Question

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

• A. $5\sigma$
• B. $6\sigma$
  
• C. $\frac{5}{6}\sigma$
• D. $2\sigma$

Answer 1

The correct option is C $\frac{5}{6}\sigma$

Let $q_1$ and $q_2$ be the initial charges on the spheres,

Since charge density is same , we can write that,

$\sigma =\frac{q_1}{4\pi R^2}=\frac{q_2}{4\pi (2R{)}^2}$

So, $q_1+q_2=\sigma 4\pi R^2+\sigma 4\pi (2R{)}^2$

$q_1+q_2=20\sigma \pi R^2....\left(1\right)$

Let, final charge on them are $q_1^{\prime }$ and $q_2^{\prime }$ respectively.

From conservation of charge principle,

$q_1^{{}^{\prime }}+q_2^{{}^{\prime }}=q_1+q_2=\sigma 20\pi R^2\dots \dots \left(2\right)$

Potentials are same after connecting. So we can write that,

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

$\Rightarrow \frac{1}{4\pi {ϵ}_0}\frac{q_1^{{}^{\prime }}}{R}=\frac{1}{4\pi {ϵ}_0}\frac{q_2^{{}^{\prime }}}{2R}$

$\Rightarrow q_1^{{}^{\prime }}=\frac{q_2^{{}^{\prime }}}{2}$

Substituting it in equation $\left(2\right)$ we get

$\frac{q_2^{{}^{\prime }}}{2}+q_2^{{}^{\prime }}=\sigma 20\pi R^2$

Let the new chrage density is ${\sigma }_2^{\prime }$. So,
$q_2^{\prime }=4\pi (2R{)}^2{\sigma }_2^{\prime }$. Substituting this in the above equation we get,

$\frac{3}{2}\times {\sigma }_2^{{}^{\prime }}\times 4\pi \times 4R^2=\sigma \times 20\times \pi \times R^2$

$\therefore {\sigma }_2^{{}^{\prime }}=\frac{5}{6}\sigma$

Hence, option (c) is correct answer.