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 | \(2\frac{C_0}{3}\) |
\[2C_0\] |
P → 2; Q → 3; R → 2; S → 1
P → 3; Q → 2; R → 2; S → 5
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: \(P → 3; Q → 2; R → 4; S → 1\).
To solve the problem, we need to find the capacitance values for different configurations of four identical thin, square metal sheets arranged parallel to each other.
1. Understanding the Capacitance of a Single Capacitor:
The capacitance of two parallel plates each of side \(a\) separated by distance \(d\) is given by:
\( C_0 = \varepsilon_0 \frac{a^2}{d} \)
2. Analyzing the Different Configurations:
(P) Capacitance between \( S_1 \) and \( S_4 \), with \( S_2 \) and \( S_3 \) not connected:
Here, three capacitors \( C_0 \) are in series between the plates:
\( C = \frac{C_0}{3} \)
This corresponds to option (3).
(Q) Capacitance between \( S_1 \) and \( S_4 \), with \( S_2 \) shorted to \( S_3 \):
Shorting \( S_2 \) and \( S_3 \) makes them a single plate, so two capacitors \( C_0 \) in series remain:
\( C = \frac{C_0}{2} \)
This corresponds to option (2).
(R) Capacitance between \( S_1 \) and \( S_3 \), with \( S_2 \) shorted to \( S_4 \):
Shorting \( S_2 \) and \( S_4 \) makes them a single conductor.
The effective capacitance is two capacitors in parallel but with adjusted distances, resulting in:
\( C = \frac{2C_0}{3} \)
This corresponds to option (4).
(S) Capacitance between \( S_1 \) and \( S_2 \), with \( S_3 \) shorted to \( S_1 \) and \( S_2 \) shorted to \( S_4 \):
Shorting these plates creates two capacitors \( C_0 \) in parallel:
\( C = 2C_0 \)
This corresponds to option (1).
Final Matching:
\( P \to 3; Q \to 2; R \to 4; S \to 1 \)
Final Answer:
Option (C) \( P \to 3; Q \to 2; R \to 4; S \to 1 \)
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