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Two metal spheres, P and Q of radii R and 2R respectively have the same surface charge density $σ$ initially. They are brought in contact and separated. What will be new surface charge density on them.

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$σ_{P} = \frac{5}{6} σ; σ_{Q} = \frac{5}{2} σ$

$σ_{P} = \frac{5}{2} σ; σ_{Q} = \frac{5}{6} σ$

$σ_{P} = \frac{5}{2} σ; σ_{Q} = \frac{5}{3} σ$

$σ_{P} = \frac{5}{3} σ; σ_{Q} = \frac{5}{6} σ$

answer is D.

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Detailed Solution

$Q = σ A$

$Q_{P} = σ (4 π R^{2})$

$Q_{Q} = σ (4 π (2 R^{2})) = σ (16 π R^{2})$

$Q_{t} = 20 π σ R^{2}$

$V = \frac{{Q_{p}}^{1}}{4 π \in_{0} R} = \frac{{Q_{Q}}^{1}}{4 π \in_{0} (2 R)}$

$\frac{{Q_{p}}^{1}}{{Q_{Q}}^{1}} = \frac{1}{2}$

$∴ Q_{P}^{1} = \frac{Q_{t}}{3}, Q_{Q}^{1} \frac{2}{3} Q_{t}$

$∴ σ_{P} = \frac{Q_{P}^{1}}{a r e a} = \frac{Q_{t}}{3 (a r e a)} = \frac{5 σ}{3}$

$σ_{Q} = \frac{Q_{Q}^{1}}{a r e a} = \frac{2 Q_{t}}{3 (a r e a)} = \frac{5}{6} σ$

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