# Electric field of a dipole

I'm trying to find the electric field due to an electric dipole $\mathbf{d}$. There are plenty of approaches to doing this online, but I want to do it "my way," which doesn't seem to be working (and I have yet to find this approach in textbooks/online). I start with the potential:

$$\phi(\mathbf{r}) = \frac{\mathbf{d}\cdot \mathbf{r}}{4\pi\epsilon_0 r^3} = \frac{d\cos\theta}{4\pi\epsilon_0 r^2}$$

And then take the negative gradient to find:

$$\mathbf{E} = -\nabla \phi = \frac{2d\cos\theta}{4\pi \epsilon_0 r^3} \hat{r} + \frac{d\sin\theta}{4\pi\epsilon_0 r^3}\hat{\theta}$$

But I don't see how to manipulate this into the form that I expect:

$$\mathbf{E} = \frac{1}{4\pi\epsilon_0}\left[ \frac{3(\mathbf{d}\cdot\hat{r})\hat{r}-\mathbf{d}}{r^3} \right] - \frac{1}{3\epsilon_0}\mathbf{d} \delta^3(\mathbf{r})$$

Since I haven't done anything mathematically illegal, I don't see why my approach shouldn't get me to the correct answer - but I can't figure out what to do (and I've also been unable to find this derivation online).

• What you did was find the electric field in spherical coordinates. The form you're talking about (written in an unfamiliar way) is the coordinate free "version". – Astrum May 29 '14 at 6:43
• See de.wikipedia.org/wiki/Dipol#Punktdipol (even if you can't read it). There you have exactly your formula with $d$ replaced by $p$. – Tobias May 29 '14 at 6:55
• @Tobias Right, I know that those are /supposed/ to be equal, but I'm trying to figure out how to show that they are. – alexvas May 30 '14 at 1:32

$\def\vE{{\mathbf E}}\def\vd{{\mathbf d}}\def\vr{{\mathbf r}}\def\hr{{\hat r}}\def\hd{{\hat d}}\def\eps{\varepsilon}\def\l{\left}\def\r{\right}\def\htheta{{\hat\theta}}$ EDIT: The expected formula is corrected now in the question. So the following canceled comment is out-of-date. The formula you expect is false (see e.g., this answer or this wikipedia-page). Replace $\vd\cdot\vr$ with $\vd\cdot\hr$ to correct it. The right formula for $\vr\neq\mathbf0$ is: $$\vE = \frac1{4\pi\eps_0}\l[\frac{3(\vd\cdot\hr)\hr-\vd}{r^3}\r].$$ There $\hr(\hr\cdot\vd)$ is the orthogonal projection of $\vd$ onto the $r$-axis. Furthermore, $\vd-\hr(\hr\cdot\vd)$ is the orthogonal projection of $\vd$ onto the $\hr$-plane (for which $\hr$ is the normal vector). Note, that you can write this projection also as $-\hr\times(\hr\times\vd)$ if this is more familiar to you.
You get the $\vE$-field in the form \begin{align} \vE &= \frac1{4\pi\eps_0}\l[\frac{2(\vd\cdot\hr)\hr+\hr(\hr\cdot\vd)-\vd}{r^3}\r]\tag{proj}\label{proj} \end{align} We have trivially $(\vd\cdot\hr)=d\cos(\theta)$ where $\theta$ is the angle between $\vd$ and $\hr$ (sign does not matter here).
The component of the orthogonal projection of $\vd$ onto the $\hr$-plane has length $|d\sin(\theta)|$. We assume here that $\hr$ and $\hd$ are not collinear. We define the direction vector $\htheta$ as $\htheta:=\frac{\hr\times(\hr\times\vd)}{|\hr\times(\hr\times\vd)|}$ such that $\theta>0$ and get \begin{align} \vE &= \frac{d}{4\pi\eps_0}\l[ \frac{2\cos(\theta)\hr+\sin(\theta)\htheta}{r^3} \r].\tag{final}\label{final} \end{align} The above question does not explicitly state the rule of measurement for $\theta$. So I fixed it with the above definition of the unit vector $\htheta$. If $\hr$ and $\vd$ are not collinear there is only the choice between $\htheta:=\frac{\hr\times(\hr\times\vd)}{|\hr\times(\hr\times\vd)|}$ and $\htheta:=-\frac{\hr\times(\hr\times\vd)}{|\hr\times(\hr\times\vd)|}$. This just follows from the term $\hr(\hr\cdot\vd)-\vd=\vr\times(\vr\times\vd)$ in \eqref{proj}. The choice $\htheta:=-\frac{\hr\times(\hr\times\vd)}{|\hr\times(\hr\times\vd)|}$ would imply $\theta<0$. The orientation change of $\htheta$ compensates the sign change of $\theta$ in equation \eqref{final}.
• Thank you for a very complete and intuitive answer. One question, though: why is $\hat{\theta} \sim \hat{r}\times(\hat{r}\times\mathbf{d})$? – alexvas May 31 '14 at 0:58
• @alexvas I added the missing $d$ in the final formula for $\vE$ and added some motivation for the choice of $\htheta$. – Tobias May 31 '14 at 4:31