Question

Two metal spheres one of radius $R$ and $2R$ 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?

Answer 1

Let,
$Q_1=\sigma 4\pi R^2{Q}_2=\sigma 4\pi (2R{)}^2=4(\sigma 4\pi R^2)$

Total charge after connecting = Total charge before connecting
$Q_1^{{}^{\prime }}+Q_2^{{}^{\prime }}=Q_1+Q_2$

$Q_1^{{}^{\prime }}+Q_2^{{}^{\prime }}=\sigma 4\pi R^2+4\left(\sigma 4\pi R^2\right)$

$Q_1^{{}^{\prime }}+Q_2^{{}^{\prime }}=5\left(\sigma 4\pi R^2\right)=5Q_1$

$Q_1^{{}^{\prime }}+Q_2^{{}^{\prime }}=5Q_1........\left(1\right)$

After connecting they will have same potential,

$V_1^{{}^{\prime }}=V_2^{{}^{\prime }}$

$\frac{1}{4\pi {\epsilon }_0}\frac{Q_1^{{}^{\prime }}}{R}=\frac{1}{4\pi {\epsilon }_0}\frac{Q_2^{{}^{\prime }}}{2R}$

$Q_1^{{}^{\prime }}=\frac{Q_2^{{}^{\prime }}}{2}{Q}_2^{{}^{\prime }}=2Q_1^{{}^{\prime }}$

From (1) we get,

$Q_1^{{}^{\prime }}+2Q_1^{{}^{\prime }}=5Q_1$

$Q_1^{{}^{\prime }}=\frac{5}{3}{Q}_1$ and

$Q_2^{\prime }=\frac{10}{3}{Q}_1$

And new surface chaege density becomes,

${\sigma }_1=\frac{Q_1^{{}^{\prime }}}{4\pi R^2}=\frac{5}{3}\frac{Q_1}{4\pi R^2}=\frac{5}{3}\sigma$

${\sigma }_2=\frac{Q_2^{{}^{\prime }}}{4\pi (2R{)}^2}=\frac{10}{3\times 4}\left[\frac{Q_1}{4\pi R^2}\right]=\frac{5}{6}\sigma$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (a) is the correct answer.