1.6E: Field on the Axis of a Uniformly Charged Disc

$\text{FIGURE I.3}$

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.