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

To find the charge on the \(20\,\mu\text{F}\) capacitor, we proceed step by step.

1. A \(10\,\mu\text{F}\) capacitor is initially connected to a \(6\,\text{V}\) battery. The charge on it when fully charged is given by:

   \(Q_1 = CV\)

2. Substituting the values:

   \(Q_1 = 10\,\mu\text{F} \times 6\,\text{V} = 60\,\mu\text{C}\)

3. The battery is then disconnected, so the \(10\,\mu\text{F}\) capacitor remains isolated with a charge of \(60\,\mu\text{C}\).
4. An uncharged \(20\,\mu\text{F}\) capacitor is connected in parallel with the charged capacitor. In a parallel connection, charge is redistributed until both capacitors reach the same potential.
5. The total capacitance of the system is:

   \(C_{\text{total}} = 10\,\mu\text{F} + 20\,\mu\text{F} = 30\,\mu\text{F}\)

6. Since the system is isolated, the total charge remains conserved:

   \(Q_{\text{total}} = 60\,\mu\text{C}\)

7. The common potential across both capacitors is:

   \(V = \dfrac{Q_{\text{total}}}{C_{\text{total}}} = \dfrac{60\,\mu\text{C}}{30\,\mu\text{F}} = 2\,\text{V}\)

8. The charge on the \(20\,\mu\text{F}\) capacitor is:

   \(Q_2 = C_2 V = 20\,\mu\text{F} \times 2\,\text{V} = 40\,\mu\text{C}\)

Hence, the charge on the \(20\,\mu\text{F}\) capacitor is \(\boxed{40\,\mu\text{C}}\).