# Potential of surface charge

I have a question about the $\hat{n}$ in this formula $\sigma = P \dot{}\hat{n}$.

Why do sometime in my book they get $\sigma = P \cos{\theta}$ for a sphere. Isn't $\hat{n} = r$ ?

And then in another problem, where $P = \cfrac{k}{r}\hat{r}$ and r = a (inner) and r = b (outer) they get $\sigma =\cfrac{k}{a}$ and $\sigma =\cfrac{k}{b}$.

$\sigma$ = surface charge density

$P$ = polarization

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Please define all the variables. –  leongz Apr 9 '12 at 5:31
@leongz: Ah, so it wasn't just me who got confused :P I always thought that $\sigma$ was conductivity and $P$ was polarization. –  Manishearth Apr 9 '12 at 8:01
Ok I just updated the question –  user101699 Apr 9 '12 at 11:45
Of course we should always define our variables, but I would note that $\sigma$ is the standard for surface charge density. –  David Z Apr 10 '12 at 0:51
@DavidZaslavsky: Aah, how stupid of me :/ –  Manishearth Apr 10 '12 at 0:52

In general, the surface charge density $\sigma$ of a polarized material with polarization $\mathbf{P}$ is indeed given by $\sigma=\mathbf{P}\cdot\hat{\mathbf{n}}$, where $\hat{\mathbf{n}}$ is the surface normal vector.

For a sphere, we have $\hat{\mathbf{n}}=\hat{\mathbf{r}}$ in spherical coordinates. So, assuming that $\mathbf{P}$ points in the $z$-axis, we obtain $$\sigma=\mathbf{P}\cdot\hat{\mathbf{r}}=P\hat{\mathbf{z}}\cdot\hat{\mathbf{r}}=P\cos{\theta}.$$

For the next case, we have a spherical shell with inner radius $a$ and outer radius $b$, with polarization $\mathbf{P}=\frac{k}{r}\hat{\mathbf{r}}$. Applying the general formula, we get

$$\sigma=\mathbf{P}\cdot\hat{\mathbf{r}}=\frac{k}{r}\hat{\mathbf{r}}\cdot\hat{\mathbf{r}}=\frac{k}{r}.$$ Letting $r=a$ and $r=b$ gives the surface charge density for the inner and outer surface respectively.

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A hat over a vector typically denotes that it is a unit vector. So, for any vector $\bf{a}$ the notation $\hat{\bf{a}}$ means $\hat{\bf{a}}=\dfrac{\bf{a}}{|\bf{a}|}$.
The term $\cos{\theta}$ probably comes from contraction of two unit vectors, as for any pair of unit vectors $\hat{\bf{n}}_1\cdot\hat{\bf{n}}_2=\cos{\theta_{12}}$ (the angle between the vectors).
In the problem that you gave $\hat{\bf{n}}$ is parallel to $\hat{\bf{r}}$, hence the cosine term is equal to unity, and the rest comes from $P=\sigma$.