Electric Field inside a Sphere (Gauss' Law)

  • Thread starter jumbogala
  • Start date
  • #1
423
2

Homework Statement


A sphere has outer radius 15 and inner radius 5.

Between r= 5 and 15, the sphere is solid and contains a total charge of 20 C.

At r < 5, the sphere is hollow.

Calculate the electric field at r = 8.

Homework Equations



Gauss' Law --> EA = (q)/epsilon zero

The Attempt at a Solution


What I'm confused about is the hollow sphere part. But ignoring that, here's how I would do the problem.

The volume of the entire sphere is (4/3)pi(15^3) = 14137.16694 m^3.
The volume of the hollow part is (4/3)pi(5^3)= 523.598 m^3.

Therefore, the volume of just the solid part is 14137 - 523 = 13613.568 m^3.

The charge density per volume is then 20 / 13613 = 0.001469.

To calculate the electric field at r = 8, I need to find the volume of such a sphere.
Same procedure as above, take the volume minus the empty part to find 1621.06.

Multiply this by the charge density... 1621.06 * 0.001469 to find 2.38133 C.

This is the internal charge.

Take a gaussian surface of radius 8. It will have surface area (4pi)(8^2)
E = (internal charge)/ (epsilon zero)(area)

E= 2.38 / (eps. zero)(804.24) = huge number, 3.34 x 10 ^ 8.

What's confusing me is whether it's ok to take the surface area of a sphere with radius 8, because of that hollow part. Do I need to subtract the hollow part's surface area or something?
 
Last edited:

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
4
With the Gaussian surface you are determining the charge inside. The surface of course is the spherical surface at 8. No other surface need apply.

As to the rest of it, that looks OK. Subtract out the hollow volume to determine charge density and then determine the volume at 8 subtracting out the void again.
 
  • #3
423
2
So it doesn't really matter how the charge is distributed inside the Gaussian sphere, then? The charge can be all on the surface or all at the center, etc. and the procedure is still the same?
 

Related Threads on Electric Field inside a Sphere (Gauss' Law)

Replies
4
Views
2K
  • Last Post
Replies
1
Views
697
Replies
2
Views
6K
  • Last Post
Replies
8
Views
1K
Replies
2
Views
1K
Replies
3
Views
3K
Replies
9
Views
6K
Replies
7
Views
2K
  • Last Post
Replies
6
Views
2K
Top