Two metallic solid spheres of radii R and 3R are charged such that both of them have same charge density $σ .$ If the spheres are located far away from each other and connected by a thin conducting wire, the new charge density on bigger sphere is

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$6 σ$

$\frac{5}{6} σ$

$2 σ$

$5 σ$

answer is C.

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Detailed Solution

$\begin{matrix} σ = \frac{q}{A} \\ q_{1} = 4 π R^{2} σ, q_{2} = 36 π R^{2} σ \\ w h e n w i r e i s c o n n e c t e d p o t e n t i a l i s s a m e, V_{1} = V_{2} \Rightarrow K \frac{q_{1}^{1}}{r_{1}} = K \frac{q_{2}^{1}}{r_{2}} \Rightarrow \frac{q_{1}^{1}}{q_{2}^{1}} = \frac{r_{1}}{r_{2}} = \frac{R}{3 R} = \frac{1}{3} \\ q_{1} + q_{2} = q_{1}^{1} + q_{2}^{1} \\ q_{1} + q_{2} = 4 π R^{2} σ + 36 π R^{2} σ = 40 π R^{2} σ = q_{1}^{1} + q_{2}^{1} \\ q_{2}^{1} = \frac{3}{4} \times 40 π R^{2} σ = 30 π R^{2} σ \\ σ^{1} = \frac{q_{2}^{1}}{36 π R^{2}} = \frac{30 π R^{2} σ}{36 π R^{2}} = \frac{5 σ}{6} \end{matrix}$