Discussion Forum

A spherical metal shell A of radius RAand 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

Answer:

Explanation:

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 $\frac{Q_{A}}{R_{A}}=\frac{Q_{B}}{R_{B}}$   $\left( \therefore V= \frac{1}{4 \pi \epsilon_{0}}.\frac{Q}{R}\right)$

 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 $\frac{ \sigma _{A}}{\sigma_{B}}= \frac{ R_{B}}{R_{A}}$

 $\therefore$  Option (c) is correct

 Electric field on surface

 $E= \frac{\sigma}{E_{0}}$ or  $E \propto \sigma$

 or $\sigma_{A} < \sigma_{B}$

 Since, $\sigma_{A} < \sigma_{B}$

 $\therefore$   $E_{A} < E_{B}$

 $\therefore$ Option (d) is also correct

 $\therefore$ Correct options are (a),(b) ,(c) and (d) 

 Analysics of Question

 (i) Question is simple

 (ii)  $E= \frac{1}{4 \pi \epsilon_{0}}\frac{Q}{R^{2}}=\frac{\sigma}{\epsilon_{0}}$ ( on surface)

 As   $\sigma$= surface charge density

 = $\frac{Q}{4 \pi R^{2}}$

 (iii)   $V= \frac{1}{4 \pi \epsilon_{0}}\frac{Q}{R^{}}=\frac{\sigma R}{\epsilon_{0}}$