Formulas/physics/Electric Charges Fields/Field on Axis of Uniformly Charged Ring

Field on Axis of Uniformly Charged Ring

Field at axial distance x from the centre of a ring of radius R and total charge Q. Maximum at x = R/√2. Zero at centre.

Class 12

Derivation

Setup

A ring of radius $R$ carries total charge $Q$ uniformly. Consider a point P on the axis at distance $x$ from the centre.

Each element $dq$ of the ring is at distance:

r = \sqrt{x^2 + R^2}

from P.

The field $d\vec{E}$ from element $dq$ has two components:

Axial ( $x$ -direction): $dE_x = dE\cos\alpha$ , where $\cos\alpha = x/r$
Transverse (perpendicular to axis): cancels by symmetry — for every element $dq$ , the diametrically opposite element contributes an equal and opposite transverse component

E = \int dE_x = \int \frac{dq}{4\pi\varepsilon_0 r^2}\cdot\frac{x}{r} = \frac{x}{4\pi\varepsilon_0 r^3}\int dq

Since $r$ and $x$ are constant over the ring:

E = \frac{Qx}{4\pi\varepsilon_0(x^2+R^2)^{3/2}}

\boxed{E = \frac{Qx}{4\pi\varepsilon_0(x^2 + R^2)^{3/2}}}

Direction: along the axis, away from the centre (for $Q > 0$ , $x > 0$ ).

At the centre ( $x = 0$ ): $E = 0$ . Every element's field is cancelled by the opposite element.

Maximum field: differentiate with respect to $x$ and set to zero:

\frac{d}{dx}\left[\frac{x}{(x^2+R^2)^{3/2}}\right] = 0 \implies x = \frac{R}{\sqrt{2}}

E_{max} = \frac{Q}{4\pi\varepsilon_0} \cdot \frac{2}{3\sqrt{3}R^2}

Far field ( $x \gg R$ ): $(x^2 + R^2)^{3/2} \approx x^3$ , so $E \approx Q/4\pi\varepsilon_0 x^2$ — the ring looks like a point charge.

Note

The ring field is the starting point for deriving the disc field, obtained by integrating over concentric rings of varying radii.