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# 1.6.6E: Field on the Axis of a Uniformly Charged Disc

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria $$\text{FIGURE I.3}$$

We suppose that we have a circular disc of radius a bearing a surface charge density of $$σ$$ coulombs per square metre, so that the total charge is $$Q = πa^2 σ$$ . We wish to calculate the field strength at a point P on the axis of the disc, at a distance $$x$$ from the centre of the disc.

Consider an elemental annulus of the disc, of radii $$r$$ and $$r + δr$$. Its area is $$2πrδr$$ and so it carries a charge $$2πσrδr$$. Using the result of subsection 1.6.4, we see that the field at P from this charge is

$\frac{2\pi\sigma r \,\delta r}{4\pi\epsilon_0}\cdot \frac{x}{(r^2+x^2)^{3/2}}=\frac{\sigma x}{2\epsilon_0}\cdot \frac{r\,\delta r}{(r^2+x^2)^{3/2}}.$

But $$r=x\tan \theta,\, \delta r=x\sec^2 \theta \delta \theta \text{ and }(r^2+x^2)^{1/2}=x\sec \theta$$. Thus the field from the elemental annulus can be written

$\frac{\sigma}{2\epsilon_0}\sin \theta \,\delta \theta .$

The field from the entire disc is found by integrating this from $$θ = 0 \text{ to }θ = α$$ to obtain

$E=\frac{\sigma}{2\epsilon_0}(1-\cos α )=\frac{\sigma}{2\epsilon_0}\left ( 1-\frac{x}{(a^2+x^2)^{1/2}}\right ).\tag{1.6.11}$

This falls off monotonically from $$σ/(2\epsilon_0)$$ just above the disc to zero at infinity.