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. $E_{Ainside}=0$
• B. $Q_A>Q_B$
• C. $\frac{{\sigma }_A}{{\sigma }_B}=\frac{R_B}{R_A}$
• D. $E_{Aonsurface}<E_{Bonsurface}$

Answer 1

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

Inside a conducting shell electric field is always zero.
Therefore, option (a) is correct. When the two are connected,their potentials become the same.
$\therefore V_A=V_B$ or $\frac{Q_A}{R_A}=\frac{Q_B}{R_B}\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R})$
Since, $R_A>R_B\therefore Q_A>Q_B$
$\therefore$ Option (b) is correct.
Potential is also equal to, $V=\frac{\sigma R}{{\epsilon }_0},V_A=V_B$
$\therefore {\sigma }_A{R}_A={\sigma }_B{R}_B$ or $\frac{{\sigma }_A}{{\sigma }_B}=\frac{R_A}{R_B}$ or ${\sigma }_A<{\sigma }_B$
$\therefore$ Option (c) is correct.
Electric field on surface, $E=\frac{\sigma }{{\epsilon }_0},$ or $E\infty \sigma$
Since, or ${\sigma }_A<{\sigma }_B\therefore E_A<E_B$
$\therefore$ Option (d) is also correct.
$\therefore$ Correct options are (a), (b), (c) and (d).