The Electric Field Due to a Charged Disk — Question 7

Question 7

Oppositely charged finite disks add in the gap and form a distant dipole

Two thin insulating disks of equal radius R are centered at $z=-h$ and $z=+h$. They carry fixed uniform densities $+\sigma$ and $-\sigma$, respectively. Define $Q=\pi R^2\sigma$ and use k.

1. Find the field at the midpoint, including direction, and compare with the infinite-sheet result.

2. Find the leading far-axis field for $z\gg R,h$ using total dipole moment.

Question 7 - Solution

Oppositely charged finite disks add in the gap and form a distant dipole

(a) Method and calculation

$$\mathbf E(0)=4\pi k\sigma\left(1-\frac h{\sqrt{h^2+R^2}}\right)\hat{\mathbf z}.$$$$E_{\mathrm{sheets}}=4\pi k\sigma,\quad E(0)/E_{\mathrm{sheets}}=1-h/\sqrt{h^2+R^2}.$$

(b) Interpretation and further calculation

$$p_z=Q(-h)+(-Q)(h)=-2Qh.$$$$z\gg R,h:\quad\mathbf E\simeq-\frac{4kQh}{z^3}\hat{\mathbf z}.$$Inside the gap the fields reinforce toward the negative disk; far above both disks, the negative dipole moment determines the opposite direction.

Physical check

These are prescribed uniform insulating charge distributions; actual conducting disks generally have nonuniform edge charge.