Two metal spheres, one of radius R and the other of radius 2R respectively have the same surface charge density σ . They are brought in contact and se - PHYSICS MCQ NEET 2019 2026 | ExamTest.live

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Two metal spheres, one of radius R and the other of radius 2 R respectively have the same surface charge density σ . They are brought in contact and separated. What will be the new surface charge densities on them?

Solution & Step-by-step Explanation

Initial charges: Q_{1} = σ (4 π R^{2}) and Q_{2} = σ (4 π (2 R)^{2}) = 4 σ (4 π R^{2}) .Total initial charge Q_{t o t a l} = Q_{1} + Q_{2} = 5 σ (4 π R^{2}) .When connected, their potentials become equal: V_{1} = V_{2} ⟹ \frac{k Q _{1}^{'}}{R} = \frac{k Q _{2}^{'}}{2 R} ⟹ Q_{2}^{'} = 2 Q_{1}^{'} .By conservation of charge: Q_{1}^{'} + Q_{2}^{'} = 3 Q_{1}^{'} = 5 σ (4 π R^{2}) ⟹ Q_{1}^{'} = \frac{5}{3} σ (4 π R^{2}) .New density σ_{1} = \frac{Q _{1}^{'}}{4 π R ^{2}} = \frac{5}{3} σ .Charge on second sphere Q_{2}^{'} = \frac{10}{3} σ (4 π R^{2}) .New density σ_{2} = \frac{Q _{2}^{'}}{4 π ( 2 R ) ^{2}} = \frac{\frac{10}{3} σ ( 4 π R ^{2} )}{16 π R ^{2}} = \frac{10}{12} σ = \frac{5}{6} σ .

Try it yourself before checking the explanation above.

Two metal spheres, one of radius R and the other of radius 2 R respectively have the same surface charge density σ . They are brought in contact and separated. What will be the new surface charge densities on them?

σ_{1} = \frac{5}{6} σ, σ_{2} = \frac{5}{6} σ

σ_{1} = \frac{5}{2} σ, σ_{2} = \frac{5}{6} σ

σ_{1} = \frac{5}{2} σ, σ_{2} = \frac{5}{3} σ

σ_{1} = \frac{5}{3} σ, σ_{2} = \frac{5}{6} σ

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