Solid Insulator Sphere Inside Hollow Sphere Conductor

[Aristotle]

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.

Homework Equations

∲E*dA = Q (inside) / ɛ0

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!

 

[mfb]

For r less than/equal to a:

The 3Q are distributed over the whole sphere, they are not all at the center ("inside").

For a<r<b:

I don't see why you use -5Q here.

However, when r is greater than c, then there would be a charge of -5Q.

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

 

[Aristotle]

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.

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?

 

[Mind_It]

Hi welcome to PF

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!

 

[Aristotle]

Hi welcome to PF
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?

 

[Mind_It]

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?

 

[mfb]

r and a should be to the third power. You have volumes here, not surfaces.