Question

An electron $(-e,m)$ is given velocity $V_o$ along a uniform electric field $\overrightarrow{E}$. After how much time, electron will return to its original position?

• A. $\frac{2m{V}_o}{eE}$
• B. $\frac{m{V}_o}{eE}$
• C. $\frac{eE}{m{V}_o}$
• D. $\frac{2eE}{m{V}_o}$

Answer 1

The correct option is A $\frac{2m{V}_o}{eE}$

Given,
charge of electron $\left(q\right)=-e$
mass of electron$=m$


As shown in figure electric force is acting in $\text{-x}$ direction and $V_o$ is given in $\text{+x}$ direction, so the charge particle will start moving in $\text{+x}$ initially but due to the opposite constant force $qE$ it will retard and achieve zero velocity.

After that particle will change its direction of motion from $\text{+x}$ to $\text{-x}$ and will come back to its original position.

Let the retardation of electron is $a$.
From the Newton's second law of motion, the force $F$, can be written as

$F=ma=qE$

Substituting the value of $q$ we get,

$a=\frac{eE}{m}$

Let the time taken by electron to achieve zero velocity or the point when it changes its direction is $t$.

Using first equation of motion,
$v=u+at$

Substituting final velocity, $v=0$ and initial velocity, $u=V_0$ .

$\Rightarrow 0=V_o-\frac{eE}{m}t$

$\therefore t=\frac{m{V}_o}{eE}$

Time taken by electron to return to its original position is,

$T=2t=\frac{2m{V}_o}{eE}$

Hence, option (a) is correct answer.


Alternate approach:
Let the time taken by particle to reach its original position is $t$. Here, its displacement is zero.
Using, $S=ut+\frac{1}{2}a{t}^2$

$0=V_{o}t+\frac{1}{2}\times \frac{-eE}{m}{t}^2$

$\therefore t=\frac{2m{V}_o}{eE}$