An electron of mass ' $m$ ' and charge ' $q$ ' is accelerated from rest in a uniform electric field of…
An electron of mass ' $m$ ' and charge ' $q$ ' is accelerated from rest in a uniform electric field of intensity ' $E$ '. The velocity acquired by it as it travels a distance ' $l$ ' is ' $v$ '. The ratio $\frac{\mathrm{q}}{\mathrm{m}}$ in terms of $\mathrm{E}, l$ and v is
$\frac{\mathrm{v}^2}{2 \mathrm{E} l}$
$\frac{\mathrm{v}^2 l}{2 \mathrm{E}}$
$\frac{2 \mathrm{E}}{\mathrm{v}^2 l}$
$\frac{\mathrm{v}^2 l}{\mathrm{E}}$
Solution
We know
$\begin{aligned}
& F=m a \text { and } F=q E \\
& \Rightarrow q E=m a \\
& \therefore \frac{\mathrm{qE}}{\mathrm{~m}}=\mathrm{a} \quad ...(i)
\end{aligned}$
According to equation of motion, $v^2-u^2=2 a L$
$\begin{aligned}
& v^2-0^2=2 a L \\
& v^2=2 a L \\
& a=\frac{v^2}{2 L} \quad ...(ii)
\end{aligned}$
From (i) and (ii),
$\begin{aligned}
& \frac{\mathrm{v}^2}{2 \mathrm{~L}}=\frac{\mathrm{qE}}{\mathrm{~m}} \\
& \frac{\mathrm{q}}{\mathrm{~m}}=\frac{\mathrm{v}^2}{2 \mathrm{EL}}=\frac{\mathrm{v}^2}{2 \mathrm{E} \ell} \quad(\because \mathrm{~L}=\ell)
\end{aligned}$