The figures below depict two situations in which two infinitely long static line charges of constant

Q: Why do we assume the displacement $x$ is small?

Assuming $x$ is small allows us to use the binomial approximation or ignore $x^2$ compared to $d^2$. This linearizes the restoring force, which is a strict requirement for Simple Harmonic Motion.

Q: Why doesn't the $-q$ charge execute SHM?

For SHM, the net force must be a *restoring* force, meaning it pushes the object back to equilibrium. For $-q$, the attractive force from the closer wire is stronger, pulling it further away. It's an unstable equilibrium!

Q: What is the formula for the electric field of a line charge?

Using Gauss's Law, the electric field at a distance $r$ from an infinitely long line charge with linear charge density $\lambda$ is $E = \frac{\lambda}{2\pi\epsilon_0 r}$.

Q: Would the result change if the line charges were negative?

Yes! If the line charges were negative, the $+q$ charge would experience attractive forces (unstable equilibrium, no SHM), and the $-q$ charge would experience repulsive forces (stable equilibrium, SHM).

Q: Can we use potential energy to solve this?

Absolutely! You can write the potential energy function $U(x)$. If $\frac{d^2U}{dx^2} > 0$ at equilibrium, it's stable (SHM). If $\frac{d^2U}{dx^2} < 0$, it's unstable.

Question

The figures below depict two situations in which two infinitely long static line charges of constant positive line charge density $\lambda$ are kept parallel to each other.
In their resulting electric field, point charges $q$ and $-q$ are kept in equilibrium between them. The point charges are confined to move in the $x$ direction only. If they are given a small displacement about their equilibrium positions, then the correct statements is/are

Options:

(A) both charges execute simple harmonic motion.

(B) both charges will continue moving in the direction of their displacement.

(C) charge $+q$ executes simple harmonic motion while charge $-q$ continues moving in the direction of its displacement.

(D) charge $-q$ executes simple harmonic motion while charge $+q$ continues moving in the direction of its displacement.

Sigma Percentile · Accepted Answer

The beauty of physics often lies in the delicate balance of forces. In this problem, we are exploring the concept of **stable and unstable equilibrium** using the electric field generated by infinitely long line charges.

Imagine you are standing exactly in the middle of two infinitely long, positively charged walls. If you are a positive charge, both walls push you away. If you are a negative charge, both walls pull you in. But what happens if you take a tiny step to the right? Let's find out!

### Analyzing the Setup

We have two infinitely long static line charges, each with a constant positive linear charge density $\lambda$. They are placed parallel to each other, separated by a distance $2d$.

Right in the middle, at a distance $d$ from both lines, the electric fields from the left and right line charges perfectly cancel each other out.

$$E_{net} = E_L - E_R = \frac{\lambda}{2\pi\epsilon_0 d} - \frac{\lambda}{2\pi\epsilon_0 d} = 0$$

This means any charge placed exactly at the center will experience zero net force. It is in a state of equilibrium. But is this equilibrium stable or unstable? To test this, we must displace the charges by a small distance $x$ and observe the net force.

### The Case of the Positive Charge

Let's first look at the positive charge $+q$. We displace it by a small distance $x$ to the right.

Now, its distance from the left wire is $(d + x)$, and its distance from the right wire is $(d - x)$. Because both the line charges and our test charge are positive, the forces are **repulsive**.

The left wire pushes the charge to the right with force $F_L$:

$$F_L = \frac{q\lambda}{2\pi\epsilon_0 (d+x)}$$

The right wire pushes the charge to the left with force $F_R$:

$$F_R = \frac{q\lambda}{2\pi\epsilon_0 (d-x)}$$

Because the charge is closer to the right wire, the denominator $(d-x)$ is smaller, making the repulsive force $F_R$ stronger than $F_L$.

Let's calculate the net force, taking the rightward direction as positive:

$$F_{net} = F_L - F_R = \frac{q\lambda}{2\pi\epsilon_0} \left[ \frac{1}{d+x} - \frac{1}{d-x} ight]$$

Taking the common denominator:

$$F_{net} = \frac{q\lambda}{2\pi\epsilon_0} \left[ \frac{d - x - (d + x)}{d^2 - x^2} ight] = -\frac{q\lambda x}{\pi\epsilon_0 (d^2 - x^2)}$$

For a very small displacement ($x \ll d$), we can approximate $d^2 - x^2 \approx d^2$.

$$F_{net} \approx -\left( \frac{q\lambda}{\pi\epsilon_0 d^2} ight) x$$

The negative sign is the hero here! It tells us that the net force is a **restoring force**, acting in the opposite direction of the displacement. Since $F_{net} \propto -x$, the positive charge will execute **Simple Harmonic Motion (SHM)**.

### The Case of the Negative Charge

Now, let's reset and place the negative charge $-q$ at the center. We displace it by the same small distance $x$ to the right.

This time, the forces are **attractive**. The left wire pulls the charge to the left with force $F_L$, and the right wire pulls it to the right with force $F_R$.

$$F_L = \frac{q\lambda}{2\pi\epsilon_0 (d+x)} \quad 	ext{(Leftwards)}$$

$$F_R = \frac{q\lambda}{2\pi\epsilon_0 (d-x)} \quad 	ext{(Rightwards)}$$

Again, because the charge is closer to the right wire, the attractive force $F_R$ is stronger. But notice the direction! The stronger force is pulling the charge *further* to the right.

Let's calculate the net force:

$$F_{net} = F_R - F_L = \frac{q\lambda}{2\pi\epsilon_0} \left[ \frac{1}{d-x} - \frac{1}{d+x} ight]$$

$$F_{net} = \frac{q\lambda}{2\pi\epsilon_0} \left[ \frac{d + x - (d - x)}{d^2 - x^2} ight] = \frac{q\lambda x}{\pi\epsilon_0 (d^2 - x^2)}$$

For small $x$:

$$F_{net} \approx \left( \frac{q\lambda}{\pi\epsilon_0 d^2} ight) x$$

Here, the net force is positive. It acts in the **same direction** as the displacement. This is the hallmark of an **unstable equilibrium**. The negative charge will not return to the center; it will accelerate and continue moving in the direction of its displacement.

### The Final Verdict

By carefully analyzing the forces, we have uncovered the physical reality of the system. The positive charge $+q$ is trapped in a stable equilibrium and will oscillate harmonically. The negative charge $-q$, however, is in an unstable equilibrium and will fly away towards the closer wire.

Therefore, the correct statement is that charge $+q$ executes simple harmonic motion while charge $-q$ continues moving in the direction of its displacement.

Select Answer:

Visualized Solution

$E = \frac{λ}{2 π ϵ _{0} r}$

$Displacement of + q$

$Forces on + q$

$Net Force on + q$

$Simplifying F_{n e t}$

$Condition for SHM$

$Equilibrium of - q$

$Displacement of - q$

$Forces on - q$

$Net Force on - q$

$Simplifying F_{n e t}$

$Conclusion for - q$

The Sigma Insight: Electric Field

Analyzing the Setup

The Case of the Positive Charge

The Case of the Negative Charge

The Final Verdict

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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; } E=2πϵ0​rλ​

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

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

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

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

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

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

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

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

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

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

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

The Sigma Insight: Electric Field

Analyzing the Setup

The Case of the Positive Charge

The Case of the Negative Charge

The Final Verdict

Similar Questions

$E = \frac{λ}{2 π ϵ _{0} r}$

$Displacement of + q$

$Forces on + q$

$Net Force on + q$

$Simplifying F_{n e t}$

$Condition for SHM$

$Equilibrium of - q$

$Displacement of - q$

$Forces on - q$

$Net Force on - q$

$Simplifying F_{n e t}$

$Conclusion for - q$