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Two metal spheres, one of radus $R$ and the other of radius $2 R$ respectively have the same surface charge density $\sigma$. They are brought in contact and separated. What will be the new surface charge densities on them?
$\sigma_{1}=\frac{5}{6} \sigma, \sigma_{2}=\frac{5}{2} \sigma$
$\sigma_{1}=\frac{5}{2} \sigma, \sigma_{2}=\frac{5}{6} \sigma$
$\sigma_{1}=\frac{5}{2} \sigma, \sigma_{2}=\frac{5}{3} \sigma$
$\sigma_{1}=\frac{5}{3} \sigma, \sigma_{2}=\frac{5}{6} \sigma$
Solution
Total charge $=\sigma \times 4 \pi R^{2}+\sigma \times 4 \pi(2 R)^{2}=20 \mathrm{\sigma \pi R}^{2}$
$\frac{\mathrm{Q}_{\mathrm{A}}}{\mathrm{Q}_{\mathrm{B}}}=\frac{1}{2}$
$\mathrm{Q}_{\mathrm{A}}=\frac{20}{3} \sigma \pi \mathrm{R}^{2}$ and $\mathrm{Q}_{\mathrm{B}}=\frac{40}{3} \sigma \pi \mathrm{R}^{2}$
$\sigma_{\mathrm{A}}=\frac{\mathrm{Q}_{\mathrm{A}}}{\text { area }}=\frac{20}{3} \frac{\sigma \pi \mathrm{R}^{2}}{4 \pi \mathrm{R}^{2}}=\frac{5 \sigma}{3}$
$\sigma_{\mathrm{B}}=\frac{40 \sigma \pi R^{2}}{4 \pi(2 R)^{2}}=\frac{5 \sigma}{6}$
