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. $\left[\frac{2Eq1}{m}\right]^{1/2}$
• B. $\left[\frac{2Eq}{ml}\right]^{1/2}$
• C. $\left[\frac{2Em}{ql}\right]^{1/2}$
• D. $\left[\frac{Eq}{ml}\right]^{1/2}$

Answer 1

Answer: (A)

$\sqrt{\frac{2qEl}{m}}$

Answer 2

When an electron is accelerated by an electric field, the work done on it is converted into kinetic energy. The work done is

$W = qEl$

and the kinetic energy acquired is

$\frac{1}{2}mv^2$.

Equate the two:

$qEl = \frac{1}{2}mv^2$.

Solving for $v$:

$v = \sqrt{\frac{2qEl}{m}}$.

Thus, the velocity acquired is $\sqrt{\frac{2qEl}{m}}$.