Two equal point charges are fixed at x = -a and x = +a on the x-axis. Another point charge Q is plac

Q: Why do we ignore the potential energy between the two fixed charges $q$?

The potential energy between the two fixed charges is constant because their distance $2a$ does not change. When we calculate the *change* in potential energy ($\Delta U$), this constant term cancels out.

Q: Why do we use the approximation $x \ll a$?

The problem states that the charge is displaced by a "small distance $x$". This implies $x$ is much smaller than $a$, allowing us to neglect $x^2$ in comparison to $a^2$ in the denominator to find the leading-order dependence.

Q: What does it mean if $\Delta U \propto x^2$?

A potential energy curve proportional to $x^2$ indicates a parabolic potential well. This is characteristic of a simple harmonic oscillator, meaning the charge will undergo SHM if released.

Q: Would the answer change if $Q$ was negative?

If $Q$ was negative, the change in potential energy would be negative, meaning the potential energy decreases as it moves away from the origin. The magnitude of the change would still be proportional to $x^2$, but the equilibrium at the origin would be unstable.

Q: Can we solve this using the electric field instead of potential energy?

Yes! You can find the electric field at distance $x$, which is approximately proportional to $x$ for small $x$. Since force $F = QE \propto x$, the work done (and thus $\Delta U$) is the integral of force, which gives $\int x \, dx \propto x^2$.

Question

Two equal point charges are fixed at $x = - a$ and $x = + a$ on the x-axis. Another point charge $Q$ is placed at the origin. The change in the electrical potential energy of $Q$, when it is displaced by a small distance $x$ along the x-axis, is approximately proportional to

Options:

(A) $x$

(B) $x^2$

(C) $x^3$

(D) $1/x$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a one-dimensional world along the x-axis. We have two identical charges, $q$, firmly anchored at $x = -a$ and $x = +a$. Right in the middle, at the origin ($x = 0$), we place a third charge, $Q$.

Because $Q$ is exactly halfway between the two $q$ charges, it feels an equal and opposite force from both, meaning it is in equilibrium. But what about its potential energy? The initial potential energy of $Q$ is simply the sum of the potential energies due to each of the fixed charges. Since both are at a distance $a$, we can write:
$$U_i = \frac{K q Q}{a} + \frac{K q Q}{a} = \frac{2 K q Q}{a}$$

### Displacing the Charge
Now, let's disturb this peaceful setup. We nudge the charge $Q$ by a very small distance $x$ along the x-axis.

Its new position changes its distance from the two fixed charges. It is now closer to one charge and further from the other. Specifically, its distance from the left charge becomes $(a + x)$, and its distance from the right charge becomes $(a - x)$.

The new potential energy of $Q$ at this displaced position is:
$$U_f = \frac{K q Q}{a+x} + \frac{K q Q}{a-x}$$

### Calculating the Change in Potential Energy
The problem asks for the *change* in electrical potential energy, $\Delta U$. This is the difference between the final and initial potential energies:
$$\Delta U = U_f - U_i$$

Let's substitute our expressions into this formula:
$$\Delta U = K q Q \left( \frac{1}{a+x} + \frac{1}{a-x} \right) - \frac{2 K q Q}{a}$$

To simplify the terms in the bracket, we take a common denominator:
$$\frac{1}{a+x} + \frac{1}{a-x} = \frac{a - x + a + x}{(a+x)(a-x)} = \frac{2a}{a^2 - x^2}$$

Substituting this back into our equation for $\Delta U$:
$$\Delta U = K q Q \left( \frac{2a}{a^2 - x^2} \right) - \frac{2 K q Q}{a}$$

Now, let's factor out $2 K q Q$:
$$\Delta U = 2 K q Q \left( \frac{a}{a^2 - x^2} - \frac{1}{a} \right)$$

Taking a common denominator again to combine these terms:
$$\Delta U = 2 K q Q \left( \frac{a^2 - (a^2 - x^2)}{a(a^2 - x^2)} \right)$$
$$\Delta U = \frac{2 K q Q x^2}{a(a^2 - x^2)}$$

### The Small Displacement Approximation
Here is the crucial step. The problem explicitly states that the displacement $x$ is "small". This means $x \ll a$.

Because $x$ is much smaller than $a$, $x^2$ will be incredibly small compared to $a^2$. Therefore, in the denominator, we can safely approximate $(a^2 - x^2) \approx a^2$.

Applying this approximation, our expression simplifies beautifully:
$$\Delta U \approx \frac{2 K q Q x^2}{a(a^2)} = \frac{2 K q Q}{a^3} x^2$$

### Final Conclusion
Looking at our final expression, we can see that $\frac{2 K q Q}{a^3}$ is entirely composed of constants. Therefore, the change in potential energy is directly proportional to the square of the displacement:
**$$\Delta U \propto x^2$$**

This parabolic relationship ($\propto x^2$) is a hallmark of simple harmonic motion. It tells us that if $Q$ and $q$ have the same sign, the origin is a point of stable equilibrium, and the charge $Q$ would oscillate back and forth if released!

Select Answer:

Visualized Solution

$x = 0$

$U_{i}$

$x \to x$

$U_{f}$

$Δ U = U_{f} - U_{i}$

$Δ U$

$Δ U$

$Δ U$

$x ≪ a$

$Δ U \propto x^{2}$

$Equilibrium$

The Sigma Insight: Electrostatic Potential Energy

Analyzing the Setup Imagine a one-dimensional world along the x-axis

Displacing the Charge Now, let's disturb this peaceful setup

Calculating the Change in Potential Energy The problem asks for the change in electrical potential energy, $Δ U$

The Small Displacement Approximation Here is the crucial step

Final Conclusion Looking at our final expression, we can see that $\frac{2 K q Q}{a ^{3}}$ is entirely composed of constants

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Select Answer:

Visualized Solution

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔU=Uf​−Ui​

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔU∝x2

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

The Sigma Insight: Electrostatic Potential Energy

Analyzing the Setup Imagine a one-dimensional world along the x-axis

Displacing the Charge Now, let's disturb this peaceful setup

Calculating the Change in Potential Energy The problem asks for the change in electrical potential energy, .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔU

The Small Displacement Approximation Here is the crucial step

Final Conclusion Looking at our final expression, we can see that .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a32KqQ​ is entirely composed of constants

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A system of three charges are placed as shown in the figure If D≫d, the potential energy of the system is best given by

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Eight point charges are placed at the corners of a cube of edge a as shown in figure. Find the work done in disassembling this system of charges.

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