# Surface-charge of a uniformly polarized sphere

1. Jun 24, 2008

### Niles

1. The problem statement, all variables and given/known data
Hi all.

Please take a look at #2 in http://www.physics.utoronto.ca/~hori/Courses/P352F/hw6.pdf [Broken]

(this is not homework, I'm just training).

Ok, I can find the electric field inside the cavity y superposition and it is E = (-1/ 3*e_o)*P. I know that the bound surface charge is P*cos(theta), where theta is the usual spherical coordinate. My question is: Is the total charge on the inner surface P*cos(theta)*4*Pi*(R_0)^2?

Last edited by a moderator: May 3, 2017
2. Jun 30, 2008

### Irid

Where is $$\theta$$ in your last sentence? In order to find the total charge, you must integrate - not multiply - since the surface charge varies with respect to $$\theta$$. The proper way to find the surface charge would be

$$Q = \int_S \sigma(\theta) \, da = \int_0^{2\pi} d\phi \int_0^{\pi} \cos \theta \sin \theta d\theta = 0$$

It's zero, because on the northern hemisphere the bound charge is positive, which is compensated by the symmetrically negative bound charge on the southern hemisphere.

In conclusion, the answer is no.