Question

A capacitor, \( C_1 = 6 \, \mu F \), is charged to a potential difference of \( V_1 = 5 \, \text{V} \) using a 5V battery. The battery is removed and another capacitor, \( C_2 = 12 \, \mu F \), is inserted in place of the battery. When the switch 'S' is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the charges \( q_1 \) and \( q_2 \) on the capacitors \( C_1 \) and \( C_2 \) when equilibrium condition is reached?

• A. \( q_1 = 15 \, \mu C, \, q_2 = 30 \, \mu C \)
• B. \( q_1 = 30 \, \mu C, \, q_2 = 15 \, \mu C \)
• C. \( q_1 = 10 \, \mu C, \, q_2 = 20 \, \mu C \)
• D. \( q_1 = 20 \, \mu C, \, q_2 = 10 \, \mu C \)

Hint:

When capacitors are in parallel and the switch is closed, the total charge is conserved, and the potential across all capacitors will be the same at equilibrium. Use charge conservation and the capacitance values to find the final charges on each capacitor.

Answer 1

The Correct Option is C

Step 1: At \( t = 0 \), the initial charge on \( C_1 \) is: \[ q_1 = C_1 \cdot V_1 = 6 \, \mu F \cdot 5 \, \text{V} = 30 \, \mu C \]

Step 2: After the switch 'S' is closed, charge flows until equilibrium is reached, and the total charge is distributed between the two capacitors. The final charge on each capacitor can be found using the conservation of charge and voltage. At equilibrium, the potential difference across both capacitors will be the same. Let \( V_c \) be the common potential difference at equilibrium. \[ q_1 = C_1 \cdot V_c \quad \text{and} \quad q_2 = C_2 \cdot V_c \] Using the total charge conservation: \[ q_1 + q_2 = 30 \, \mu C \quad \text{(total charge is conserved)} \] Substitute the expressions for \( q_1 \) and \( q_2 \): \[ C_1 \cdot V_c + C_2 \cdot V_c = 30 \, \mu C \] \[ V_c \cdot (C_1 + C_2) = 30 \, \mu C \] Now, solve for \( V_c \): \[ V_c = \frac{30 \, \mu C}{C_1 + C_2} = \frac{30 \, \mu C}{6 \, \mu F + 12 \, \mu F} = \frac{30 \, \mu C}{18 \, \mu F} = 1.67 \, \text{V} \]

Step 3: Finally, the charges on the capacitors are: \[ q_1 = C_1 \cdot V_c = 6 \, \mu F \cdot 1.67 \, \text{V} = 10 \, \mu C \] \[ q_2 = C_2 \cdot V_c = 12 \, \mu F \cdot 1.67 \, \text{V} = 20 \, \mu C \] Thus, the charges are \( q_1 = 10 \, \mu C \) and \( q_2 = 20 \, \mu C \), so the correct answer is option (3).

Answer 2

Step 1: Given data.
Capacitor \( C_1 = 6 \, \mu F \) is charged to a potential difference of \( V_1 = 5 \, V \).
Another capacitor \( C_2 = 12 \, \mu F \) is uncharged and connected in parallel (after removing the battery).
We have:
\[ C_1 = 6 \, \mu F, \quad C_2 = 12 \, \mu F, \quad V_1 = 5 \, V. \]

Step 2: Initial charge on the first capacitor.
\[ Q_1 = C_1 V_1 = 6 \times 5 = 30 \, \mu C. \] Initially, \( Q_2 = 0 \) (since \( C_2 \) is uncharged).
When connected, the total charge is conserved:
\[ Q_{\text{total}} = Q_1 + Q_2 = 30 \, \mu C. \]

Step 3: When switch S is closed, both capacitors are connected in parallel.
In parallel connection, both capacitors have the same potential difference after equilibrium is reached.
Let that common potential be \( V_f \).
By charge conservation:
\[ Q_{\text{total}} = (C_1 + C_2)V_f. \] Hence,
\[ 30 = (6 + 12)V_f. \] \[ 30 = 18V_f \Rightarrow V_f = \frac{30}{18} = \frac{5}{3} \, V. \]

Step 4: Final charges on each capacitor.
For capacitor \( C_1 \):
\[ q_1 = C_1 V_f = 6 \times \frac{5}{3} = 10 \, \mu C. \] For capacitor \( C_2 \):
\[ q_2 = C_2 V_f = 12 \times \frac{5}{3} = 20 \, \mu C. \]

Step 5: Verification.
Total charge after connection:
\[ q_1 + q_2 = 10 + 20 = 30 \, \mu C, \] which equals the initial total charge. Hence, the solution is consistent.

Final Answer:
\[ \boxed{q_1 = 10 \, \mu C, \, q_2 = 20 \, \mu C.} \]