Question

A particle of mass $m$ and charge $-q$ is projected from the origin with a horizontal speed $v$ into a space having electric field of magnitude $E$ directed downwards as shown in the figure. If the effect due to gravity can be neglected, then choose the correct statement(s) among the following.

• 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 $y=\frac{1}{2}\left(\frac{q{E}_x^2}{m{v}^2}\right)$
• 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: (B), (C), (D)

The horizontal and vertical displacements $x$ and $y$ after a time $t$ are $x=vt$ and $y=\frac{1}{2}{a}_y{t}^2$

Since, there is no force acting on the mass in the $x$ direction, $F_x=0$

$\therefore$ $a_x=0$

Now, since the electric field acts downwards, the force acting on the particle in the $y$ direction, $F_y=qE$

$\therefore$ Using $F_y=m{a}_y$

$a_y=\frac{qE}{m}$

So, option $\left(b\right)$ is the correct answer.

Now considering the second equation of motion, $x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

$\because$ $u_x=v$ and $a_x=0$

$\Rightarrow x=vt.....\left(1\right)$
Similarily, for $y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$\because u_y=0$ and $a_y=\frac{qE}{m}$

$y=\frac{1}{2}{a}_y{t}^2=\frac{1}{2}\left(\frac{qE}{m}\right)t^2.....\left(2\right)$

So, option $\left(d\right)$ is the correct answer.

We, have from $\left(1\right),t=\frac{x}{v}$

Substituting $t=\frac{x}{v}$ in $\left(2\right)$, we get

$y=\frac{1}{2}\left(\frac{qE{x}^2}{m{v}^2}\right)$

So, option $\left(c\right)$ is the correct answer.

Now, kinetic energy $\text{KE}$ is,

$\text{KE}=\frac{1}{2}m(v_x^2+v_y^2)$

where, $v_x=v$ and $v_y=a_{y}t=\frac{qE}{m}t$

$\text{KE}=\frac{1}{2}m(v^2+\frac{q^2{E}^2{t}^2}{m^2})$

So, option $\left(a\right)$ is the incorrect answer.

Hence, options (b) , (c) and (d) are the correct answers.