Question

Two capacitors \(C_1\) and \(C_2 = 2C_1\) are connected in a circuit with a switch between them as shown in the figure. Initially the switch is open and \(C_1\) holds charge Q. The switch is closed. At steady state, the charge on each capacitors will be

• A. Q, 2Q
• B. Q/3, 2Q/3
• C. 3Q/2, 3Q
• D. 2Q/3, 4Q/3

Hint:

In parallel, capacitors share charge proportional to capacitance.

Answer 1

The Correct Option is B

To determine the charge on each capacitor at steady state after the switch is closed, let's analyze the circuit and apply the principles of charge distribution in capacitors.

Understanding the Setup:

• Initially, when the switch is open, capacitor \( C_1 \) holds charge \( Q \).
• Capacitor \( C_2 \) is uncharged, and \( C_2 = 2C_1 \).

Key Concept:

When capacitors are connected in parallel with a switch closed, they will both reach the same potential difference (voltage).

Step-by-Step Solution:

1. Total charge \( Q \) is conserved. Thus, the initial charge \( Q \) on \( C_1 \) must be redistributed across both capacitors.
2. Let the final charges on \( C_1 \) and \( C_2 \) be \( q_1 \) and \( q_2 \), respectively.
3. Since they are in parallel: \(V_1 = V_2\)
4. The voltage across a capacitor is given by: \(V = \frac{q}{C}\)
5. Thus, \[ \frac{q_1}{C_1} = \frac{q_2}{2C_1} \]
6. Solving this, we get: \[ q_2 = 2q_1 \]
7. Using charge conservation: \[ q_1 + q_2 = Q \] Substituting for \( q_2 \), we have: \[ q_1 + 2q_1 = Q \Rightarrow 3q_1 = Q \Rightarrow q_1 = \frac{Q}{3} \]
8. The charge on \( C_2 \) becomes: \[ q_2 = 2q_1 = 2 \left(\frac{Q}{3}\right) = \frac{2Q}{3} \]

Conclusion:

At steady state, the charge on capacitor \( C_1 \) is \(\frac{Q}{3}\) and on capacitor \( C_2 \) is \(\frac{2Q}{3}\).

This matches the given correct answer option: Q/3, 2Q/3.