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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 $\frac{-p}{4\pi\epsilon r^{3}}$

But this link says: enter image description here

Then,why do not work this equation in case of the equatorial plane of the dipole?

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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.

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  • $\begingroup$ But can the integration of -Eds be 0? $\endgroup$ – user106015 Sep 18 '16 at 8:56
  • $\begingroup$ @ItachíUchiha I don't understand your question. Integral over what path? $\endgroup$ – tparker Sep 18 '16 at 8:58
  • $\begingroup$ How can -\int_{a}^{b} Eds=\int_{a}^{b}dv=0 in the equatorial plane of dipole? $\endgroup$ – user106015 Sep 18 '16 at 9:01
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    $\begingroup$ @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. $\endgroup$ – tparker Sep 18 '16 at 9:06

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