How Do You Determine the Electric Potential Outside a Charged Sphere?

[defineNormal]

Here's the question:

A sphere has a charge distributed throughout it such that the charge density goes as p(r)=Ar^(1/2). A=0.0200C/m^(7/12). The sphere has a radius of 2.00m, determine the electric potentioal at 3.00m from the center of the sphere?

well I got the charge density which is (0.0200C/m^(7/12))(2.00m)^(1/2)=0.0282 <---( not sure what unit would be C/m^(?)).

now E=(Q/a^3)(r), I solved for Q and ended up with Q=(E/r)(a^3), not 1) I am not sure I'm even on the right track, and 2) If I am do I use the charge density for E?, I'm really confused can someone help me? I don't where to start! ?

 

[estel]

OK - to get potential, in this case, since it's got spherical symmetry, you can exploit Guass' law

Take the total charge inside the sphere by integrating the charge density over the whole volume.

Then, by Guass' Law, the potential outside the sphere is the same as that for a point charge of the same total charge as the sphere, which is located at the centre of the sphere.