# 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". Commented May 29, 2014 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 $p$. Commented May 29, 2014 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. Commented May 30, 2014 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 So the following canceled comment is out-of-date.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 $\hat{\theta} \sim \hat{r}\times(\hat{r}\times\mathbf{d})$? Commented May 31, 2014 at 0:58
• @alexvas I added the missing $d$ in the final formula for $\vE$ and added some motivation for the choice of $\htheta$. Commented May 31, 2014 at 4:31