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Electrostatic energy with gauss' law

  • Thread starter bishy
  • Start date
13
0
1. 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?


2. Homework Equations
https://www.physicsforums.com/latex_images/13/1397427-0.png [Broken]
Qenclosed = r^3/R^3
flux= 4pi*r^2*E
3. 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.
 
Last edited by a moderator:

Answers and Replies

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.
 
13
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Can you explain why? I don't think thats 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 :)
 
Last edited:
208
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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.
 
208
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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.
 
Last edited:
13
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Thanks man. Solved it with your advice.
 

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