# Non-conductve sphere with cavity -- find Electric field

I have a non conducting sphere with a charge ρ=A/r per uni vollume A is constant. suppose there is a cavity in the centre and within a particle of charge q. i want to find the E inside the sphere in respect with r.

## The Attempt at a Solution

for radius equal of the cavity i get ##E=kq/r^2## for r bigger than the radius of the sphere ##E=k(Q+q)/r^2## now only one case left
if r is bigger than the radius of the cavity and smaller than the radius of the sphere.
##E4πr^2=\frac{q+ρV}{ε_0} ##
where ρ is the density per unit vollume and V is the vollume
is my enclosed Q right? and my V is 4π(r^3-a^3) where a^3 is the small radius.

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gneill
Mentor
is my enclosed Q right?
Since the charge density varies with radius you'll need to take that into account. Hint: an integration is required.

Since the charge density varies with radius you'll need to take that into account. Hint: an integration is required.
???

gneill
Mentor
Think of the sphere being comprised of many thin spherical shells, each having a slightly different charge density according to their radii. No two shells have the same charge density due to your given expression for the charge density, ρ=A/r. The enclosed charge will be the sum of the charges in those shells "below" the radius of interest.

Think of the sphere being comprised of many thin spherical shells, each having a slightly different charge density according to their radii. No two shells have the same charge density due to your given expression for the charge density, ρ=A/r. The enclosed charge will be the sum of the charges in those shells "below" the radius of interest.
##E4πr^2=\frac{q+ \int ρVdr}{ε_0} ## is the right one?

gneill
Mentor
##E4πr^2=\frac{q+ \int ρVdr}{ε_0} ## is the right one?
That's the idea, yes.

That's the idea, yes.
so i Keep the vollume inside the integral!

gneill
Mentor
You want to integrate over the volume of interest, summing up the charges contained in each differential volume element. So really that V should be a dV(r). I assumed (my bad) that that was what you were implying with that V.