A small point mass carrying some positive charge on it, is released from the edge of a table. There is a uniform electric field in this region in the horizontal direction. Which of the following options then correctly describe the trajectory of the mass? (Curves are drawn schematically and are not to scale).

Electrostatics PYQ — JEE Main 2020 · 2 Sep · Shift 2 · Q47 with Solution

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Correct answer: A

A positively charged particle is released from the edge of a table.

After release, two constant forces act on it:

Since the electric field is uniform and horizontal, the horizontal acceleration is constant:

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

The vertical acceleration is due to gravity:

$a_y = g$

So the particle has:

The particle is released from rest, so initially:

$u_x = 0, \qquad u_y = 0$

Thus, taking the point of release as origin,

$x = \frac{1}{2}a_x t^2 = \frac{1}{2}\frac{qE}{m}t^2$

$y = \frac{1}{2}gt^2$

(where $y$ is measured downward).

From the above equations,

$x \propto t^2, \qquad y \propto t^2$

Hence,

$\frac{x}{y} = \frac{qE/m}{g} = \text{constant}$

So,

$x = \frac{qE}{mg} \, y$

This is the equation of a straight line.

Since the charge is positive, the electric force is in the direction of the electric field. At the same time, gravity pulls it downward.

Therefore the particle moves along a straight line inclined from the vertical/horizontal depending on the relative magnitudes of $qE$ and $mg$ .

So the correct schematic must be the one showing a straight-line trajectory slanting downward in the direction of the electric field.

Hence the correct option is:

$\boxed{A}$

This matches the stored correct answer.