Two concentric spherical shells have radii R and 2R. The outer shell is grounded and the inner one is given a charge +Q. A small particle having mass m and charge –q enters the outer shell through a small hole in it. The speed of the charge entering the shell was u and its initial line of motion was at a distance $a=\sqrt{2}R$ from the centre. Assume that distribution of charge on the spheres do not change due to presence of charged particle. Choose the CORRECT statement(s):

 

Charge –Q is induced on the inner surface of the outer shell. There is no charge on the outer surface of the outer shell as it is grounded.
An electric field exists in the space between the two shells.
$E=K\frac{Q}{x^2}$ for $R<x<2R$  
Just after the charge enters, it experience a force due to this electric field directed towards the center $F=K\frac{Qq}{{\left(2R\right)}^2}$
.
Component of this force perpendicular to the direction of instantaneous velocity u is
$F_{\perp }=F\sin \theta =F\frac{\sqrt{2}R}{2R}=\frac{F}{\sqrt{2}}=K\frac{Qq}{4\sqrt{2}{R}^2}$ 
If radius of curvature of the path is r
$\frac{m{u}^2}{r}=F_{\perp }$ 
$\therefore \frac{m{u}^2}{r}=\frac{KqQ}{4\sqrt{2}{R}^2}$ 
$\therefore r=\frac{16\sqrt{2}\pi {\epsilon }_0{R}^{2}m{u}^2}{Qq}$ 
b) The potential difference between the two spheres is
$V_{inner}-V_{outer}=(K\frac{Q}{R}-\frac{KQ}{2R})-(\frac{KQ}{2R}-\frac{KQ}{2R})=\frac{KQ}{2R}$ 
Energy conservation for the charge entering the shell gives:
$\frac{1}{2}m{v}^2+(-q)v_{inner}=\frac{1}{2}m{u}^2+(-q)V_{outer}$ 
$\therefore \frac{1}{2}m{v}^2=\frac{1}{2}m{u}^2+q(V_{inner}-V_{outer})$ 
$\frac{1}{2}m{v}^2=\frac{1}{2}m{u}^2+\frac{Qq}{8\pi {\epsilon }_{0}R}$ 
$\therefore v=\sqrt{u^2+\frac{Qq}{4\pi {\epsilon }_{0}mR}}$