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. $E_{A\text{inside}} = 0$
• B. $Q_A > Q_B$
• C. $\frac{\sigma_A}{\sigma_B}\, = \frac{R_B}{R_A}$
• D. $E_{A\text{on surface}} < E _{B\text{on surface}}$

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 $\frac{Q_A}{R_A}= \frac{Q_B}{R_B} \frac{1}{4 \pi \varepsilon_0}\, \frac{Q}{R}$
Since, $R_A > R_B \therefore Q_A > Q_B $
$\therefore $ Option (b) is correct.
Potential is also equal to, $V=\frac{\sigma R}{\varepsilon_0},\, V_A=V_B$
$\therefore \sigma_A R_A=\sigma_B R_B$ 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, $E=\frac{\sigma }{\varepsilon_0},$ or $E \infty \sigma$
Since, or $ \sigma_A < \sigma_B \therefore E_A < E_B $