In the Feynman Lectures on Physics, Volume 2, Chapter 6, Section 6-4 Feynman derives the electric potential due to a sphere with a surface charge distribution proportional to the cosine of the polar angle given by,

$$\sigma = \sigma_0 \cos \theta$$

Here is a link for reference.


Here he argues that we can use the superposition principle and say that the electric potential of such a sphere can be approximated by considering two spheres of equal radius $a$ and uniform charge density with opposite signs almost nearly overlapping each other.

Since the potential due to a single sphere of uniform charge density, outside the sphere, is same as if all the charge was concentrated at the center, he concludes that the potential due to this arrangement at very huge distances can be approximated as the potential of a dipole by the superposition principle.

$$\phi = \frac{1}{4\pi \epsilon_0}\frac{p\cos \theta}{r^2}$$

(If you don't understand my wording, please refer to the link above to see what I am talking about)

Here is my question,

He states that the dipole moment of this dipole is given by,

$$p=\frac{4}{3}\pi \sigma_0 a^3$$

I don't understand how he arrived at this expression for the dipole moment. Can someone please help me understand?


1 Answer 1


Feynman states this without explaining it. From what he's told you so far, it isn't obvious. (At least it isn’t obvious to me.)

In the next section, he defines the dipole moment for a collection of point charges as

$$\mathbf p=\sum_i q_i\mathbf d_i.\tag{6.26}$$

Physical "moments" are quantities that are weighted by their spatial positions. In this case, we are weighting each charge by how far ($\mathbf d$) it is from whatever origin we choose. This is similar to the definition of the center of mass in mechanics.

But the sphere with the non-uniform surface charge isn't a collection of point charges, so you need to understand the generalization of (6.26) to a continuous charge distribution. (He presumably gets around to this in a later lecture.)

The most general charge distribution is a volume charge density $\rho(\mathbf r)$. Every infinitesimal volume $dV$ contains an infinitesimal charge $dq=\rho\,dV$. So the general definition of the dipole moment is

$$\mathbf p=\int_V dq\,\mathbf d=\int_V \mathbf d\,\rho\,dV$$

where we are doing a volume integral over some volume $V$ where the charge distribution is. Remember that, to a physicist, integration is just summing up an infinite number of infinitesimal quantities, so you should think of this integral as the physically "obvious" generalization of the sum in (6.26).

(In other books, you'll usually see this written with an $\mathbf r$ instead of a $\mathbf d$, but it's the same thing.)

In the case of a surface charge, instead of a volume charge density $\rho$ we have a surface charge density $\sigma$. Since an infinitesimal area $dA$ contains charge $dq=\sigma dA$, the appropriate generalization is

$$\mathbf p=\int_S dq\,\mathbf d=\int_S \mathbf d\,\sigma\,dA$$

where we are integrating over some surface $S$. In this case it is a spherical surface, and the integration is most easily done in spherical polar coordinates. The $z$-component of the dipole moment is

$$\begin{align}p_z&=\int_\text{sphere}z\,\sigma\,dA =\int_\text{sphere}(a\cos\theta)(\sigma_0\cos\theta)(a^2\sin\theta\,d\theta\,d\phi)\\ &=2\pi\sigma_0a^3\int_0^\pi\cos^2\theta\sin\theta\,d\theta =2\pi\sigma_0a^3\int_{-1}^1u^2du =2\pi\sigma_0a^3\left(\frac23\right)\\ &=\frac43\pi\sigma_0a^3 \end{align}$$

while $p_x$ and $p_y$ can be shown by a similar calculation to be zero.

  • $\begingroup$ Thanks G.Smith. After reading the following sections, this approach makes a lot more sense. My confusion arose because I was not considering the generalised definition of dipole moment but rather I was thinking in terms of a dipole with two point charges. $\endgroup$
    – Ethan
    Dec 30, 2020 at 12:41

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