# Dipole moment clarification

I'm reading Griffith's E&M book, and he presents the following formulae for dipole moment $$\mathbf{p}=\iiint \mathbf{r}'\rho(\mathbf{r}')\ d^3\mathbf{r'}$$ where $\mathbf{r'}$ is the vector pointing from the origin to our charge distribution. What I don't understand about this equation is determining the $\mathbf{r'}$ term. For example, if we have a spherical shell of charge density $\sigma=\cos\theta$ with radius $R$ then $$\mathbf{p}=\iint \mathbf{r}'\cos\theta\ d^2\mathbf{r'}$$ where $d^2\mathbf{r'}=R^2\sin\theta\ d\theta\ d\phi$. My intuition was to say $\mathbf{r'}=R\hat{\mathbf{r}}$ but this doesn't seem to work since I know by intuition $\mathbf{p}$ should point in the $\hat{\mathbf{z}}$ direction. I guess the logical jump is then to say $\mathbf{r'}=R\cos\theta\hat{\mathbf{z}}$ but this seems wrong since $\mathbf{r'}$ should be the vector pointing to all the charges, not just the 'net' effective charge? Where is my flaw in understanding?

• It's worth noting that the dipole moment in this case depends upon the placement of the origin. see physics.stackexchange.com/questions/281477/… – R. Rankin Sep 3 '17 at 9:25
• 'doesn't seem to work' - have you actually done the integral and verified that it doesn't work? – Emilio Pisanty Sep 3 '17 at 11:27

If I got it correctly, you are creating a surface charge distribution on a sphere with radius $R$, concentrated around the equator. So the surface densitiy is $\sigma = k \cos(\vartheta)$, where $k$ is a constant with the correct unit of measure (C/m²). This distribution exists only on a thin layer d$r$, so that the volume density is: $$\rho = \begin{cases}\dfrac{k \cos(\vartheta)}{\mbox{d}r} & R \ge r \ge R+\mbox{d}r \\ 0 & \mbox{otherwise} \end{cases}$$
The complete transformation from cartesian to spherical coordinates is: $$\begin{cases} x = r \sin(\vartheta) \cos(\varphi) \\ y = r \sin(\vartheta) \sin(\varphi) \\ z = r \cos(\vartheta) \end{cases}$$ where $\vartheta$ is the angle between the radius and the $z$ axis and $\varphi$ is the angle between the projection of the radius on the $xy$-plane and the $x$ axis, as usual. These are also the components or the vector $\vec{r}$.
For d$^3\vec{r}$ we have: $$\mbox{d}^3\vec{r} = \mbox{d}V = \mbox{d}x\, \mbox{d}y\, \mbox{d}z = r^2 \sin(\vartheta)\, \mbox{d}r\, \mbox{d}\vartheta\, \mbox{d}\varphi$$ We can then fix $r = R$ and the dipole moment becomes: $$\vec{p} = \int \limits_0^{\pi} \mbox{d}\vartheta \int \limits_0^{2\pi} \mbox{d}\varphi\, \begin{pmatrix} R \sin(\vartheta) \cos(\varphi) \\ R \sin(\vartheta) \sin(\varphi) \\ R \cos(\vartheta) \end{pmatrix} k\, R^2 \cos(\vartheta) \sin(\vartheta)$$ Solving the $\varphi$ integral for the $x$ and $y$ components gives 0, so all that is left is the $z$ component: $$p_z = 2\pi k\, R^3 \int \limits_0^\pi \mbox{d}\vartheta\, \cos^2(\vartheta) \sin(\vartheta) = -\frac{2}{3} \pi k\, R^3\, \left. \cos^3(\vartheta) \right \vert_0^\pi = \frac{4}{3} \pi k\, R^3$$ In conclusion: $$\vec{p} = \frac{4}{3} \pi k\, R^3 \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$$ so your intuition was correct.