Question

Two isolated metallic solid spheres of radii $R$ and $2 R$ 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

Radius of the smaller sphere is $R$,
$\therefore q _{1}=\sigma\left(4 \pi R ^{2}\right)$
Similarly, for bigger sphere,
$q _{2}=\sigma(4 \pi)(2 R )^{2}=16 \sigma \pi R ^{2}$
Total charge (q) on both the spheres is,
$q = q _{1}+ q _{2}=20 \sigma \pi R ^{2}$
As, capacitance for spherical capacitors,
$C =4 \pi \varepsilon_{0} R$
$\therefore \frac{C_{1}}{C_{2}}=\frac{R}{2 R}$
$\therefore \frac{C_{1}}{C_{2}}=\frac{1}{2}$
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 $
$\therefore q _{2}^{\prime}=\frac{2 q }{3}=\frac{40}{3} \sigma \pi R ^{2}$
Now,
$\sigma^{\prime}=\frac{ q _{2}^{\prime}}{4 \pi(2 R )^{2}}=\frac{\frac{40}{3} \sigma \pi R ^{2}}{16 \pi R ^{2}}=\frac{5}{6} \sigma $
$\frac{\sigma^{\prime}}{\sigma}=\frac{5}{6}=0.83$