Q.Total charge $-Q$ is uniformly spread along the length of a ring of radius $R$. A small test charge $+q$ of mass $m$ is kept at the centre of the ring and is given a gentle push along the axis of the ring.

Concept: Coulomb Force Superposition — the net force on the test charge is the vector sum of forces from all infinitesimal ring elements. By symmetry, only the axial component survives.

Reasoning:

1. For a ring of radius $R$ with total charge $-Q$, the linear charge density is $\lambda = -Q/(2\pi R)$.

   A small element $dq = \lambda R\,d\theta$ at an angle $\theta$ exerts a force on $+q$ at a distance $x$ along the axis. The distance from $dq$ to $q$ is $\sqrt{R^2 + x^2}$.

2. The axial component of force from $dq$ is

Since $dq$ is negative, the force is attractive (toward the ring centre). Integrating over the ring gives

3. For small displacements $x \ll R$, use the binomial approximation: $(R^2 + x^2)^{-3/2} \approx R^{-3} \left(1 - \frac{3x^2}{2R^2}\right) \approx R^{-3}$ …

The key insight here is that superposition (Coulomb’s law integrated over the ring) gives the net force on the test charge. For a ring, symmetry cancels all perpendicular components, leaving only the axial component. And for small displacements, that axial field turns out to be linear in $x$ — exactly what we need for simple harmonic motion.

Let’s work through it.

1. Set up the geometry and the field expression

   Place the ring in the $yz$-plane, centred at the origin. The ring has radius $R$ and total charge $-Q$ uniformly distributed, so its linear charge density is $\lambda = -\frac{Q}{2\pi R}$.

   A test charge $+q$ is at a point on the $x$-axis at distance $x$ from the centre. Consider a tiny element of the ring at angle $\theta$, carrying charge $dq = \lambda R\,d\theta$. The distance from this element to the test charge is $\sqrt{R^2 + x^2}$.

   By Coulomb’s law, the magnitude of the force from $dq$ on $+q$ is

But force is a vector. The radial (perpendicular) components from opposite elements cancel in pairs — only the component along the $x$-axis survives. For each $dq$, the axial component is $dF \cos\phi$, where $\phi$ is the angle between the line joining the element to the test charge and the $x$-axis. From the geometry,

So the net axial force is

The integral $\int dq$ is just the total charge on the ring, $-Q$. Therefore

The negative sign tells us the force is restoring — it points back toward the centre (since $Q$ is positive in magnitude here; the ring’s charge is $-Q$, so the product $q(-Q)$ is negative, making $F_x$ opposite to $x$).

2. Approximate for small oscillations

   For the particle to execute SHM, the restoring force must be proportional to displacement (and opposite in direction). The expression above is not linear in $x$ for arbitrary $x$, but for small $x$ compared to $R$ (i.e. $|x| \ll R$), we can expand.

   Write

For $|x| \ll R$, use the binomial approximation $(1 + \epsilon)^{-3/2} \approx 1 - \frac{3}{2}\epsilon$:

Keeping only the leading term (the constant 1), we get

$F_x \approx -\frac{1}{4\pi\varepsilon_0} \frac{q Q}{R^3}\, x.$ …

Concept: Electric Field of a Charged Ring & SHM Condition

When a charge moves under a restoring force proportional to displacement ($F \propto -x$), it executes simple harmonic motion (SHM). For a charged ring, the electric field along its axis is not uniform — but near the centre, it varies linearly with distance, giving the required restoring force.

Method: Taylor Expansion of Electric Field near Equilibrium

Steps:

1. Electric field of a uniformly charged ring on its axis

For a ring of radius $R$ with total charge $-Q$, the electric field at a distance $x$ from the centre along the axis is:

The negative sign indicates the field points toward the ring (since the ring is negatively charged).

2. Force on test charge $+q$

3. Small displacement approximation ($x \ll R$)

Use the binomial expansion:

For very small $x$, keep only the first term:

4. Identify SHM condition

The force is:

• Proportional to displacement $x$
• Opposite in direction (negative sign) …

Common Mistakes & How to Avoid Them

Mistake 1: Treating the ring as a point charge at its centre

The error: Students often assume the electric field at a point on the axis is simply $E = \frac{kQ}{x^2}$, treating the entire ring as a point charge at the centre.

Why it's wrong: The ring is a continuous charge distribution. The field at a point on the axis is not the same as that of a point charge at the centre — the symmetry and geometry matter.

How to avoid: Always derive the field for a ring using integration:

• Take a small element $dq$ on the ring
• Resolve the field into axial and perpendicular components
• The perpendicular components cancel due to symmetry
• Only the axial component survives

The correct expression is:

Mistake 2: Forgetting the direction of force on the test charge

The error: Students write $F = qE$ without checking sign, leading to wrong conclusions about whether the motion is oscillatory.

Why it's wrong: The ring has negative charge $-Q$. For a positive test charge $+q$:

• When displaced along the axis, the field points toward the ring (since the ring is negative)
• The force on $+q$ is toward the centre — this is a restoring force

How to avoid: Always check:

1. Sign of the source charge
2. Sign of the test charge
3. Direction of field → direction of force

For $-Q$ and $+q$:

The negative sign confirms restoring nature.

Mistake 3: Using the exact expression instead of approximating for small $x$

The error: Students try to solve the equation of motion using the full expression $F = -\frac{kqQx}{(R^2 + x^2)^{3/2}}$ without approximation.

Why it's wrong: This gives a non-linear differential equation that is not simple harmonic. SHM requires force ∝ displacement.

How to avoid: For small oscillations ($x \ll R$), use the binomial approximation:

Keeping only the constant term:

This is of the form $F = -kx$, confirming SHM.

Mistake 4: Incorrectly identifying the spring constant $k$ …