An electron is projected from a distance d and with initial velocity u parallel to a uniformly charge flat conducting plate as shown. It strikes the plate after travelling a distance $\ell$ along the direction of projection. The surface charge density of the conducting plate is equal to

$\frac{2 d \varepsilon_0 m u^2}{e \ell^2}$

For the motion along Y axis of the electron

$d=\frac{1}{2} \frac{eE}{m} . t^2$

Where E is the electric field due to the charged plate given by

E = $\frac{\sigma}{\varepsilon_0}, \quad \sigma$ is the surface density of charge of the plate.

The time taken by the electron is

t = $\frac{\ell}{u}$          . . . (3)

From (1), (2) and (3) we obtain,

$d=\frac{1}{2} \frac{e \sigma}{\varepsilon_0 m}\left(\frac{\ell}{u}\right)^2$

$\Rightarrow \sigma=\frac{2 \mathrm{~d} \varepsilon_0 mu^2}{e \ell^2}$

∴ (A) is correct