Question

A particle of mass m and charge $-q$ is projected from the origin with a horizontal speed $v$ into an electric field of intensity $E$ directed downward. Choose the correct statement. Neglect gravity :

• A. The kinetic energy after a displacement $y$ is $qEy$
  
• B. The horizontal and vertical components of acceleration are $a_x=0,a_y=\frac{qE}{m}$
• C. The equation of trajectory is $y=\frac{1}{2}\left(\frac{qE{x}^2}{m{v}^2}\right)$
• D. The horizontal and vertical displacements $x$ and $y$ after a time $t$ are $x=vt$ and $y=\frac{1}{2}{a}_y{t}^2$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A The kinetic energy after a displacement $y$ is $qEy$
B The horizontal and vertical components of acceleration are $a_x=0,a_y=\frac{qE}{m}$
C The equation of trajectory is $y=\frac{1}{2}\left(\frac{qE{x}^2}{m{v}^2}\right)$
D The horizontal and vertical displacements $x$ and $y$ after a time $t$ are $x=vt$ and $y=\frac{1}{2}{a}_y{t}^2$

Here, $\overrightarrow{E}=E(-\hat{j})$
Force , $\overrightarrow{F}=(-q)\overrightarrow{E}=qE\hat{j}$
so, $F_x=0$ or $m{a}_x=0\Rightarrow a_x=0...\left(1\right)$ and
$F_y=qE$ or $m{a}_y=qE\Rightarrow a_y=qE/m...\left(2\right)$
using (1), $\frac{d{v}_x}{dt}=0\Rightarrow v_x=C$
when $t=0,v_x=v$ so,$C=v$
thus, $v_x=v$ or $\frac{dx}{dt}=v$
integrating, $x=vt+C_1$ where $C_1$, integrating constant.
at $t=0,x=0,$ so, $C_1=0$
hence, $x=vt...\left(3\right)$
using (2), $\frac{d{v}_y}{dt}=qE/m$
integrating, $v_y=\left(qE/m\right)t+C_2$
at $t=0,v_y=0,$ so $C_2=0$
thus, $v_y=\left(qE/m\right)t=\frac{dy}{dt}$
integrating, $y=\left(qE/m\right)\frac{t^2}{2}+C_3$
at $t=0,y=0$, so $C_3=0$
hence, $y=\left(qE/m\right)\frac{t^2}{2}=a_y\frac{t^2}{2}...\left(4\right)$
The equation of trajectory is $y=\left(qE/m\right)\frac{x^2}{2v^2}=\frac{1}{2}\left(\frac{qE{x}^2}{m{v}^2}\right)$ using (3)
The kinetic energy after a displacement y is $KE\left(y\right)=\frac{1}{2}m{v}_y^2=\frac{1}{2}m(qE/m{)}^2{t}^2=m(qE/m{)}^2\left(my/qE\right)=qEy$