# Electric Field

1. Apr 10, 2008

### tronter

A thin spherical shell of radius $$R$$ has charge density $$+ \sigma$$ on the upper half and $$- \sigma$$ on the bottom half. Determine the electric field both inside and outside the sphere.

So its an area charge density. So I tried using Gauss's law: $$\oint \bold{E} \cdot d \bold{a} = \frac{Q_\text_{int}}{\epsilon_{0}}$$.

$$E(\pi r^2) = \frac{\sigma}{\epsilon_{0}}$$.

2. Apr 10, 2008

### Staff: Mentor

OK. What did you get for the field?

3. Apr 10, 2008

### Gear300

I'm not completely sure of this answer but...since its a surface charge density, the sphere seems to match the conditions of a conductor in electrostatic equilibrium. So the net electric field inside the sphere should simply be 0N/C. Outside the sphere, you'll get that the net electric flux is 0...but that doesn't mean that the electric field outside the sphere is 0. Near the surface of the sphere, the electric field should be $$E(\pi r^2) = \frac{\sigma}{\epsilon_{0}}$$ (what you basically did) and the further away you are, it should come close enough to 0N/C since the net charge on the sphere is 0C.

4. Apr 10, 2008

### Staff: Mentor

Note that the problem specifies upper half and lower half, not inner surface and outer surface. (The latter would indeed be an easy problem to solve using Gauss's law; the former, not so easy.)