Animated Solution for Physics - Electrostatics: A small point mass carrying some positive charge on it, is released from the edge of a table. There is an 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)

The problem of a charged particle falling under the influence of both gravity and an electric field is a classic in physics. At first glance, your intuition might scream "Parabola!" after all, isn't that what happens when things fall off tables? But physics is a discipline of rigorous truth, and the math tells a beautifully different story.

Let's dive deep into the mechanics of this system and uncover why the universe behaves the way it does.

Imagine you are standing at the edge of a table. You hold a tiny point mass, carrying a positive charge $+q$, between your fingers. The moment you let go, the particle is released from rest. This is our most critical initial condition: the initial velocity $u=0$.

We can set up a coordinate system right at the edge of the table. Let the horizontal direction to the right be the positive $x$-axis, and the vertical direction upwards be the positive $y$-axis.

The instant the particle leaves your fingers, it becomes a slave to two fundamental forces of nature.

First, there is the relentless pull of Earth's gravity. It exerts a downward force, $F_g=-mg$, acting entirely along the $y$-axis.

Second, the problem states there is a uniform horizontal electric field $\vec{E}$. Because our particle carries a positive charge, it experiences an electric force in the exact direction of the field. This gives us a horizontal force, $F_e=qE$, acting entirely along the $x$-axis.

By Newton's Second Law, these constant forces produce constant accelerations:

Now, we bring in the heavy artillery of kinematics. Since the accelerations are constant, we can use the classic equation of motion:

Because the particle was released from rest ($u=0$), the equations for the $x$ and $y$ coordinates simplify beautifully. The distance the particle travels in any direction is purely a function of its acceleration and the square of the time elapsed.

For the horizontal motion:

For the vertical motion:

To understand the shape of the particle's path—its trajectory—we need to see how $y$ changes with respect to $x$. This means we must eliminate the hidden variable, time $t$, from our equations.

From the $x$-equation, we can isolate $t^2$:

Now, we substitute this expression for $t^2$ directly into our $y$-equation:

Notice how the factor of $\frac{1}{2}$ and the $2$ perfectly cancel each other out. We are left with a pristine, elegant relationship:

Let's substitute our actual acceleration values back into this relationship:

Look closely at this equation. The mass $m$, the gravitational acceleration $g$, the charge $q$, and the electric field $E$ are all constants. Therefore, the entire term in the parentheses is just a constant number.

This equation takes the exact form of $y=-mx$, which is the undeniable mathematical signature of a straight line passing through the origin with a negative slope!

As the particle moves to the right (increasing $x$), it drops downwards (negative $y$) at a perfectly constant rate. There is no curve, no parabola—just a direct, linear plunge. Therefore, the correct visual representation is a straight line going downwards, matching option (d).

If you read the reference solution provided in some textbooks, you might see a statement like: "Because the gravitational force is considerably smaller than the electric force, the path is nearly a straight line."

While it might be true that $mg$ is numerically smaller than $qE$ for an electron or a proton, the trajectory is not nearly a straight line—it is exactly a straight line. Even if you performed this experiment on Jupiter, where gravity is massive, the path would still be perfectly straight. The strength of gravity only changes the steepness of the slope, not the shape of the curve. The linearity comes from the fact that both $x$ and $y$ scale with $t^2$, keeping their ratio locked in a constant proportion.

So, when does a particle follow a parabola?

A parabolic trajectory requires an initial velocity that is perpendicular to the acceleration. If you had flicked the particle horizontally off the table with some initial velocity $u_x$, the $x$-equation would become $x=u_{x}t+\frac{1}{2}{a}_x{t}^2$. When you eliminate $t$ in that scenario, you end up with an $x^2$ term in your final equation, bending the straight line into a beautiful parabola.

Always respect the initial conditions—they dictate the geometry of the universe!