An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet…

The value of the electric field due to the uniformly charged plane sheet is given by

$E=\frac{\sigma }{2{\epsilon }_0} ...\left(1\right)$

Thus, the force on the electron is given by

$\begin{array}{rcl}F& =& -eE\\ & =& -\frac{\sigma e}{2{\epsilon }_0} ...\left(2\right)\end{array}$

So, the acceleration of the electron is given by

$a=-\frac{\sigma e}{2{\epsilon }_0 m}$

Given, $u=1 \mathrm{m} {\mathrm{s}}^{-1}$, $t=1 \mathrm{s}$ and $S=-1 \mathrm{m}$

Using the equation of motion, it can be written that

$$\begin{gathered} -1=1\times 1-\frac{1}{2}\times \frac{\sigma e}{2{\epsilon }_0 m}\times (1{)}^2 \\ \Rightarrow \sigma =8\frac{{\epsilon }_0 m}{e} \mathrm{C} {\mathrm{m}}^{-2} \end{gathered}$$$$

$\therefore \alpha =8$.