Animated Solution for Physics - Electrostatics: Two equal negative charges $-q$ are fixed at points $(0,-a)$ and $(0,a)$ on $y$-axis. A positive charge $Q$ is released from rest at the point $(2a,0)$ on the $x$-axis. The charge $Q$ will

Welcome to one of the most classic and conceptually rich problems in Electrostatics! This question beautifully bridges the gap between Coulomb's Law and the mechanics of Simple Harmonic Motion (SHM). It is a perfect example of why we must never blindly assume that every back-and-forth motion is Simple Harmonic.

Imagine you are standing at the origin of a vast coordinate system. On the $y$-axis, at a distance $a$ above and below you, two negative charges, each of magnitude $-q$, are firmly anchored in place. Now, a positive charge $+Q$ is placed on the $x$-axis at a distance $x=2a$ and released from rest.

Because opposite charges attract, the positive charge $+Q$ immediately feels an electrostatic pull towards both of the negative charges. Let's call these forces ${\vec{F}}_1$ and ${\vec{F}}_2$.

This is where the magic of symmetry comes into play. If we resolve these two force vectors into their $x$ and $y$ components, something elegant happens. The $y$-components are perfectly equal in magnitude but opposite in direction, meaning they cancel each other out completely. The charge $+Q$ will not drift up or down. However, the $x$-components of both forces point in the exact same direction: straight towards the origin. They add up, creating a net restoring force.

Let's quantify this restoring force. According to Coulomb's Law, the magnitude of the force exerted by each $-q$ charge on $+Q$ is:

Using the Pythagorean theorem, the distance $r$ between $+Q$ and either $-q$ is $r=\sqrt{x^2+a^2}$. The net force is the sum of the $x$-components:

From the geometry of our setup, $\cos \theta =\frac{x}{r}=\frac{x}{\sqrt{x^2+a^2}}$. Substituting this into our net force equation, we get:

This is our master equation. Notice the negative sign? It mathematically confirms that the force is a restoring force—it always points opposite to the displacement $x$, trying to pull the charge back to the origin.

Because the force is restoring, the charge will accelerate towards the origin, overshoot it due to inertia, slow down on the negative $x$-axis, stop, and be pulled back again. The motion is undeniably oscillatory.

But here is the million-dollar question: Is this motion Simple Harmonic?

To answer this, we must consult the strict mathematical definition of SHM. For a motion to be Simple Harmonic, the restoring force must be strictly and linearly proportional to the displacement. In other words, it must look like Hooke's Law: $F=-kx$, where $k$ is a constant.

Let's look at our master equation. Is $F_{net}$ directly proportional to $x$?

At first glance, there is an $x$ in the numerator. But wait! There is also an $x$ trapped inside the $(x^2+a^2{)}^{3/2}$ term in the denominator. Because the denominator changes as $x$ changes, the overall force is not linearly proportional to $x$.

There is only one special case where this setup produces SHM. If the charge $+Q$ is released from a distance $x$ that is extremely small compared to $a$ ($x\ll a$), then $x^2$ becomes negligible compared to $a^2$. The denominator simplifies to $(a^2{)}^{3/2}=a^3$, which is a constant. Under this strict approximation, the force becomes $F_{net}\approx -\left(\frac{2qQ}{4\pi {ϵ}_0{a}^3}\right)x$, which perfectly matches the $F=-kx$ condition.

In our specific problem, the charge $+Q$ is released from $x=2a$. This is definitely not small compared to $a$. The approximation $x\ll a$ completely fails.

Therefore, the restoring force is not linearly proportional to the displacement. The charge will dance back and forth across the origin, but its rhythm will be complex, not simple. The motion is oscillatory, but not simple harmonic.