Question

A spherical metal shell $A$ of radius $R_A$ and a solid metal sphere $B$ of radius $R_B(<R_A)$ are kept far apart and each is given charge $Q.$ Now they are connected by a thin metal wire.
Then

• A. ${\left(E_A\right)}_{\text{inside}}=0$
• B. $Q_A>Q_B$
• C. $\frac{{\sigma }_A}{{\sigma }_B}=\frac{R_B}{R_A}$
• D. $\frac{{\sigma }_A}{{\sigma }_B}=\frac{R_A}{R_B}$

Answer 1

Answer: (A), (B), (C)

$\frac{{\sigma }_A}{{\sigma }_B}=\frac{R_B}{R_A}$

Given that,
Charge on shell, $Q_A=Q$
Charge on solid sphere, $Q_B=Q$
Radius of the shell $=R_A$
Radius of the solid sphere $=R_B$
and $R_B<R_A$

The electric field inside the conductor remains zero, hence the electric field inside the metal shell will be zero.
So option $\left(a\right)$ is correct.

Now after connecting them by a thin wire, the charge will get distributed on them such that both attains equal potential.

Let the charge on shell is $Q_A$ and solid sphere is $Q_B$ after connecting them by wire.
$Q_A+Q_B=2Q...\left(1\right)$
Now
Potential of the shell = Potential of the solid sphere
$\Rightarrow \frac{k{Q}_A}{R_A}=\frac{k{Q}_B}{R_B}$
$\Rightarrow \frac{Q_A}{Q_B}=\frac{R_A}{R_B}...\left(2\right)$

As it is given that $R_A>R_B$
$\Rightarrow Q_A>Q_B$
So option $\left(b\right)$ is correct.

The surface charge density of the shell is
${\sigma }_A=\frac{Q_A}{4\pi R_A^2}$
Surface charge density of solid sphere is
${\sigma }_B=\frac{Q_B}{4\pi R_B^2}$
$\Rightarrow \frac{{\sigma }_A}{{\sigma }_B}=\frac{Q_A}{Q_B}\times \frac{R_B^2}{R_A^2}$

By using equation $\left(2\right)$
$\frac{{\sigma }_A}{{\sigma }_B}=\frac{R_B}{R_A}$
So option $\left(c\right)$ is correct.
All the options are right. So the correct option is $\left(d\right)$.