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**An insulating spherical shell of inner radius r**

ρ(r) = 0 for 0 ≤ r < r

p(r) = A/r for r

p(r) = 0 for r > r

Where A is a constant and r is the radial distance from the center of the shell. Find the electric field due to the shell for all values of r.

_{1}and outer radius r_{2}is charged so that its volume density is given by:ρ(r) = 0 for 0 ≤ r < r

_{1}p(r) = A/r for r

_{1}≤ r ≤ r_{2}p(r) = 0 for r > r

_{2}Where A is a constant and r is the radial distance from the center of the shell. Find the electric field due to the shell for all values of r.

## Homework Equations

Eqn 1 - ∫E⋅ds=Q

_{enc}/ε

_{0}

Eqn 2 - ρ=Q

_{enc}/V

## The Attempt at a Solution

First I got an equation for ρ(r) in terms of the spherical volume

ρ(r) = 3Q

_{enc}/4πεr

^{3}

I then equated this in terms of Q

_{enc}and plugged into Eqn 1

Q

_{enc}= 4ρπr

^{3}/3

so knowing the area of a sphere is 4πr

^{2}

∫E⋅ds = 4ρπr

^{3}/3ε

_{0}

From here, knowing that ρ(r) = 0 for two case, this means that in both cases Eqn 2 = 0, so I basically set:

E⋅4πr

^{2}= 0 so in cases 1 and 3, E = 1/4πr

^{2}

In the case where ρ(r) = A/r, is set:

E⋅4πr

^{2}= 4Aπr

^{3}/3ε

_{0}r

Did a bit of Algebra-kedabra and ended up with:

E = A/ε

_{0}

Intuitively, this seems to tell me that the electric field is constant anywhere in the within the two radii, whilst outside it is inversely proportional to the radius.

Basically, I want to know have I done anything stupid or have I gone the right way?

Many thanks

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