The Electric Field Due to a Charged Disk — Question 3

Question 3

The infinite-sheet approximation is restricted by height relative to disk radius

For a uniform disk, use $E(z)=E_s[1-z/\sqrt{z^2+R^2}]$ with $E_s=2\pi k\sigma$. Treat $E_s$ as the infinite-sheet approximation above the disk.

1. Find the maximum $z/R$ for the disk field to differ from $E_s$ by at most 5 percent of $E_s$.

2. Find the minimum disk radius needed at a height of 2.00 mm. Explain whether increasing density helps meet this relative-error criterion.

Question 3 - Solution

The infinite-sheet approximation is restricted by height relative to disk radius

(a) Method and calculation

$$\frac{E_s-E}{E_s}=\frac z{\sqrt{z^2+R^2}}\le.0500.$$$$z/R\le\frac{.0500}{\sqrt{1-.0500^2}}\approx.0500626.$$

(b) Interpretation and further calculation

$$R\ge(.00200)\frac{\sqrt{1-.0500^2}}{.0500}\approx.0399500\,\mathrm m.$$$$\text{relative error is independent of }\sigma.$$Changing density scales the exact and approximate fields together; extending the radius reduces the geometric error.

Physical check

This bound concerns the symmetry axis; near the edge a small height alone does not ensure the same approximation.