Question

A capacitor of capacitance 10 μF is connected with a battery of emf 6V. Now the battery is disconnected and another uncharged capacitor of capacitance 20 μF is connected to the capacitor. Find the charge on the 20 μF capacitor.

• A. \( \frac{30}{4} \) μC
• B. 10 μC
• C. \( \frac{20}{3} \) μC
• D. 40 μC

Hint:

When capacitors are connected in parallel, the total charge is shared according to the capacitance values.

Answer 1

The Correct Option is D

Step 1: Initial charge on the first capacitor.
The initial charge \( Q_1 \) on the first capacitor is given by: \[ Q_1 = C_1 V = 10 \, \mu\text{F} \times 6 \, \text{V} = 60 \, \mu\text{C}. \] Step 2: When the second capacitor is connected.
When the second capacitor \( C_2 = 20 \, \mu\text{F} \) is connected in parallel, the total capacitance becomes: \[ C_{\text{total}} = C_1 + C_2 = 10 \, \mu\text{F} + 20 \, \mu\text{F} = 30 \, \mu\text{F}. \] Step 3: Final charge distribution.
The charge is distributed across the two capacitors. The charge on the second capacitor is: \[ Q_2 = \frac{C_2}{C_{\text{total}}} \times Q_1 = \frac{20}{30} \times 60 = 40 \, \mu\text{C}. \] Final Answer: \[ \boxed{40 \, \mu\text{C}}. \]