Electric field at a point on axis of uniformly charged ring

Question

Find the electric field at a point on the axis of a uniformly charged ring of radius $R$ and total charge $Q$ , at a distance $x$ from the centre of the ring.

Solution — Step by Step

Place the ring in the $xy$ -plane with its centre at the origin. The axis is the $z$ -axis (or $x$ -axis — we’ll call it the axial direction).

Point $P$ is at distance $x$ along the axis from the centre.

Consider a small element of the ring with charge $dq$ . This element is at the rim of the ring, at distance $R$ from the centre.

Distance from $dq$ to point $P$ : $r = \sqrt{R^2 + x^2}$

The small electric field $dE$ due to $dq$ at point $P$ :

dE = \frac{kq}{r^2} = \frac{k \, dq}{R^2 + x^2}

This field $d\vec{E}$ points from $dq$ toward $P$ (for positive $dq$ ), making an angle $\theta$ with the axis, where:

\cos\theta = \frac{x}{\sqrt{R^2 + x^2}}, \quad \sin\theta = \frac{R}{\sqrt{R^2 + x^2}}

The ring is symmetric. For every charge element $dq$ at one point on the ring, there is an equal element directly opposite.

The transverse (perpendicular to axis) components of the two opposite elements cancel: one points in the $+y$ direction, the other in the $-y$ direction.

Only the axial components ( $z$ -direction, toward $P$ ) add up.

dE_{axial} = dE \cos\theta = \frac{k \, dq}{R^2 + x^2} \cdot \frac{x}{\sqrt{R^2 + x^2}}

E = \int dE_{axial} = \frac{kx}{(R^2 + x^2)^{3/2}} \int dq = \frac{kQx}{(R^2 + x^2)^{3/2}}

Since $\frac{kx}{(R^2 + x^2)^{3/2}}$ is constant for all elements (the geometry is the same for every $dq$ ).

\boxed{E = \frac{kQx}{(R^2 + x^2)^{3/2}}}

Direction: along the axis, away from the ring (for positive $Q$ and positive $x$ ).

Why This Works

The key insight is symmetry: the ring’s uniform charge distribution ensures all transverse components cancel, leaving only the axial contribution. This is the standard technique in Gauss’s law and superposition problems — identify symmetry to simplify before calculating.

Special Cases

At the centre ( $x = 0$ ): $E = 0$ . The field contributions from all parts of the ring cancel (every element has an equal one directly opposite).

Far from the ring ( $x >> R$ ): $E \approx \frac{kQ}{x^2}$ (the ring looks like a point charge from far away).

Maximum field: Taking $\frac{dE}{dx} = 0$ and solving gives $x = \frac{R}{\sqrt{2}}$ . Maximum field $E_{max} = \frac{kQ}{R^2} \cdot \frac{2}{3\sqrt{3}}$ .

Common Mistake

Students forget to account for the direction of each $dE$ and try to integrate the full $dE$ without resolving into components. The magnitude of each $dE$ contribution is the same, but they point in different directions — directly adding the magnitudes overcounts. The transverse components cancel by symmetry; only the axial components ( $dE \cos\theta$ ) contribute to the net field at $P$ .

Question

Solution — Step by Step

Why This Works

Special Cases

Common Mistake

Try These Next