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Two isolated metallic solid spheres of radii $R$ and $2 R$ are charged such that both have same charge density $\sigma$. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is $\sigma^{\prime}$. The ratio $\frac{\sigma^{\prime}}{\sigma}$ is
$\frac{9}{4}$
$\frac{4}{3}$
$\frac{5}{3}$
$\frac{5}{6}$
Solution
$\frac{ Q _1^{\prime}}{4 \pi \varepsilon_0 R }=\frac{ Q _2^{\prime}}{4 \pi \varepsilon_0(2 R )}$
$\therefore Q _2^{\prime}=2 Q _1^{\prime}$
$Q _1^{\prime}+ Q _2^{\prime}= Q _1+ Q _2$
$\therefore \frac{ Q _2^{\prime}}{2}+ Q _2^{\prime}=20 \pi R ^2 \sigma$
$\frac{3}{2} Q _2^{\prime}=20 \pi R ^2 \sigma$
$\therefore \frac{ Q _2^{\prime}}{4 \pi(2 R )^2}=\frac{2}{3} \cdot \frac{20 \pi R ^2 \sigma}{16 \pi R ^2}$
$\therefore \frac{\sigma^{\prime}}{\sigma}=\frac{5}{6}$
