The Electric Field Due to a Charged Disk — Question 8

Question 8

Two equally charged disks have an axial field that changes sign through the midpoint

Two disks of radius R at $z=\pm h$ both carry fixed uniform positive density $\sigma$. Observe only on their common axis between them. Let $f(s)=1-s/\sqrt{s^2+R^2}$ for $s>0$.

1. Write $E_z(z)$ for $|z|<h$ and find its derivative at the midpoint.

2. Give the leading field for $|z|\ll h$ and the force direction on a positive test charge displaced slightly above the midpoint.

Question 8 - Solution

Two equally charged disks have an axial field that changes sign through the midpoint

(a) Method and calculation

$$E_z(z)=2\pi k\sigma[f(h+z)-f(h-z)],\quad E_z(0)=0.$$$$f\prime(s)=-\frac{R^2}{(s^2+R^2)^{3/2}},\quad E_z\prime(0)=-\frac{4\pi k\sigma R^2}{(h^2+R^2)^{3/2}}.$$

(b) Interpretation and further calculation

$$E_z(z)\simeq-\frac{4\pi k\sigma R^2}{(h^2+R^2)^{3/2}}z.$$$$z>0,q>0:\quad F_z=qE_z<0.$$The nearer upper disk pushes more strongly downward, creating an axial restoring tendency. This alone says nothing about transverse stability.

Physical check

One-dimensional restoring behavior is insufficient to establish a freely stable electrostatic trap.