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Formulas/physics/Electrostatic Potential/Potential on Equatorial Line of Dipole

Potential on Equatorial Line of Dipole

Potential is zero at all points on the equatorial plane (θ = 90°). The contributions from +q and −q are equal in magnitude and opposite in sign.
Class 11Class 12
Derivation

From the general formula

At θ=90°\theta = 90°: cos⁡90°=0\cos 90° = 0, so:

Veq=p⋅04πε0r2=0V_{eq} = \frac{p \cdot 0}{4\pi\varepsilon_0 r^2} = 0

Direct verification

On the equatorial plane, any point P is equidistant from +q+q and −q-q — both at distance r2+a2\sqrt{r^2 + a^2}:

V=q4πε0r2+a2+−q4πε0r2+a2=0V = \frac{q}{4\pi\varepsilon_0\sqrt{r^2+a^2}} + \frac{-q}{4\pi\varepsilon_0\sqrt{r^2+a^2}} = 0 Veq=0\boxed{V_{eq} = 0}

Physical interpretation

The equatorial plane is an equipotential surface at zero potential. No work is done in moving a charge along this plane. The field, however, is not zero on this plane — the field is antiparallel to p⃗\vec{p} with magnitude p/4πε0r3p/4\pi\varepsilon_0 r^3.

Note
$V = 0$ does not imply $\vec{E} = 0$. The field is the negative gradient of potential — a uniform non-zero potential implies zero field, but $V = 0$ at isolated points or planes says nothing about $\vec{E}$.