⇒Electrostatics

Correct Option is (A)

The electric field at the centre due to a small element ds of the ring is given by

$\mathrm{dE}=\frac{1}{4\pi {ϵ}_0}\frac{\lambda \mathrm{ds}}{{\mathrm{r}}^2}$

The field can be resolved along $\mathrm{X}$ and $\mathrm{Y}$ directions as shown in the figure. Similarly field due to symmetrically situated another element can also be resolved along $\mathrm{X}$ and $\mathrm{Y}$ directions. The X-components are in the same direction and get added while the Y-components are in opposite directions and cancel each other. Hence, the total field due to the range is

$\begin{array}{rl}\mathrm{E}& =\int \mathrm{dE}\cos \theta \\ \therefore \mathrm{E}& =\int \frac{1}{4\pi {ϵ}_0}\frac{\lambda ds}{r^2}\cos \theta \\ & =\int \frac{1}{4\pi {ϵ}_0}\frac{\lambda d\theta \cos \theta }{{\mathrm{r}}^2}=\frac{1}{4\pi {ϵ}_0}\frac{\lambda }{\mathrm{r}}{\int }_{-\pi /2}^{\pi /2}\cos \theta d\theta \\ & =\frac{1}{4\pi {ϵ}_0}\frac{\lambda }{\mathrm{r}}[\sin \theta {]}_{0\pi /2}^{\pi /2}=\frac{1}{4\pi {ϵ}_0}\frac{\lambda }{\mathrm{r}}[1-(-1)]\\ & =\frac{1}{4\pi {ϵ}_0}\frac{\lambda }{r}2=\frac{\lambda }{2\pi {ϵ}_{0}r}\end{array}$

Correct Option is (A)

Due to charge $-$q at the centre of the shell, a charge q will be induced on the inner surface and $-$q on the outer surface. The charge on outer surface will become Q $-$ q. Hence, surface change densities will be $\frac{q}{4\pi r_1^2}$ and $\frac{\mathrm{Q}-\mathrm{q}}{4\pi {\mathrm{r}}_2^2}$

Correct Option is (B)

Force, F = Charge $\times$ Electric field = 2eE

Acceleration $=\frac{\mathrm{Force}}{\mathrm{Mass}}=\frac{2\mathrm{eE}}{4\mathrm{m}}=\frac{\mathrm{eE}}{2\mathrm{m}}$