Potential Due to an Electric Dipole — Question 9

Question 9

The first finite-size correction has its own angular zeros

A dipole has charges $\pm q$ at z=$\pm a$. For r>a use the expansion $1/|\mathbf r-a\hat{\mathbf z}|=r^{-1}\sum_{\ell=0}^{\infty}(a/r)^\ell P_\ell(\cos\theta)$, with $P_1(c)=c$ and $P_3(c)=(5c^3-3c)/2$.

1. Use charge antisymmetry to retain the first two nonzero terms in V.

2. Find angles where the cubic correction vanishes and discuss the next error order at the non-equatorial zeros.

Question 9 - Solution

The first finite-size correction has its own angular zeros

(a) Method and calculation

$$V=\frac{2kq}{r}\left[\frac ar P_1(\cos\theta)+\frac{a^3}{r^3}P_3(\cos\theta)+\cdots\right].$$$$V=\frac{kp\cos\theta}{r^2}+\frac{2kqa^3}{r^4}P_3(\cos\theta)+O(kqa^5/r^6),\quad p=2qa.$$

(b) Interpretation and further calculation

$$P_3(c)=0\Rightarrow c=0,\ \pm\sqrt{3/5}.$$$$c=\pm\sqrt{3/5}:\quad\frac{V-V_{\mathrm{dip}}}{V_{\mathrm{dip}}}=O((a/r)^4).$$At the equator every odd term vanishes and V is exactly zero.

Physical check

A relative error referenced to the dipole voltage is undefined on the equator, so that exact-zero case must be treated separately.