The Electric Field Due to a Charged Disk — Question 2

Question 2

A field reading at half the surface-limit value fixes the observation height

A uniform disk has $E(z)=E_s[1-z/\sqrt{z^2+R^2}]$ for $z>0$, with $E_s=2\pi k\sigma$. Its radius is .300 m.

1. Find the height where $E=E_s/2$.

2. Differentiate E to show the positive-axis solution is unique and determine how a small height increase changes the field there.

Question 2 - Solution

A field reading at half the surface-limit value fixes the observation height

(a) Method and calculation

$$\frac z{\sqrt{z^2+R^2}}=\frac12\Rightarrow4z^2=z^2+R^2.$$$$z=R/\sqrt3\approx.173205\,\mathrm m.$$

(b) Interpretation and further calculation

$$\frac{dE}{dz}=-\frac{E_sR^2}{(z^2+R^2)^{3/2}}<0.$$$$z=R/\sqrt3:\quad\Delta E\simeq-\frac{3\sqrt3}{8}\frac{E_s}{R}\Delta z.$$Strict decrease from $E_s$ to zero permits exactly one positive height for each intermediate field magnitude.

Physical check

The radius controls the distance scale, while the density controls the overall field scale.