Question

An electron of mass ‘$m$’ and charge ‘$q$’ is accelerated from rest in a uniform electric field of strength ‘$E$’. The velocity acquired by it as it travels a distance ‘$l$’ is

• A. $\left[\frac{2Eql}{m}\right]^{\frac{1}{2}}$
• B. $\left[\frac{2Eq}{m}\right]^{\frac{1}{2}}$
• C. $\left[\frac{2Em}{ql}\right]^{\frac{1}{2}}$
• D. $\left[\frac{Eq}{ml}\right]^{\frac{1}{2}}$

Answer 1

Answer: (A)

The magnitude of force on a charge $q$ in an electric field is given by
$F = qE \,\,\,\,\,..(i)$
From Newton’s second law, we have
$F = ma \,\,\,\,\,..(ii)$
From Eqs. (i) and (ii), we get
$a=\frac{q E}{m}$
Since, electron starts accelerating from rest, therefore, initial velocity $(u) $ of electron is zero. Using.equation of motion,
$v^{2}-u^{2}=2 a l$
or $v^{2}-0=2\left(\frac{q E}{m}\right) I$ on $=\sqrt{\frac{2 E q l}{m}}=\left[\frac{2 E q l}{m}\right]^{1 / 2}$