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## Homework Statement

A) Use Gauss's Law to derive the electric field in all space for a non-conducting sphere with volumetric charge distribution ρ=ρ

_{0}r

^{3}and radius, R.

B) Repeat when there is a concentric spherical cavity within the non conducting sphere with radius, A.

## Homework Equations

∫E⋅dA=Q

_{enclosed}/ε

_{0}

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=ρ

_{0}r

^{3}and dV=(4πr

^{2})(dr) lets plug those into our dQ:

dQ=(ρ

_{0}r

^{3})(4πr

^{2})(dr)

We can now find our Q

_{enclosed}of gaussian surface.

My below integral's limits go from 0→a (a being the radius of my gaussian surface)

Q

_{enclosed}=∫dQ=∫ (ρ

_{0}r

^{3})(4πr

^{2})(dr)=4πρ

_{0}∫r

^{4}→once evaluated= (4πρ

_{0}a

^{5})/5 (this is our Q

_{enclosed})

Now we can use gauss's law to find electric field

E(4πa

^{2})=(4πρ

_{0}a

^{5})/(5ε

_{0})

and finally:

E=(ρa

^{3})/(5ε

_{0})→E(r)=ρr

^{3})/(5ε

_{0})r(hat)

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 Q

_{enclosed}and just change the limits, instead of integrating from 0→a, we can integrate from 0→R. If we do this our Q

_{enclosed}becomes:

Q

_{enclosed}=(4πρ

_{0}R

^{5})/5

My new gaussian surface has radius a'

E(4πa'

^{2})=(4πρ

_{0}R

^{5})/(5ε

_{0})

solving for E:

E=(ρ

_{0}R

^{5})/(5a'

^{2}ε

_{0})

Now for part B

Again, new limits for Q

_{enclosed}, the limits now are going from A→a

our new Q

_{enclosed}=[(4πρ

_{0}a

^{5})/(5)]-[(4πρ

_{0}A

^{5})/(5)]

and our new E=[(ρ

_{0}a

^{3})/(5ε

_{0})]-[(ρ

_{0}A

^{5})/(5a

^{2}ε

_{0})

and outside our Q

_{enclosed}=[(4πρ

_{0}R

^{5})/(5)]-[(4πρ

_{0}A

^{5})/(5)]

and our new E=[(ρ

_{0}R

^{5})/(5a

^{2}ε

_{0})]-[(ρ

_{0}A

^{5})/(5a

^{2}ε

_{0})

Hope that was clear enough, please ask for more steps if Im wrong.

thanks!