Q.Calculate potential energy of a point charge $-q$ placed along the axis due to a charge $+Q$ uniformly distributed along a ring of radius $R$. Sketch P.E. as a function of axial distance $z$ from the centre of the ring. Looking at the graph, can you see what would happen if $-q$ is displaced slightly from the centre of the ring (along the axis)?

Concept: Electric Potential — The potential energy of a charge in an external field is $U = qV$, where $V$ is the potential due to all other charges.

Reasoning:

1. For a ring of radius $R$ with total charge $+Q$, the potential at a point on the axis at distance $z$ from the centre is

(All points on the ring are equidistant from the axial point.)

2. The potential energy of a point charge $-q$ placed at that point is

3. Sketch: $U(z)$ is negative, symmetric about $z=0$, and rises from a minimum at $z=0$ (value $U(0) = -\frac{1}{4\pi\varepsilon_0}\frac{qQ}{R}$) towards zero as $|z| \to \infty$. …

Why potential energy, not force, is the natural starting point

When a charge moves in an electrostatic field, its potential energy tells you the work the field can do. For a point charge $-q$ placed near a fixed charge distribution, the potential energy is simply $U = (-q) \times V$, where $V$ is the electric potential created by the fixed charges. This is far easier than integrating forces directly, because potential is a scalar — no vector components to wrestle with.

The ring has total charge $+Q$ spread uniformly. By symmetry, every point on the ring is at the same distance from any point on the axis. That single fact makes the calculation almost trivial.

Step-by-step

1. Potential of the ring on the axis

Take the ring in the $xy$-plane, centred at the origin, radius $R$. A point on the axis at height $z$ has coordinates $(0,0,z)$. Every infinitesimal charge $dq$ on the ring is at distance

from that point. Since all $dq$ are at the same distance, the total potential is just

2. Potential energy of the test charge

The test charge is $-q$. Potential energy is charge times potential:

That’s the entire expression. No approximations, no series — exact for any $z$.

3. Shape of the graph

The function $U(z)$ is even (depends only on $z^2$), negative everywhere, and has its minimum at $z=0$:

• At $z=0$: $U(0) = -\dfrac{kQq}{R}$ (deepest point).
• As $|z| \to \infty$: $U(z) \to 0$ from below (the well flattens out).
• The curve is symmetric, bowl-shaped, with no other extrema.

Sketch it: a smooth U-shaped curve centred at $z=0$, approaching zero as $z$ grows large, with the minimum at $-kQq/R$.

4. What happens for a small axial displacement …

Method: Potential Energy on the Axis of a Symmetric Distribution, and Testing Equilibrium

Use this method whenever a question gives (or asks you to first find) the potential $V(z)$ due to a symmetric charge distribution along an axis, then asks about the potential energy of a test charge there and its equilibrium behaviour.

Steps

Step 1: Get the potential due to the distribution along the axis

Use the symmetry of the distribution (e.g. every element of a ring is equidistant from an axial point) to write $V(z)$ in closed form.

Step 2: Convert potential to potential energy of the test charge

Potential energy of a point charge $q'$ sitting in an external potential $V$ is simply

Substitute the sign of the test charge carefully — a negative test charge $-q$ gives $U(z) = -q\,V(z)$ if the source distribution's potential was written for magnitude $q$.

Step 3: Locate extrema of $U(z)$ and classify them …