A) Use Gauss's Law to derive the electric field in all space for a non-conducting sphere with volumetric charge distribution ρ=ρ0r3 and radius, R.
B) Repeat when there is a concentric spherical cavity within the non conducting sphere with radius, A.
ρ=Q/V (charge density)
The Attempt at a Solution
Im struggling with part B more than A, but since B uses part A's answer (in a way) I will show both my attempts
A) I drew a picture (sphere.jpg attached) and will use words to explain what Im doing.
My sphere has radius, R.
My Gaussian Surface has radius, A.
I will be finding dQ of a small spherical piece of the sphere, the radius of this small spherical piece is, r, and has thickness, dr.
First find dQ
we know that ρ=Q/V →dQ=ρdV since ρ=ρ0r3 and dV=(4πr2)(dr) lets plug those into our dQ:
We can now find our Qenclosed of gaussian surface.
My below integral's limits go from 0→a (a being the radius of my gaussian surface)
Qenclosed=∫dQ=∫ (ρ0r3)(4πr2)(dr)=4πρ0∫r4→once evaluated= (4πρ0a5)/5 (this is our Qenclosed)
Now we can use gauss's law to find electric field
This is the answer if you are inside the sphere at some distance r<R
Now for outside the sphere:
We can go to the Qenclosed and just change the limits, instead of integrating from 0→a, we can integrate from 0→R. If we do this our Qenclosed becomes:
My new gaussian surface has radius a'
solving for E:
Now for part B
Again, new limits for Qenclosed, the limits now are going from A→a
our new Qenclosed=[(4πρ0a5)/(5)]-[(4πρ0A5)/(5)]
and our new E=[(ρ0a3)/(5ε0)]-[(ρ0A5)/(5a2ε0)
and outside our Qenclosed=[(4πρ0R5)/(5)]-[(4πρ0A5)/(5)]
and our new E=[(ρ0R5)/(5a2ε0)]-[(ρ0A5)/(5a2ε0)
Hope that was clear enough, please ask for more steps if Im wrong.