Question:
A capacitor of capacitance \(10\mu\text{F}\) is charged to potential 50 V with a battery. The battery is now disconnected and an additional charge \(200\mu\text{C}\) is given to the positive plate of the capacitor. The potential difference across the capacitor will be
A capacitor of capacitance \(10\mu\text{F}\) is charged to potential 50 V with a battery. The battery is now disconnected and an additional charge \(200\mu\text{C}\) is given to the positive plate of the capacitor. The potential difference across the capacitor will be
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When battery is disconnected, charge on each plate remains fixed unless externally added.
Updated On: Apr 20, 2026
- 50 V
- 80 V
- 100 V
- 60 V
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
Initial charge: \(Q_0 = CV = 10\times10^{-6} \times 50 = 500\,\mu C\). Extra charge given = \(200\,\mu C\).
Step 2: Detailed Explanation:
Total charge on system: \(= 500 + 200 = 700\,\mu C\). For an isolated capacitor, effective charge producing potential difference: \(Q = \frac{700 + 500}{2} = 600\,\mu C\). \(V = \frac{Q}{C} = \frac{600}{10} = 60\,V\).
Step 3: Final Answer:
\(\boxed{60\ \text{V}}\)
Initial charge: \(Q_0 = CV = 10\times10^{-6} \times 50 = 500\,\mu C\). Extra charge given = \(200\,\mu C\).
Step 2: Detailed Explanation:
Total charge on system: \(= 500 + 200 = 700\,\mu C\). For an isolated capacitor, effective charge producing potential difference: \(Q = \frac{700 + 500}{2} = 600\,\mu C\). \(V = \frac{Q}{C} = \frac{600}{10} = 60\,V\).
Step 3: Final Answer:
\(\boxed{60\ \text{V}}\)
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