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Physics LibreTexts

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

  • Page ID
    6476
  • [ "article:topic", "authorname:tatumj" ]

     


    \(\text{FIGURE I.3}\)


     

    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.