I want to calculate the dipole potential in spherical coordinates. I know that the potential can be calculated with $$ \phi = - \int \mathbf E \cdot\mathrm d\mathbf r,$$ but I don't know the electric field. I would say $$ \mathbf E = \frac{1}{4 \pi \epsilon_0 r^3} ( 3p\,\hat{\mathbf {r}} \cos(\theta) -\mathbf p) $$ is the electric field in spherical coordinates but I'm not sure because I didn't calculate this formula, in fact I got it from a book and don't understand the way it's calculated.


The easiest derivation is probably to start with the potential, and then calculate the electric field as the gradient of that potential.

Consider a dipole at the origin aligned along the z-axis, with two point charges of charge $ \pm q $ positioned at $z = \pm\frac{a}{2}$. By symmetry, the potential will be independent of the azimuthal angle, so we only need to consider the potential at the coordinates $\left( r, \theta \right)$. Since we are trying to find the field around a point dipole, we can assume $r >> a$.

The distance of this point to the two charges is given by the cosine rule as $r_\pm^2 = \frac{a^2}{4} + r^2 \mp ar \cos \theta$.

The potential at this point is then given by summing the potential from the two point charges, which is $V\left(r, \theta\right) = \frac{q}{4\pi\varepsilon_0}\left(\left(\frac{a^2}{4} + r^2-ar\cos\theta\right)^{-\frac{1}{2}}-\left(\frac{a^2}{4} + r^2+ar\cos\theta\right)^{-\frac{1}{2}}\right)$. The approximation $r >> a$ suggests that we can Taylor expand this in $\frac{a}{r}$, and that is what we will do.

Taylor expanding gives $V\left(r, \theta\right) = \frac{qa \cos \theta}{4\pi\varepsilon_0 r^2}$. Any higher-order terms can always be ignored, because this is an ideal point dipole.

Now all that's left is finding the electric field from this potental. We know that $ \vec E = - \nabla V$. Taking the gradient of the dipole potential gives $\vec E = \frac{qa}{4 \pi \varepsilon_0 r^3} \left( 2 \cos \theta \hat r + \sin \theta \hat \theta \right) $.

This is nice, but we'd like an expression in terms of the dipole moment; there aren't really any point charges separated by a physical distance, that was just scaffolding to get us this far. Fortunately, the way we've defined the coordinate axes gives us a nice expression for the dipole moment: $\vec p = qa \hat z$, or, in spherical coordinates, $\vec p = qa\left( \cos \theta \hat r - \sin \theta \hat \theta\right)$. So $ \vec p . \hat r = p \cos \theta = qa \cos \theta $.

Putting this all into the expression for the electric field gives us $ \frac{1}{4 \pi \varepsilon_0 r^3} \left( 3p \cos \theta \hat r - \vec p \right)$, just as planned.

  • $\begingroup$ Thanks for the detailed answer. The charge of the point charges is -2Q and Q. So this will change the second term in the potential by the factor of 2. I am a little bit confused by the cosin rule. What is the difference between $r_\pm$ and r and isn't the cosin rule: $c^2 = a^2 +b^2 - 2 * a*b*cos(\theta)$? $\endgroup$ – gamma Sep 5 '18 at 14:13
  • $\begingroup$ $r$ is the distance of the point at which we're measuring the potential from the origin, and $r_\pm$ is the distance of that point from the $\pm q$ charges respectively. For the cosine rule, the side that we're trying to find is $r_+$ or $r_-$, and the two sides we know are $r$ and either $+ \frac{a}{2}$ or $-\frac{a}{2}$. $\endgroup$ – Diamond-Grouping Sep 5 '18 at 14:19

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see above is a dipole with electric dipole moment $\mathbf{\vec{p}}$

since you know potential due to dipole it means $$\phi(r,\theta)=\dfrac { \mathbf{\vec{p}}\cdot\mathbf {\hat r}}{4\pi \epsilon_{0}r^2}=\dfrac{p\cos\theta}{4\pi\epsilon_{0}r^2}$$

To find radial component of field $$E_{r}=-\dfrac{\partial \phi}{\partial r}=\dfrac{2\ p \cos\theta}{4\pi\epsilon_{0}\ r^3}$$

To find angular component of electric field
$$E_{\theta }=-\frac{1}{r}\cdot\frac{\partial \phi}{\partial \theta}=\dfrac{p\sin\theta}{4\pi\epsilon_{0}\ r^3}$$


$$\mathbf{E}_\text{dip}(r,\theta)=\dfrac{p}{4\pi\epsilon_{0}\ r^3}\left(2\cos\theta\ \mathbf {{\hat r}}+\sin \theta\ \mathbf {\hat \theta}\right)$$ (in polar cordinate form)

$$\mathbf{E}_\text{dip}=\dfrac{1}{4\pi\epsilon_{0}\ r^3}\left(3(\mathbf {\vec p} \cdot \mathbf {{\hat r}})\mathbf{\hat r}-\mathbf{\vec p }\right)$$ (in coordinate free form)

$$\mathbf{E}_\text{dip}(r,\theta)=\dfrac{1}{4\pi\epsilon_{0}\ r^3}\left(3(p\cos\theta)\mathbf{\hat r}-\mathbf{\vec p }\right)$$


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