The Electric Field Due to an Electric Dipole — Question 1

Question 1

The exact axial dipole field reveals cancellation of the inverse-square terms

A finite dipole has $-q$ at $x=-a$ and $+q$ at $x=a$, with $q,a>0$. Define $\mathbf p=2qa\hat{\mathbf x}$ and use Coulomb constant k. Observe at $x>a$.

1. Derive the exact axial field by adding the signed point-charge fields.

2. Find its leading form for $x\gg a$ at fixed p and explain the faster decay than a single charge.

Question 1 - Solution

The exact axial dipole field reveals cancellation of the inverse-square terms

(a) Method and calculation

$$E_x=kq[(x-a)^{-2}-(x+a)^{-2}]=\frac{4kqax}{(x^2-a^2)^2}.$$$$E_x=\frac{2kpx}{(x^2-a^2)^2}=\frac{2kp}{x^3}(1-a^2/x^2)^{-2}.$$

(b) Interpretation and further calculation

$$x\gg a:\quad E_x\simeq2kp/x^3.$$$$Q_{\mathrm{total}}=q-q=0,\quad p=2qa\ne0.$$The leading inverse-square contributions cancel, leaving a field controlled by the separation-weighted charge.

Physical check

The finite formula is valid only outside the positive charge on the specified axial side; it diverges at the point source $x=a$.