Electric field on the equatorial line of a dipole

[DrBanana]

$\vec{E}$ on the line that perpendicularly bisects the segment that joins two equal and opposite charges is non-zero, as it should be. But the potential of any point along that line is zero. But we know know that $E=-\frac{dV}{dr}$, where V is approximately $\frac{1}{4\pi \epsilon} \frac{pcos\theta}{r}$ (if the charges are close together) where p is the magnitude of the dipole moment . If I differentiate that with respect to r and set $\theta=\frac{\pi}{2}$, I still get E=0. What gives?

 

[kuruman]

What gives?

Nothing gives. The electric field is a three dimensional vector. You only calculated its radial component in the equatorial plane.

 

[renormalize]

$E=-\frac{dV}{dr}$

That's wrong. It should be:$$\overrightarrow{E}=-\overrightarrow{\nabla}V=-\hat{r}\frac{\partial V}{\partial r}-\hat{\theta}\frac{1}{r}\frac{\partial V}{\partial\theta}-\hat{\phi}\frac{1}{r\sin\theta}\frac{\partial V}{\partial\phi}$$

 

[DrBanana]

That's wrong. It should be:$$\overrightarrow{E}=-\overrightarrow{\nabla}V=-\hat{r}\frac{\partial V}{\partial r}-\hat{\theta}\frac{1}{r}\frac{\partial V}{\partial\theta}-\hat{\phi}\frac{1}{r\sin\theta}\frac{\partial V}{\partial\phi}$$

Ok I think I understand what happened. My physics book doesn't touch on vector calculus and only mentions $E=-\frac{dV}{dr}$, however most of the forces in the book only have radial components anyway so it didn't matter. But that broke down here.

 

[kuruman]

The dipole potential in spherical coordinates is $$V=\frac{1}{4\pi \epsilon_0}\frac{p~\cos\!\theta}{r^2}.$$Note the correct power of $r$ in the denominator. Also note that with $z=r\cos\!\theta$, you have $$V=\frac{1}{4\pi \epsilon_0}\frac{p~z}{r^3}.$$This last expression can be considered to be the dipole potential in cylindrical coordinates. In the equatorial plane ($z=0$) the potential vanishes but not its derivative with respect to $z$ which you can easily calculate.