Electrostatic energy with gauss' law

[bishy]

Homework Statement

A solid sphere contains a uniform volume charge density (charge Q, radius R).
(a) Use Gauss’s law to find the electric field inside the sphere.
(b) Integrate
E^2 over spherical shells over the volumes inside and outside the sphere.
(c) What fraction of the total electrostatic energy of this configuration is contained within the sphere?

Homework Equations

https://www.physicsforums.com/latex_images/13/1397427-0.png
Qenclosed = r^3/R^3
flux= 4pi*r^2*E

The Attempt at a Solution

a) E=(Q*r)/(4*pi*(epsilon0)*R^3)
b) So I am thinking for this one that I need to integrate E^2 with upper limits being inside and lower limits being the outside of the sphere. what I'm not sure is if its intergral(E^2 dE) or if a value inside of E is being integrated. R or r would make sense to intergrate as well hence intergral(E dr)
c) since I can't solve b, I can't solve c either.

 

[Mike Cookson]

I'm not entirely sure how to prove it with gauss's law as it was a common assumption made when I use it but there is no electrical field inside a solid conductor.

 

[bishy]

Can you explain why? I don't think that's right. Heres why. A solid sphere has a point of symmetry where the distance from this point gives you a charge density called r. In the question it states that the charge is uniform so I expect the electric field at a distance r within a sphere to be Q (as given with the problem) given that its symmetrical. if R is the full radius of the sphere than R-r does not affect electric field due to symmetry.

Also I've found an example problem similar to this one with a given E inside a solid uniformly charged sphere within Essential University Physics by Richard Wolfson? Now I'm confused :)

 

[EngageEngage]

it doesn't have a field only if its a conductor in which case all charge resides on the surface. This is easy to see if you consider the following: say you put some charge on one point of a sphere. The charge will immediately flow due to coulomb forces and it will flow until all charges are furthest from each-other and all forces are balanced. This happens when you have charge distributed evenly on the surface of the conductor.

However, you can have a sphere with charge density that is an insulator which will have a field in its interior.

 

[EngageEngage]

Forget what I wrote here earlier (if you read it) i was thinking capacitors. But if they want you to integrate over volume using spherical shells the generic set up for something spherically symmetrical will be:

Int(4*pi r^2)dr from 0 to R.

 

[bishy]

Thanks man. Solved it with your advice.