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
Answer & explanation
Correct answer: option 1
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