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Physics · Ch 1 — Electric Charges and Fields

Electric Field of a Dipole at a General Point

1.11

Electric Field of a Dipole at a General Point

The axial result of Section 1.9 and the equatorial result of Section 1.10 are, in fact, two special cases of a single general formula for the field of a short dipole at a point PP whose position vector (measured from the dipole's centre OO) makes an angle θ\theta with the dipole axis.

Resolving the field at such a general point into a component along the radial direction (the line OPOP extended) and a component along the direction perpendicular to OPOP (tangential), the two components work out to be

Er=2kpcos⁡θr3,Eθ=kpsin⁡θr3E_r = \frac{2kp\cos\theta}{r^3}, \qquad E_\theta = \frac{kp\sin\theta}{r^3}

Combining these two perpendicular components by the Pythagorean rule gives the magnitude of the total field:

E=Er2+Eθ2=kpr34cos⁡2θ+sin⁡2θ=kpr31+3cos⁡2θE = \sqrt{E_r^2+E_\theta^2} = \frac{kp}{r^3}\sqrt{4\cos^2\theta+\sin^2\theta} = \frac{kp}{r^3}\sqrt{1+3\cos^2\theta}

(using sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta, so 4cos⁡2θ+sin⁡2θ=4cos⁡2θ+1−cos⁡2θ=1+3cos⁡2θ4\cos^2\theta+\sin^2\theta = 4\cos^2\theta+1-\cos^2\theta = 1+3\cos^2\theta).

Checking the two special cases already derived:

At θ=0∘\theta=0^\circ (a point on the dipole axis, Section 1.9): cos⁡θ=1\cos\theta=1, so E=(kp/r3)1+3=(kp/r3)×2=2kp/r3E=(kp/r^3)\sqrt{1+3}=(kp/r^3)\times2 = 2kp/r^3 -- exactly matching EaxialE_{\text{axial}}.

At θ=90∘\theta=90^\circ (a point on the equatorial line, Section 1.10): cos⁡θ=0\cos\theta=0, so E=(kp/r3)1+0=kp/r3E=(kp/r^3)\sqrt{1+0}=kp/r^3 -- exactly matching EeqE_{\text{eq}}. …