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.
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?