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Surface Charge Density

  • #1

Homework Statement


A point charge +Q is placed at the center of a spherical insulator of radius a. The insulator completely fills three cavity of a spherical conducting shell of radius b. Find the inner and outer surface charge density of the conductor and the bound surface charge density of the conductor.

I've the electric field for each region Im having a hard time figuring out how to find the charge densities though.
I know how to find the total charge on each surface but again, Im hazy on how EXACTLY it relates to charge density.


Relevant equations

##E_{sphere}={\frac{Q_{enc}}{4πε_0r^2}}{\hat{r}}##
Field in conductor= 0

The Attempt at a Solution


My thought is that it should just be ##Q_{enc}=σA## Where A is the surface area. But I cant find anything in my book to confirm this.

Another thought was to use Gaussian pillboxes, but I don't think those apply to spherical systems.


I have a picture of the problem but I'm having trouble uploading it.
 
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Answers and Replies

  • #2
rude man
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Homework Statement


A point charge +Q is placed at the center of a spherical insulator of radius a. The insulator completely fills three cavity of a spherical conducting shell of radius b.
Please clarify.

Another thought was to use Gaussian pillboxes, but I don't think those apply to spherical systems.
Yes they do. Very nicely.

I have a picture of the problem but I'm having trouble uploading it.[/QUOTE]

Try to upload it. Or clarify the problem verbally.
 
  • #3
Upload still not working.

Its a spherical insulator with radius a, inside of a spherical conducting shell with inner radius a, and outer radius b. So the insulator is touching the conducting shell.

I will try the Gaussian Pillbox and get back to you.
 
  • #4
So I for the outer edge of the conductor I got:

##{\epsilon}_0({\frac{Q}{4{\pi}r^2{\epsilon}_0}})={\sigma}_{c,out}##
##={\frac{Q}{4{\pi}r^2}}##

But it doesn't seem right to have the r in there since that refers to the radius of the Gaussian surface. I feel like I might be going at this all wrong.
 
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  • #5
rude man
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Homework Statement


A point charge +Q is placed at the center of a spherical insulator of radius a. The insulator completely fills three cavity of a spherical conducting shell of radius b. Find the inner and outer surface charge density of the conductor and the bound surface charge density of the conductor.
Should that be "insulator"?
 
  • #6
rude man
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You don't need to upload the picture. I think we have it.

Use Gaussian spherical shells for everything.
 
  • #7
I see how Gaussian surfaces would help to find the surface charge but I'm still unclear on how exactly that relates to the charge density with spherical systems.
 
  • #8
rude man
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Respond to post #5 please?
 
  • #9
Yes, sorry I was interuptted in the middle of my first attempt at a response. It was supposed to be insulator.
 
  • #10
rude man
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Let's first get the bound charge qb and from that the bound surface charge density.

Wrap a gaussian spherical surface around Q, the charge at the center, and radius just shy of "a" (so that the insulator's surface charge is not included). What equation invokig Gauss' law can you write for the E field just below the insulator's surface charge?

Hint : since there is surface charge and since the insulator is charge-free overall, there must be equal and opposite extra charge (besides Q) in its volume below the surface charge layer. Write Gauss' law using ε0 and also using ε.

Then, the surface charge density.is obviously qb/4πa2.

By wrapping one more Gaussian spherical surfaces at a larger radius you can figure out what the shell's inner and outer surface charge densities are also.

BTW the dielectric constant of the insulator is part of the answer.
 
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  • #11
rude man said:
Then, the surface charge density.is obviously q b /4πa 2 .

So then what I said here is correct?

With the specification that ##Q_{enc}## refers to the charge enclosed by the Gaussian surface.

The Attempt at a Solution


My thought is that it should just be ##Q_{enc}=σA## Where A is the surface area. But I cant find anything in my book to confirm this.
 
  • #12
Also, you mentioned that pillboxes would fit nicely. Are we scrapping that?
 
  • #13
rude man
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Also, you mentioned that pillboxes would fit nicely. Are we scrapping that?
No. I described two "pillboxes" for you. Both are spherical surfaces.

Using each of those surfaces, determine E just below the insulator's surface charge layer. Do this using ε0 and ε = kε0.

Maybe I don't know what exactly you mean by a 'pillbox'. I assume it's a Gaussian closed surface. As in kε0∫∫E * dA = Q or ∫∫D * dA = Q. BTW that's a hint for you in how to obtain qb.
 
  • #14
Pillbox was described to use as an "conveniently" small cylindrical surface. I think it was used to calculate flux of a curved surface, the idea being, if the surface is small enough then the flux out of the sides will be zero and you'll have equal and opposite flux out of the ends.
 
  • #15
rude man
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Pillbox was described to use as an "conveniently" small cylindrical surface. I think it was used to calculate flux of a curved surface, the idea being, if the surface is small enough then the flux out of the sides will be zero and you'll have equal and opposite flux out of the ends.
That's fine for a cylindrical gaussian surface, but here you want a spherical surface. So let's drop the appellation 'pillbox'.

But the idea is exactly the same: you determine the total E (or D) flux thru the surface based on the charge(s) within that surface.
 
  • #16
Ok, I got that part, and I apologize if I'm misunderstanding, but before I solicit anymore help, can you respond to post #11? That seems to follow what you said in the quoted text, I've not had a chance to put pencil to paper to try and work through this problem again(more hw assignments due). But if you could just answer the aforementioned post, I'll be on my until I do attempt it once more.
 
  • #17
rude man
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Ok, I got that part, and I apologize if I'm misunderstanding, but before I solicit anymore help, can you respond to post #11? That seems to follow what you said in the quoted text, I've not had a chance to put pencil to paper to try and work through this problem again(more hw assignments due). But if you could just answer the aforementioned post, I'll be on my until I do attempt it once more.
You said Qenc = σA which is correct. But in order to get σ you need to know Qenc, which is the total charge within the surface, which you don't know because the total charge inside the surface = free charge - bound charge = Qenc = Q - qb and you don't know the bound charge qb.

By using gauss' law twice, once in terms of ε and another in terms of ε0, you get two equations which allow you to solve for qb, then
σ = (Q - qb)/A.
 
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  • #18
Right, I thought I mentioned that I knew how/had already found the surface charges, I just couldn't verify that finding charge density was that simple.


But thank you nonetheless for all your time!
 

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