Question

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

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

Answer 1

The correct option is C increases with time

The force on charge $\overrightarrow{F}=q{E}_0\hat{j}$

so, $F_x=0\Rightarrow a_x=0$

$F_y=q{E}_0$ or $m\frac{d{v}_y}{dt}=q{E}_0\Rightarrow a_y=\left(q{E}_0/m\right)$

also, $\frac{d{v}_x}{dt}=0$

integrating , $v_x=C$ at $t=0,v_x=0$ so $C=0$

hence, $v_x=0$

and $\frac{d{v}_y}{dt}=\left(q{E}_0/m\right)$

integrating, $v_y=\left(q{E}_0/m\right)t+C_1$

at $t=0,v_y=0$ so, $C_1=0$

hence, $v_y=\left(q{E}_0/m\right)t$

Thus, velocity of particle is $v=\sqrt{v_x^2+v_y^2}=\left(q{E}_0/m\right)t$

Angular momentum about origin is $L=mv{x}_0=m\left(q{E}_0/m\right)t{x}_0=q{E}_0{x}_{0}t$

so, $L\propto t$