Two isolated metallic solid spheres of radii R and 2R are charged such that both have same charge density $σ$ . The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is $σ^{'}$ . The radio $\frac{σ^{'}}{σ}$ is:

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$\frac{9}{4}$

$\frac{5}{6}$

$\frac{5}{3}$

$\frac{4}{3}$

answer is C.

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Detailed Solution

After spheres are connected by a metal wire, charges will flow till their potentials become same
$V = \frac{1}{4 π \in 0} \frac{σ (4 {πr}^{2})}{r} = constant$
$\begin{array}{l} \Rightarrow σr = constant \\ σ^{'} (2 R) = σ_{2} (R) \\ \Rightarrow σ_{2} = 2 σ^{'} \end{array}$
Using charge conservation
$\begin{array}{l} σ 4 {πR}^{2} + σ (4 π {(2 R)}^{2}) = 2 σ^{'} (4 {πR}^{2}) + σ^{'} 4 π {(2 R)}^{2} \\ σ^{'} / σ = 5 / 6 \end{array}$