Question 182 of 311

Two isolated metallic solid spheres of radii $R$ and $2R$ are charged such that both have same charge density $\sigma$ . The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is ${\sigma }^{'}$. The ratio $\dfrac{{\sigma }^{'}}{\sigma }$ is :

A$\dfrac{9}{4}$B$\dfrac{4}{3}$C$\dfrac{5}{3}$D$\dfrac{5}{6}$

Show answer and solution▾

Answer: (D) $\dfrac{5}{6}$

Charge and charge density on both isolated metallic solid spheres are show below.

Let the charges and the charge densities on both the spheres after connected by thin conducting wire be ${Q}_{1}^{'},{Q}_{2}^{'},\sigma \text{and}{\sigma }^{'}$ as shown below.

Once connected, charge on both the spheres will redistribute such that potentials at the surface of both spheres is the same.
$∴\dfrac{k{Q}_{1}^{'}}{R}=\dfrac{k{Q}_{2}^{'}}{2R}$
$\Rightarrow \dfrac{{Q}_{1}^{'}}{4{\pi \epsilon }_{0}R}=\dfrac{{Q}_{2}^{'}}{4\pi {\epsilon }_{0}\left(2R\right)}$
$∴{Q}_{2}^{'}=2{Q}_{1}^{'}$
Using charge conservation,
${Q}_{1}^{'}+{Q}_{2}^{'}={Q}_{1}+{Q}_{2}$
$∴\dfrac{{Q}_{2}^{'}}{2}+{Q}_{2}^{'}=20{\pi R}^{2}\sigma$
$\dfrac{3}{2}{Q}_{2}^{'}=20{\pi R}^{2}\sigma$
$∴\dfrac{{Q}_{2}^{'}}{4\pi {\left(2R\right)}^{2}}=\dfrac{2}{3}\times \dfrac{20{\pi R}^{2}\sigma }{16{\pi R}^{2}}$
$∴\dfrac{\sigma '}{\sigma }=\dfrac{5}{6}$