Gauss Law problem

Where does it say in the question that you must only use Gauss' Law to make a statement about the charge ditribution ?

Anyway, saying that the charge distribution is uniform would be incorrect. The right answer should be that :
[tex]\rho (r, \theta , \phi) = \rho _0 f(r) [/itex]

 

I think you're making this problem more difficult than it is. There is no charge on the surface of the sphere...we don't even know whether it's a conductor or a dielectric, and such considerations are irrelevant, because it need not even be a physical surface! It's just a Gaussian surface, one at which we are measuring the properties of the local electric field in order to determine some information about the distribution of charge enclosed by said surface. Gauss's law can also used to solve the opposite problem (given some info about the charge enclosed, what can you deduce about the flux of the electric field, and therefore the electric field itself?)

In this case, given the properties of the local electric field (more specifically, the field at all points on the surface of the sphere), what information does Gauss's Law (below) tell us about the charge enclosed?

[tex] \oint{\vec{E} \cdot d\vec{a}} = \frac{Q_{enc}}{\epsilon_0} [/tex]

It gives us two pieces of information:

1. The field is radially inward...an inward field suggests that Q_enc is some cluster of negative charge. More rigorously, since [itex] \vec{E} \ \ \text{and} \ \ d\vec{a} [/itex] point in opposite directions, (radially inward and outward, respectively) it follows that [itex] \vec{E} \cdot d\vec{a} [/itex] is negative: The net flux through the surface is negative, therefore so is Q_enc.

2. The electric field is radial and its strength is constant at all points on the sphere. Since all of these points are equidistant from the centre, it seems that the magnitude of the electric field, depends only on radius i.e. it is a function of r. What would produce such a nice, radially inward E(r)? A point charge at the centre of the Gaussian sphere certainly would, and IIRC, so would any uniform volume charge distribution centred in the Gaussian sphere.

There are your two answers...that the charge is negative, and that it is distributed uniformly.