Question:
Two separated conducting spheres \(S_1\) and \(S_2\), of radii \(3R/4\) and \(R/4\) having \(15 \mu C\) and \(-3 \mu C\) charges respectively are at a large distance. They are now connected by a conducting wire. After a long time, the charges on \(S_1\) and \(S_2\) respectively are:
Two separated conducting spheres \(S_1\) and \(S_2\), of radii \(3R/4\) and \(R/4\) having \(15 \mu C\) and \(-3 \mu C\) charges respectively are at a large distance. They are now connected by a conducting wire. After a long time, the charges on \(S_1\) and \(S_2\) respectively are:
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The charge on conductors in equilibrium distributes according to their capacitance, which is directly proportional to their radius.
Updated On: Mar 13, 2025
- \(2 \mu C, 10 \mu C\)
- \(4 \mu C, 8 \mu C\)
- \(6 \mu C, 6 \mu C\)
- \(9 \mu C, 3 \mu C\)
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The Correct Option is D
Solution and Explanation
Step 1: Calculate the total charge before connection. Total charge \( Q_{total} = 15 \mu C - 3 \mu C = 12 \mu C\).
Step 2: Determine the final charges when connected by a wire. Since the potential becomes equal and the charge is distributed proportional to the radius of each sphere, the larger sphere \(S_1\) (radius \(3R/4\)) will acquire a larger fraction of the total charge: \[ Q_1 = \frac{3}{4} \times 12 \mu C = 9 \mu C, \] \[ Q_2 = \frac{1}{4} \times 12 \mu C = 3 \mu C. \]
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