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
$a=\frac{qE}{m}$
Since, electron starts accelerating from rest, therefore, initial velocity $(u)$ of electron is zero. Using.equation of motion,
$v^2-u^2=2al$
or $v^2-0=2\left(\frac{qE}{m}\right)I$ on $=\sqrt{\frac{2Eql}{m}}={\left[\frac{2Eql}{m}\right]}^{1/2}$