Question

A positive charged particle of mass $m$ and charge $q$ is released from rest from the position $(x_0,0)$ in a uniform electric field $E_0\hat{j}$. The angular momentum of the particle about origin is

• A. zero
• B. constant
• C. increases with time
• D. decreases with time

Answer 1

The correct option is C increases with time

Here, charge is placed in a constant electric field, so it experienced a constant electric force $F$.

$F=q{E}_0$

Due to force $F$, charge start moving under constant acceleration $a$.

$a=\frac{F}{m}=\frac{q{E}_0}{m}$

Now using kinematics relation,

$v=u+at$
$\Rightarrow v=0+\frac{q{E}_0}{m}t(\because u=0)$

$\Rightarrow v=\frac{q{E}_0}{m}t$

$\Rightarrow \overrightarrow{v}=\frac{q{E}_0}{m}t\hat{j}$

$\overrightarrow{p}=m\overrightarrow{v}=m\frac{q{E}_0}{m}t\hat{j}.......\left(1\right)$

Now using kinematics relation,

$s=ut+\frac{1}{2}a{t}^2$

$y=0+\frac{1}{2}a{t}^2=\frac{1}{2}(\frac{q{E}_0}{m}-g)t^2$

$\overrightarrow{r}=x_0\hat{i}+\frac{1}{2}(\frac{q{E}_0}{m}-g)t^2\hat{j}.....\left(2\right)$

We know that angular momentum about a point,

$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$

from Eqs.(1) and (2)

$\Rightarrow \overrightarrow{L}=[x_0\hat{i}+\frac{1}{2}(\frac{q{E}_0}{m}-g)t^2\hat{j}]\times [m\frac{q{E}_0}{m}t\hat{j}]$

$\Rightarrow \overrightarrow{L}=x_{0}m\frac{q{E}_0}{m}t\hat{k}=x_{0}q{E}_{0}t\hat{k}$

$\Rightarrow |\overrightarrow{L}|\propto t$

$[x_0,q,E_0=\text{constant}]$

Therefore, the angular momentum of the particle about origin increases with time.

Hence, option $\left(c\right)$ is correct.