The Electric Field Due to a Charged Disk — Question 4

Question 4

A far-away disk approaches a point charge from below

A uniformly charged disk has total charge Q and radius R. On its positive axis, $E=2kQ[1-z/\sqrt{z^2+R^2}]/R^2$. Compare with $E_p=kQ/z^2$.

1. Find the exact ratio $E/E_p$ and evaluate it at $z=5R$.

2. Use $(1+u)^{-1/2}=1-u/2+3u^2/8+O(u^3)$ to find the leading relative correction.

Question 4 - Solution

A far-away disk approaches a point charge from below

(a) Method and calculation

$$\frac E{E_p}=\frac{2z^2}{R^2}\left(1-\frac z{\sqrt{z^2+R^2}}\right).$$$$z=5R:\quad E/E_p=50(1-5/\sqrt{26})\approx.970966.$$

(b) Interpretation and further calculation

$$u=R^2/z^2:\quad1-(1+u)^{-1/2}=u/2-3u^2/8+O(u^3).$$$$E/E_p=1-\frac{3R^2}{4z^2}+O(R^4/z^4).$$The point-charge estimate is too large because distributed charge has greater distance and smaller axial projection than charge concentrated at the center.

Physical check

The expansion must retain one extra term before division by $u$ to capture the first field correction.