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An electron of mass ' $m$ ' and charge ' $q$ ' is accelerated from rest in a uniform electric field of strength ' $E$ '. The velocity acquired by the electron when it travels a distance ' $L$ ' is
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The correct answer is:
$\sqrt{\frac{2 \mathrm{qEL}}{\mathrm{m}}}$
Force on the electron $\mathrm{F}=\mathrm{qE}$
Work done by the force $\mathrm{W}=\mathrm{qEL}$
Work done is equal to gain in kinetic energy
$$
\begin{aligned}
& \therefore \frac{1}{2} \mathrm{mv}^2=\mathrm{qEL} \\
& \therefore \mathrm{v}=\sqrt{\frac{2 \mathrm{qEL}}{\mathrm{m}}}
\end{aligned}
$$
Work done by the force $\mathrm{W}=\mathrm{qEL}$
Work done is equal to gain in kinetic energy
$$
\begin{aligned}
& \therefore \frac{1}{2} \mathrm{mv}^2=\mathrm{qEL} \\
& \therefore \mathrm{v}=\sqrt{\frac{2 \mathrm{qEL}}{\mathrm{m}}}
\end{aligned}
$$
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