Question

Two isolated metallic solid spheres of radii $R$ and $2R$ are charged such that both of these have same charge density $\rho .$ The spheres are located far away from each other, and connected by a thin conducting wire. Then the new charge density on the bigger sphere is:

Answer 1

Charge of sphere of radius R.

$Q_1=4\pi R^2\sigma$

Charge of sphere of $2R=4\pi {\left(2R\right)}^2\sigma$

$C_1=4\pi {\epsilon }_{0}R$

$C_2=4\pi {\epsilon }_0\left(2R\right)$

So, common potential

$=\frac{C_1{V}_1+C_2{V}_2}{CH{C}_2}$

$=\frac{Q_1+Q_2}{C_1+C_2}$

$=\frac{4\pi R^2\sigma +4\pi {\left(2R\right)}^2\sigma }{4\pi {\epsilon }_{0}R+4\pi {\epsilon }_0\left(2R\right)}$

$=\frac{5\sigma R}{3{\epsilon }_0}$

So, new charge on Bigger sphere

$Q_2^1=C_{2}V$

$=4\pi {\epsilon }_0\left(2R\right)\times \text{common potential}$

$=4\pi {\epsilon }_0\left(2R\right)\times \frac{5\sigma R}{3{\epsilon }_0}$

$Q_2^1=\frac{40\pi R^2\sigma }{3}$

New charge density on bigger one $=\frac{\frac{40\pi R^2\sigma }{3}}{A_2}$

$=\frac{4\pi R^2\sigma }{3}\times \frac{1}{4\pi {\left(2R\right)}^2}=\frac{5}{6}\sigma$