# Dipole's electric field and potential at the equatorial plane

The potential at the equatorial plane of the dipole is $0$

Again,The E at point at the the equatorial plane of the dipole is $\frac{-p}{4\pi\epsilon r^{3}}$

First of all, $-p/(4 \pi \epsilon_0)$ is $E_z$, not $\vec{E}$. $\vec{E}$ is a vector so you need to specific which component you're talking about.
At the equatorial plane, $V = 0$ but $E_z = -dV/dz \neq 0$. You can have a function be zero at a point but its derivative not be 0.
• @ItachíUchiha It's a line integral $\vec{E} \cdot \vec{ds}$. If the path of integration lies in the equatorial plane, then $\vec{E}$ is parallel to $\hat{z}$ and $\vec{ds}$ is perpendicular to $\hat{z}$, so their dot product is always zero, and the integral will be as well. – tparker Sep 18 '16 at 9:06