Academy
Formulas/physics/Electrostatic Potential/Potential due to Dipole (General)

Potential due to Dipole (General)

Potential at distance r from the centre of a dipole of moment p, at angle θ from the dipole axis. Approximation valid for r ≫ a.
Class 11Class 12
Derivation

Setup

Dipole: +q+q at (a,0)(a, 0), −q-q at (−a,0)(-a, 0), dipole moment p=2qap = 2qa along +x+x.

Point P at distance rr from centre, at angle θ\theta to the dipole axis. Let r+r_+ and r−r_- be distances from +q+q and −q-q to P.

Superposition

V=q4πε0r++−q4πε0r−=q4πε0(1r+−1r−)V = \frac{q}{4\pi\varepsilon_0 r_+} + \frac{-q}{4\pi\varepsilon_0 r_-} = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{r_+} - \frac{1}{r_-}\right)

Far-field approximation (r≫ar \gg a)

Using the cosine rule and binomial approximation:

r+≈r−acos⁡θ,r−≈r+acos⁡θr_+ \approx r - a\cos\theta, \qquad r_- \approx r + a\cos\theta 1r+−1r−=r−−r+r+r−≈2acos⁡θr2\frac{1}{r_+} - \frac{1}{r_-} = \frac{r_- - r_+}{r_+ r_-} \approx \frac{2a\cos\theta}{r^2}

Therefore:

V=q4πε0⋅2acos⁡θr2=pcos⁡θ4πε0r2V = \frac{q}{4\pi\varepsilon_0} \cdot \frac{2a\cos\theta}{r^2} = \frac{p\cos\theta}{4\pi\varepsilon_0 r^2} V=pcos⁡θ4πε0r2\boxed{V = \frac{p\cos\theta}{4\pi\varepsilon_0 r^2}}

Key features

  • V∝1/r2V \propto 1/r^2 — falls faster than a point charge (1/r1/r)
  • V=0V = 0 on the equatorial plane (θ=90°\theta = 90°)
  • Maximum VV on the axis (θ=0°\theta = 0°, 180°180°)
Note
The potential depends on both $r$ and $\theta$. Equipotential surfaces are not spheres — they are more complex closed surfaces symmetric about the dipole axis.