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Finding the electric field of insulated shell

  • #1
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=Qenc0
Eqn 2 - ρ=Qenc/V

The Attempt at a Solution



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

ρ(r) = 3Qenc/4πεr3

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

Qenc = 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
 
Last edited:

Answers and Replies

  • #2
BvU
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  • #3
Yes, fair enough, I meant equated Qenc in terms of ρ.

Is there anything else wrong with my attempt?

Many thanks
 
  • #4
haruspex
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Qenc = 4ρπr3/3
So what would be the formula if the density were constant?
 
  • #5
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
 
  • #6
haruspex
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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?
 
  • #7
BvU
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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##
 

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