Two electrons each are fixed at a distance 2d. A third charge proton placed at the mid-point is disp

Q: Why do we only consider the vertical components of the force?

Because of the symmetry of the setup, the horizontal components of the forces from the two electrons are equal in magnitude but opposite in direction. They perfectly cancel each other out, leaving only the vertical components to act as the net restoring force.

Q: What does the approximation $x \ll d$ actually do?

It simplifies the math by allowing us to ignore $x^2$ when it's added to $d^2$. Physically, it means we are only looking at very small oscillations near the equilibrium point, where the restoring force is approximately linear (proportional to $x$).

Q: Why is the restoring force proportional to $x$ important?

A restoring force that is directly proportional to the displacement ($F \propto -x$) is the defining characteristic of Simple Harmonic Motion (SHM). It allows us to compare our force equation with the standard SHM equation $F = -m\omega^2 x$ to find the angular frequency.

Q: What if the proton was displaced along the x-axis instead?

If displaced along the x-axis, the proton would move closer to one electron and further from the other. The net force would pull it towards the closer electron, away from the center. This is an unstable equilibrium, and it would not execute SHM.

Q: Does the mass of the electrons matter in this problem?

No, the problem states that the electrons are 'fixed' in place. This means we assume they are held there by some external forces and don't move, regardless of their mass. We only care about the mass of the proton that is oscillating.

Question

Two electrons each are fixed at a distance $2d$. A third charge proton placed at the mid-point is displaced slightly by a distance $x (x \ll d)$ perpendicular to the line joining the two fixed charges. Proton will execute simple harmonic motion having angular frequency? ($m =$ mass of charged particle)

Options:

(A) $\left(\frac{2q^2}{\pi\varepsilon_0 md^3}\right)^{1/2}$

(B) $\left(\frac{\pi\varepsilon_0 md^3}{2q^2}\right)^{1/2}$

(C) $\left(\frac{q^2}{2\pi\varepsilon_0 md^3}\right)^{1/2}$

(D) $\left(\frac{2\pi\varepsilon_0 md^3}{q^2}\right)^{1/2}$

Sigma Percentile · Accepted Answer

### The Setup: A Delicate Balance

Imagine a microscopic tug-of-war. We have two electrons, each carrying a charge of $-q$, firmly anchored in space at a distance of $2d$ from each other. Right in the middle of this invisible line, we place a proton, carrying a charge of $+q$.

If the proton stays exactly in the middle, the attractive pulls from both electrons cancel out perfectly. It's in a state of equilibrium. But what happens if we give it a tiny nudge? We displace the proton by a small distance $x$ along the perpendicular bisector of the line joining the electrons.

Suddenly, the balance is broken. The proton is now closer to the electrons than it would be if it kept moving away, and the forces start to pull it back towards the center. Let's break down these forces.

### Coulomb's Law and Symmetry

According to **Coulomb's Law**, the electrostatic force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

For our displaced proton, the distance $r$ to each electron forms the hypotenuse of a right-angled triangle with sides $d$ and $x$. Using the Pythagorean theorem, we know that $r^2 = d^2 + x^2$. Therefore, the magnitude of the attractive force $F$ from each electron is:

$$F = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{d^2 + x^2}$$

Because the setup is perfectly symmetric, the forces from the two electrons mirror each other. If we resolve these forces into horizontal and vertical components, a beautiful thing happens. The horizontal components ($F \cos	heta$) pull in opposite directions and perfectly cancel each other out.

However, the vertical components ($F \sin	heta$) both point straight down towards the equilibrium position. They add up to create the net restoring force that pulls the proton back.

### The Restoring Force

The total restoring force acting on the proton is simply the sum of these vertical components:

$$F_{net} = 2F \sin	heta$$

From our right-angled triangle, we can see that $\sin	heta$ is the ratio of the opposite side ($x$) to the hypotenuse ($r = \sqrt{d^2 + x^2}$). Substituting this and our expression for $F$ into the net force equation, we get:

$$F_{net} = 2 \left( \frac{1}{4\pi\varepsilon_0} \frac{q^2}{d^2 + x^2} ight) \left( \frac{x}{\sqrt{d^2 + x^2}} ight)$$

Combining the terms in the denominator, we arrive at the exact expression for the restoring force:

$$F_{net} = \frac{1}{2\pi\varepsilon_0} \frac{q^2 x}{(d^2 + x^2)^{3/2}}$$

### The Small Displacement Approximation

Now we apply a crucial piece of information from the problem: the displacement $x$ is much, much smaller than the distance $d$ ($x \ll d$).

In the grand scheme of the denominator, adding a tiny $x^2$ to a much larger $d^2$ barely makes a difference. We can safely approximate $d^2 + x^2 \approx d^2$. This simplifies our denominator significantly:

$$(d^2 + x^2)^{3/2} \approx (d^2)^{3/2} = d^3$$

Substituting this back into our force equation, we get a beautifully simple, linearized restoring force:

$$F_{net} \approx \left( \frac{q^2}{2\pi\varepsilon_0 d^3} ight) x$$

### Simple Harmonic Motion and Angular Frequency

Notice the structure of this equation. The restoring force is directly proportional to the displacement $x$, multiplied by a constant term. This is the exact mathematical signature of **Simple Harmonic Motion (SHM)**!

For any object undergoing SHM, the restoring force is given by $F = m\omega^2 x$, where $m$ is the mass and $\omega$ is the angular frequency. By comparing our derived force with the standard SHM equation, we can isolate $\omega$:

$$m\omega^2 x = \left( \frac{q^2}{2\pi\varepsilon_0 d^3} ight) x$$

Dividing both sides by $mx$ and taking the square root, we find the angular frequency of the proton's oscillation:

$$\omega = \sqrt{\frac{q^2}{2\pi\varepsilon_0 md^3}}$$

And there we have it! The proton dances back and forth across the center line, driven by the elegant interplay of electrostatic attraction and geometric symmetry.

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The Way Forward

The Sigma Insight: Coulomb's Law

The Setup

Coulomb's Law and Symmetry

The Restoring Force

The Small Displacement Approximation

Simple Harmonic Motion and Angular Frequency

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