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Solid Insulator Sphere Inside Hollow Sphere Conductor

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1. Homework Statement
I was looking for some practice problems in my textbook and found this problem that I was just a little stuck on. I drew the diagram from my textbook with the givens of the problem.

Screen shot 2015-02-27 at 10.14.00 AM.png

2. Homework Equations
∲E*dA = Q (inside) / ɛ0

3. The Attempt at a Solution

For r less than/equal to a:

E(4pir^2) = 3Q/ɛ ---> E = 3Q / 4*pi*ɛ(r^2)

For a<r<b:

E= -5Q/ (4*pi*r^2)

Can somebody verify with me that I've done the first two above correctly?

Also the part that I got stuck on was the r is greater than/equal to c...
I see that if r was equal to c there would be no charge, hence e-field will be zero inside the conductor. However, when r is greater than c, then there would be a charge of -5Q. How would I usually deal with the case of "greater than/equal to"?

Thank you Physics community!
 
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The 3Q are distributed over the whole sphere, they are not all at the center ("inside").
I don't see why you use -5Q here.

Not only the -5Q.
Ignore the case of r exactly equal to c, that does not have a reasonable answer in this simplified model.

Screen shot 2015-02-27 at 6.25.50 PM.png


Sorry this was the actually image from the textbook I was working on. There was a Q in the middle, but the statement of the problem still applies.

So for r is less than/equal to a, I would draw a gaussian surface at r=a and one for r<a. I notice that at r=a, the only charge enclose in the middle of the insulator is the whole +3Q and +Q. But for the r<a, the only charge inside is the +Q. Is that correct?
 
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Hi welcome to PF:smile:

View attachment 79694

But for the r<a, the only charge inside is the +Q. Is that correct?
No,for r<a the charge inside is not just +Q.As it is a insulating solid sphere the charge +3Q is distributed uniformly over its total volume.So for exampleif you are considering a Gaussian surface of radius a/2 then you also have to consider the portion of +3Q distributed in the volume inside the Gaussian surface!
 
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Hi welcome to PF:smile:



No,for r<a the charge inside is not just +Q.As it is a insulating solid sphere the charge +3Q is distributed uniformly over its total volume.So for exampleif you are considering a Gaussian surface of radius a/2 then you also have to consider the portion of +3Q distributed in the volume inside the Gaussian surface!
ah I see...so in that case to find Q inside , it would be Q= (4/3*pi*r^2 / 4/3*pi*a^2) * Q => Qinside = (r^2/a^2)*Q?
 
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r and a should be to the third power. You have volumes here, not surfaces.
 

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