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A Hollow Conducting Sphere Gauss' Law

  • Thread starter JerryWakai
  • Start date
  • #1
I am in general physics III (E&M) at Cal Poly and I have a midterm tomorrow... my professor didn't go over Gauss' Law too much, so I'm having issues with this problem that is on his review. I will write the question exactly as it appears on the sheet.

Homework Statement



"You have a hollow conducting sphere with inner radius R1 and outer Radius R2. You place a point charge (-q) at the center of the sphere. What is the charge/unit area on the inner radius? On the outer radius?"

Homework Equations



gauss's law

The Attempt at a Solution



I used gauss's law and found the electric field of the charge, but I read through the text book and through my notes and I can't seem to find anything on finding the "charge/unit area" on the hollow sphere. I understand how to use Gauss' law, but I just don't understand how I can find this charge...

Any help would be greatly appreciated

Thank!

Jerry
 

Answers and Replies

  • #2
ehild
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Can you determine the induced charges on the surfaces of the hollow sphere?

ehild
 
  • #3
That is where the trouble comes in. I am not sure how to find the induced charges, as my professor didn't go over them. I am thinking that the induced charge would simply be the charge of the point charge inside of the hollow sphere, but I am not entirely sure because I cannot find it in the book and my professor didn't cover it.

I am thinking that the induced charge would be -q, therefore the charge/unit area would be -q/4piR^2, but I just don't know why the induced charge would be -q. I feel that I mainly need help in determining induced charges.
 
  • #4
SammyS
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I am in general physics III (E&M) at Cal Poly and I have a midterm tomorrow... my professor didn't go over Gauss' Law too much, so I'm having issues with this problem that is on his review. I will write the question exactly as it appears on the sheet.

Homework Statement



"You have a hollow conducting sphere with inner radius R1 and outer Radius R2. You place a point charge (-q) at the center of the sphere. What is the charge/unit area on the inner radius? On the outer radius?"

Homework Equations



gauss's law

The Attempt at a Solution



I used gauss's law and found the electric field of the charge, but I read through the text book and through my notes and I can't seem to find anything on finding the "charge/unit area" on the hollow sphere. I understand how to use Gauss' law, but I just don't understand how I can find this charge...

Any help would be greatly appreciated

Thank!

Jerry
Do you know what the electric field is within the conducting material?

If so, then you can determine the electric flux passing through a Gaussian surface embedded in the conducting material.
 
  • #5
ehild
Homework Helper
15,498
1,878
That is where the trouble comes in. I am not sure how to find the induced charges, as my professor didn't go over them. I am thinking that the induced charge would simply be the charge of the point charge inside of the hollow sphere, but I am not entirely sure because I cannot find it in the book and my professor didn't cover it.

I am thinking that the induced charge would be -q, therefore the charge/unit area would be -q/4piR^2, but I just don't know why the induced charge would be -q. I feel that I mainly need help in determining induced charges.
There is a relationship between the electric field at a metallic surface and the surface charge density σ : E=σ/ε0. If you haven't heard about that, think: The electrons in the metal move freely. The central -q charge repels them so a positive charge equal to q accumulates on the inner surface. As the metal shell is neutral, equal negative charge appears on the outer surface, just as you suggested.

ehild
 
  • #6
There is a relationship between the electric field at a metallic surface and the surface charge density σ : E=σ/ε0. If you haven't heard about that, think: The electrons in the metal move freely. The central -q charge repels them so a positive charge equal to q accumulates on the inner surface. As the metal shell is neutral, equal negative charge appears on the outer surface, just as you suggested.

ehild
this helps tremendously, thanks! that relationship between electric field and surface charge density helps tremendously, and it was not introduced to me priot. much appreciated!

Jerry
 
  • #7
ehild
Homework Helper
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You know that the electric field lines originate and end at charges. The electric field inside the metal shell is zero, so the field lines end at the surfaces: There must be surface charge there.

The electric field at distance R from a point charge Q is E=1/(4πε0R2). Integrating it over the sphere of radius R you get the number of field lines emerging from the point charge: 4πR2E= Q/ε0. It is true in general: Q charge means Q/ε0 field lines. So σ, the charge on unit area of the metal surface produces σ/ε0 field lines, perpendicular to the surface. And the electric field intensity is equal to the number of field lines crossing perpendicularly a unit area.

ehild
 

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