Finding the electric field of insulated shell

[Physics Dad]

An insulating spherical shell of inner radius r₁ and outer radius r₂ is charged so that its volume density is given by:

ρ(r) = 0 for 0 ≤ r < r₁
p(r) = A/r for r₁ ≤ r ≤ r₂
p(r) = 0 for r > r₂

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/ε₀
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/4πεr³

I then equated this in terms of Q_enc and plugged into Eqn 1

Q_enc = 4ρπr³/3

so knowing the area of a sphere is 4πr²

∫E⋅ds = 4ρπr³/3ε₀

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

E⋅4πr² = 0 so in cases 1 and 3, E = 1/4πr²

In the case where ρ(r) = A/r, is set:

E⋅4πr² = 4Aπr³/3ε₀r

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

E = A/ε₀

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

 

[BvU]

First I calculated the value of ρ(r)

$\rho(r)$ was given !

 

[Physics Dad]

Yes, fair enough, I meant equated Q_enc in terms of ρ.

Is there anything else wrong with my attempt?

Many thanks

 

[haruspex]

Q_enc = 4ρπr³/3

So what would be the formula if the density were constant?

 

[Physics Dad]

So what would be the formula if the density were constant?

So Q = p dv = p 4πr^2 Dr

Again, for case 1 and 3 this would still be zero.

For case 2 this would be integrated from r2 to r1?

So Q = 2Aπ(r2^2 - r1^2)

So E = A(r2^2 - r1^2)/2€0r^2

Sorry, doing this on my phone, have no symbols.

Thanks

 

[haruspex]

So Q = p dv = p 4πr^2 Dr

Again, for case 1 and 3 this would still be zero.

For case 2 this would be integrated from r2 to r1?

So Q = 2Aπ(r2^2 - r1^2)

So E = A(r2^2 - r1^2)/2€0r^2

Sorry, doing this on my phone, have no symbols.

Thanks

That is right for r>r₂. What about r₁<r<r₂?

 

[BvU]

So Q = p dv = p 4πr^2 Dr

No. On the left you have a finite Q, the others are infinitesimals ('d').

So if you want to write something like this very casually (as is usual in physics), you write $$
dQ = \rho \; dV = \rho \;4πr^2 \; dr$$ where you have done two integrations in the last step ($dV = r^2 \sin\theta\; d\theta \,d\phi\, dr$) which you can do because $\rho(\vec r)=\rho(r)$.

But $dQ$ still depends on $r$, both in the $\rho$ factor and in the $r^2$
$\mathstrut$