The Electric Field Due to an Electric Dipole — Question 4

Question 4

Equatorial finite-size corrections have the opposite approximation bias

At distance $r$ on the perpendicular bisector of a finite dipole, $E_{\mathrm{exact}}=kp/(r^2+a^2)^{3/2}$ in magnitude. The point-dipole estimate is $E_d=kp/r^3$.

1. Derive $(E_d-E_{\mathrm{exact}})/E_{\mathrm{exact}}$ and evaluate it at $r=10a$.

2. Find its leading small-$a/r$ form and compare the sign of the approximation bias with the axial case.

Question 4 - Solution

Equatorial finite-size corrections have the opposite approximation bias

(a) Method and calculation

$$\frac{E_d}{E_{\mathrm{exact}}}=(1+a^2/r^2)^{3/2}.$$$$r=10a:\quad\epsilon=(1.01)^{3/2}-1\approx.0150374.$$

(b) Interpretation and further calculation

$$u=a/r\ll1:\quad(1+u^2)^{3/2}=1+\tfrac32u^2+O(u^4).$$$$E_d>E_{\mathrm{exact}}\quad\text{on the perpendicular bisector}.$$Here each real source is farther from P than the center distance r, so the point-dipole estimate is too large.

Physical check

The axial and equatorial corrections need not have the same sign even for the same dipole.