Question

Two isolated metallic solid spheres of radii $R$ and $2R$ are charged such that both of these have same density $\sigma$. The spheres are located far away from each other and connected by a thin connecting wire. The ratio of new charge density to the initial charge density on the bigger sphere is______.

Answer 1

Answer: 0.83

Radius of the smaller sphere is $R$,
$\therefore q_1=\sigma \left(4\pi R^2\right)$
Similarly, for bigger sphere,
$q_2=\sigma (4\pi )(2R{)}^2=16\sigma \pi R^2$
Total charge (q) on both the spheres is,
$q=q_1+q_2=20\sigma \pi R^2$
As, capacitance for spherical capacitors,
$C=4\pi {\epsilon }_{0}R$
$\therefore \frac{C_1}{C_2}=\frac{R}{2R}$
$\therefore \frac{C_1}{C_2}=\frac{1}{2}$
Now after connecting, charge is distributed in the ratio of their capacities.
$\therefore \frac{q_1^{\prime }}{q_2^{\prime }}=\frac{1}{2}$
$\therefore \frac{q_1^{\prime }+q_2^{\prime }}{q_2^{\prime }}=\frac{1+2}{2}=\frac{3}{2}$
As total charge is conserved,
$q_1^{\prime }+q_2^{\prime }=q_1+q_2=q$
$\therefore q_2^{\prime }=\frac{2q}{3}=\frac{40}{3}\sigma \pi R^2$
Now,
${\sigma }^{\prime }=\frac{q_2^{\prime }}{4\pi (2R{)}^2}=\frac{\frac{40}{3}\sigma \pi R^2}{16\pi R^2}=\frac{5}{6}\sigma$
$\frac{{\sigma }^{\prime }}{\sigma }=\frac{5}{6}=0.83$