Two metallic solid spheres of radii R and 2R 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 :

A
5σ
B
6σ
C
$\frac{5}{6} σ$
D
2σ

Solution:

$σ = \frac{q_{1}}{4 {πR}^{2}} = \frac{q_{2}}{4 π (2 R)^{2}} \Rightarrow \frac{q_{1}}{q_{2}} = \frac{1}{4}$ . . . . . (1)

when connected by thin connecting wire

$\begin{array}{l} V_{1} = V_{2} \frac{1}{4 π \in_{0}} \frac{q_{1}^{'}}{R} = \frac{1}{4 {πϵ}_{0}} \frac{q_{2}^{'}}{(2 R)} \\ \Rightarrow \frac{q_{1}^{'}}{q_{2}^{'}} = \frac{1}{2} \\ As we know q'_{1} + q'_{2} = q_{1} + q_{2} \\ \Rightarrow \frac{3 q'_{2}}{2} = \frac{5 q_{2}}{4} \\ \Rightarrow q'_{2} = \frac{5}{6} q_{2} \\ \Rightarrow σ^{'} = \frac{5}{6} σ \end{array}$