Two isolated metallic solid spheres of radii $\mathrm{R}$ and $2 \mathrm{R}$ 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 :
Solution
<p>$\sigma = {{{Q_1}} \over {4\pi {R^2}}} = {{{Q_2}} \over {4\pi {{(2R)}^2}}}$</p>
<p>Now $Q{'_2} = {Q_{total}}\left[ {{{{R_2}} \over {{R_1} + {R_2}}}} \right]$</p>
<p>$= ({Q_1} + {Q_2})\left[ {{{2R} \over {3R}}} \right]$</p>
<p>$= \sigma (20\pi {R^2}){2 \over 3}$</p>
<p>$\therefore$ $$\sigma {'_2} = {{Q{'_2}} \over {4\pi {{(2R)}^2}}} = {{\sigma (20\pi {R^2}){2 \over 3}} \over {16\pi {R^2}}}$$</p>
<p>$= {5 \over 4} \times {2 \over 3}\sigma$</p>
<p>$= {5 \over 6}\sigma$</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Potential and Capacitance
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