A particle of mass 1 mg and charge q is lying at the mid-point of two stationary particles kept at a

Q: Why do we ignore $x^2$ in the denominator but keep $x$ in the numerator?

In the numerator, $x$ is multiplied by $r$, so it represents the leading order term of the restoring force. If we ignored it, the force would be zero! In the denominator, $x^2$ is subtracted from $r^2$. Since $x \ll r$, $x^2$ is infinitesimally small compared to $r^2$, making its subtraction negligible.

Q: Why did we convert the mass from milligrams to kilograms?

Coulomb's constant $k$ is given in standard SI units ($\text{N m}^2/\text{C}^2$), which inherently uses kilograms for mass (since $1\text{ N} = 1\text{ kg m}/\text{s}^2$). If you leave the mass in milligrams, your final frequency will be off by a massive factor of $10^3$!

Q: What does the negative sign in the net force indicate?

The negative sign is the heartbeat of Simple Harmonic Motion. It mathematically guarantees that the force is a *restoring* force—meaning it always acts in the opposite direction of the displacement, constantly trying to pull the particle back to equilibrium.

Q: What would happen if the middle charge was negative instead of positive?

If the middle charge was negative, the forces from the fixed charges would be attractive. Displacing it slightly towards one charge would cause that attractive force to overpower the other, pulling it entirely away from the center. This is called unstable equilibrium, and SHM would not occur along this axis.

Q: Can we solve this using the potential energy method instead of forces?

Absolutely! You can write the total electrostatic potential energy $U(x)$, find its second derivative with respect to $x$ at $x=0$ to get the effective spring constant $k_{\text{eff}} = \frac{d^2U}{dx^2}$, and then use $\omega = \sqrt{k_{\text{eff}}/m}$. Both methods yield the exact same result.

Question

A particle of mass $1	ext{ mg}$ and charge $q$ is lying at the mid-point of two stationary particles kept at a distance $2	ext{ m}$ when each is carrying same charge $q$. If the free charged particle is displaced from its equilibrium position through distance $x$ ($x \ll 1	ext{ m}$), the particle executes SHM. Its angular frequency of oscillation will be $\dots 	imes 10^5	ext{ rad/s}$, if $q^2 = 10	ext{ C}^2$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are standing right in the middle of two powerful, stationary charges. This is exactly the situation our free particle finds itself in. We have two fixed charges, each with a charge of $q$, placed $2	ext{ m}$ apart. This means our free particle, which also has a mass $m$ and charge $q$, sits comfortably at the origin, exactly $1	ext{ m}$ away from both fixed charges.

At this exact midpoint, the repulsive electrostatic forces from both sides perfectly cancel each other out. The particle is in a state of equilibrium. But what happens if we disturb this delicate balance?

### The Master Equation
Let's nudge the middle charge slightly to the right by a tiny distance $x$. Suddenly, the symmetry is broken! The particle is now closer to the right charge (at a distance of $r - x$) and further from the left charge (at a distance of $r + x$).

Because all the charges are positive, they repel. The left charge pushes our particle to the right with a force $F_1$, and the right charge pushes it to the left with a force $F_2$. Since the particle is closer to the right charge, the leftward push $F_2$ is stronger. This creates a net restoring force trying to push the particle back to the center:
$$F_{	ext{net}} = F_1 - F_2 = \frac{kq^2}{(r+x)^2} - \frac{kq^2}{(r-x)^2}$$

By taking $kq^2$ common and cross-multiplying, we get:
$$F_{	ext{net}} = kq^2 \left[ \frac{(r-x)^2 - (r+x)^2}{(r^2-x^2)^2} ight]$$

Here is where the magic of algebra comes in. The numerator is a classic $(a-b)^2 - (a+b)^2$ expansion, which beautifully simplifies to $-4ab$. Thus, the numerator becomes $-4rx$.

Now, we use the crucial constraint given in the problem: $x \ll r$. Because $x$ is incredibly small, $x^2$ is practically zero compared to $r^2$. The denominator $(r^2 - x^2)^2$ simply becomes $r^4$. 
$$F_{	ext{net}} \approx -\frac{4kq^2 r}{r^4} x = -\left( \frac{4kq^2}{r^3} ight) x$$

### Final Calculation
Did you get the feel of it? We have just proven that the restoring force is directly proportional to the negative of the displacement. This is the absolute hallmark of Simple Harmonic Motion (SHM)! By comparing our result with the standard SHM force equation $F = -m\omega^2 x$, we can extract the angular frequency:
$$m\omega^2 = \frac{4kq^2}{r^3} \implies \omega = \sqrt{\frac{4kq^2}{mr^3}}$$

Now, it's just a matter of plugging in the numbers. We know Coulomb's constant $k = 9 	imes 10^9	ext{ N m}^2/	ext{C}^2$, $q^2 = 10	ext{ C}^2$, $m = 1	ext{ mg} = 10^{-6}	ext{ kg}$, and $r = 1	ext{ m}$.
$$\omega = \sqrt{\frac{4 	imes (9 	imes 10^9) 	imes 10}{10^{-6} 	imes 1^3}}$$
$$\omega = \sqrt{\frac{360 	imes 10^9}{10^{-6}}} = \sqrt{36 	imes 10^{16}}$$
$$\omega = 6 	imes 10^8	ext{ rad/s}$$

The question asks for the answer in the format of $	ext{Value} 	imes 10^5	ext{ rad/s}$. Therefore, we write $6 	imes 10^8$ as $6000 	imes 10^5$.

**The final answer is 6000.**

Enter Numerical Value:

Visualized Solution

$System Setup$

$Small Displacement$

$Electrostatic Forces$

$Net Restoring Force$

$Simplifying the Expression$

$Applying the Approximation$

$Angular Frequency of SHM$

$Substituting Values$

$Final Calculation$

$What if the charge was negative?$

The Sigma Insight: Coulomb's Law

Analyzing the Setup Imagine you are standing right in the middle of two powerful, stationary charges

The Master Equation Let's nudge the middle charge slightly to the right by a tiny distance $x$

Final Calculation Did you get the feel of it? We have just proven that the restoring force is directly proportional to the negative of the displacement

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Small Displacement

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electrostatic Forces

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Net Restoring Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the Approximation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Frequency of SHM

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if the charge was negative?