacitance between S1 and S3, with S2 shorted to S4:

When S2 and S4 are shorted, it becomes a system of capacitors in series equivalent to \(\frac{C_0}{3}\).

S. The capacitance between S1 and S2, with S3 shorted to S1, and S2 shorted to S4:

In this configuration, S3 to S1 shorted, S2 to S4 shorted forms two parallel configurations resulting in \(2C_0\).

Based on calculations, match each case:

P	The capacitance between S1 and S4, with S2 and S3 not connected, is	III	\(\frac{2C_0}{3}\)
Q	The capacitance between S1 and S4, with S2 shorted to S3, is	II	\(\frac{C_0}{2}\)
R	The capacitance between S1 and S3, with S2 shorted to S4, is	IV	\(\frac{C_0}{3}\)
S	The capacitance between S1 and S2, with S3 shorted to S1, and S2 shorted to S4, is	I	\(2C_0\)

Thus, the correct option is: \(P → 3; Q → 2; R → 4; S → 1\).