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 '$+Q$'. Now they are connected by thin metal wire. Then:

• A. $E_A^{inside}=0$
• B. $Q_A>Q_B$
• C. $\frac{{\sigma }_A}{{\sigma }_B}=\frac{R_B}{R_A}$
• D. $E_A^{onsurface}<E_B^{onsurface}$

Answer 1

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

The potential for both the objecs is same so, $\frac{K{Q}_A}{R_A}$ $=\frac{K{Q}_B}{R_B}$
or
$\frac{Q_A}{Q_B}=\frac{R_A}{R_B}$
As $R_B<R_A$ we get $Q_A>Q_B$
Above equation represents $B$ is correct option.
From Gauss law the electric field inside a spherical shell is zero so option A is correct.
Now ${\sigma }_A=\frac{Q_A}{4\pi R_A^2}$ and ${\sigma }_B=\frac{Q_B}{4\pi R_B^2}$
Thus $\frac{{\sigma }_A}{{\sigma }_B}=\frac{Q_A}{Q_B}\times (\frac{R_B}{R_A}{)}^2=\frac{R_A}{R_B}\times (\frac{R_B}{R_A}{)}^2=\frac{R_B}{R_A}$
Thus option C is also correct.
Electric fields on the surface of shell and sphere are
$E_A=\frac{{\sigma }_A}{{ϵ}_o}$
and
$E_B=\frac{{\sigma }_B}{{ϵ}_o}$
Thus $\frac{E_A}{E_B}=\frac{{\sigma }_A}{{\sigma }_B}<1$
or
$E_A<E_B$
So option D is also correct.