The Electric Field Due to an Electric Dipole — Question 2

Question 2

On the perpendicular bisector the dipole field points opposite its moment

The same finite dipole has $-q$ at $(-a,0)$ and $+q$ at $(a,0)$. Find its field at $(0,y)$ with $y>0$, defining $p=2qa$.

1. Resolve and sum both source fields to obtain the exact vector.

2. Compare its far-field magnitude at $y=r\gg a$ with the axial magnitude at the same center distance.

Question 2 - Solution

On the perpendicular bisector the dipole field points opposite its moment

(a) Method and calculation

$$\mathbf E_+=\frac{kq(-a,y)}{(a^2+y^2)^{3/2}},\quad\mathbf E_-=\frac{-kq(a,y)}{(a^2+y^2)^{3/2}}.$$$$\mathbf E_{\mathrm{eq}}=-\frac{kp}{(a^2+y^2)^{3/2}}\hat{\mathbf x}.$$

(b) Interpretation and further calculation

$$r\gg a:\quad|\mathbf E_{\mathrm{eq}}|\simeq kp/r^3,\quad|\mathbf E_{\mathrm{axial}}|\simeq2kp/r^3.$$$$|\mathbf E_{\mathrm{axial}}|/|\mathbf E_{\mathrm{eq}}|\simeq2.$$The vertical components cancel while both horizontal contributions point from the positive side toward the negative side.

Physical check

The field on the perpendicular bisector is not zero despite equal distances to the two charges.