The Electric Field Due to a Charged Disk — Question 10

Question 10

At fixed total charge a smaller disk gives a larger axial field

A uniform disk has fixed total positive charge Q but variable radius R. Observe at a fixed height $z>0$. Its density is constrained by $\sigma=Q/(\pi R^2)\le\sigma_{\max}$.

1. Rationalize $E=2kQ[1-z/\sqrt{z^2+R^2}]/R^2$ to show directly how it varies with R.

2. Find the radius maximizing E under the density constraint and compare the unconstrained $R\to0$ limit with a point charge.

Question 10 - Solution

At fixed total charge a smaller disk gives a larger axial field

(a) Method and calculation

$$s=\sqrt{z^2+R^2},\quad1-z/s=\frac{R^2}{s(s+z)}.$$$$E=\frac{2kQ}{s(s+z)},\quad ds/dR=R/s>0.$$Both positive denominator factors increase with R, so E decreases strictly for $R>0$.

(b) Interpretation and further calculation

$$R\ge\sqrt{\frac{Q}{\pi\sigma_{\max}}}\Rightarrow R_{\mathrm{opt}}=\sqrt{\frac{Q}{\pi\sigma_{\max}}}.$$$$R\to0:\quad E\to\frac{2kQ}{z(2z)}=kQ/z^2.$$The unconstrained limiting field is finite, but that limit would require unbounded density at fixed Q.

Physical check

Keeping Q fixed and keeping density fixed are different design constraints and lead to different radius dependence.