A Dipole in an Electric Field — Question 6

Question 6

A cubic field exposes the first finite-size correction to dipole force

A rigid axial dipole has $-q$ at $x-d/2$ and $+q$ at $x+d/2$, with $p=qd$. The external field on the x axis is $E_x(X)=AX^3$, where $A>0$. Its orientation is held fixed. The usual point-dipole approximation is $F_x=p\,dE_x/dx$.

1. Calculate the exact finite-dipole force and compare it with the point-dipole result.

2. For $x\ne0$, find the correction relative to the point-dipole result. Explain the force at $x=0$, where the local first derivative vanishes.

Question 6 - Solution

A cubic field exposes the first finite-size correction to dipole force

(a) Method and calculation

$$F_x=qA[(x+d/2)^3-(x-d/2)^3].$$$$F_x=pA\left(3x^2+\frac{d^2}{4}\right),\quad F_{\mathrm{point}}=3pAx^2.$$

(b) Interpretation and further calculation

$$\frac{F_x-F_{\mathrm{point}}}{F_{\mathrm{point}}}=\frac{d^2}{12x^2}\quad(x\ne0).$$$$F_x(0)=pAd^2/4>0.$$At the origin both charges feel positive-x forces: the negative charge is in a negative field and the positive charge is in a positive field. The local gradient alone misses this finite-separation effect.

Physical check

With p fixed and $d\to0$, the finite-size correction vanishes as $d^2$; for fixed d it dominates sufficiently close to the origin.