An electron of mass ' $\mathrm{m}$ ' and charge ' $\mathrm{q}$ ' is accelerated from rest in a uniform…
An electron of mass ' $\mathrm{m}$ ' and charge ' $\mathrm{q}$ ' is accelerated from rest in a uniform electric field of strength ' $E$ '. The velocity acquired by the electron, when it travels a distance ' $\mathrm{L}$ ', is
$\sqrt{\frac{2 q E}{m L}}$
$\sqrt{\frac{2 \mathrm{qEL}}{\mathrm{m}}}$
$\sqrt{\frac{2 \mathrm{Em}}{\mathrm{qL}}}$
$\sqrt{\frac{\mathrm{qE}}{\mathrm{mL}}}$
Solution
We know
$\mathrm{F}=\mathrm{ma}$ and $\mathrm{F}=\mathrm{qE}$
$\Rightarrow q E=m a$
$\therefore \quad \frac{\mathrm{qE}}{\mathrm{m}}=\mathrm{a}$
According to equation of motion,
$\begin{aligned}
& v^2-u^2=2 a L \\
& v^2-0^2=2 a L \\
& v^2=2 a L \\
& v=\sqrt{2 a L} \\
\therefore \quad & v=\sqrt{\frac{2 q E L}{m}}
\end{aligned}$