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Free surface charges on concentric cylinders

  • #1

Homework Statement


Consider an infinitely long cylindrical rod with radius a carrying a uniform charge density ##\rho##. The rod is surrounded by a co-axial cylindrical metal-sheet with radius b that is connected to ground. The volume between the sheet and the rod is filled with a dielectric, ##\epsilon##.

Calculate the free and bound surface charges at r=a and r=b

Homework Equations





The Attempt at a Solution


[/B]
I tried to use discontinuity in E.

From Gauss. ##\nabla \cdot E = \frac{\rho}{\epsilon_0} \Rightarrow \oint E \cdot ds = \frac{Q}{\epsilon_0} ##

##\Rightarrow (E_{above}-E_{below}) = \frac{\sigma_b}{\epsilon_0}##.

For r= a, ##\sigma_b = \frac{Q}{2 \pi s*l} (\frac{1}{\epsilon_r} - 1)##

So now we have the bound charge per unit area.

By integrating we ca get the total bound surface charge, ##\sigma_{bt} = Q(\frac{1}{\epsilon_r}-1)##.

I'm not sure this is right. But if it is, how do I now get the ##free## surface charge?

And I do remember that either the bound charge or free charge should be zero for a grounded material, but not which one.
 

Answers and Replies

  • #2
Charles Link
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I think you omitted a step: ## \nabla \cdot D=\rho_{free}=0 ## at and around ## r=a ##, so that ## D_{above}=D_{below} ## which gives ## \epsilon_o \epsilon_r E_{above}=\epsilon_o E_{below} ##. (And then ## E_{below} ## is computed from knowing ## \rho ## which was given). ## \\ ## To get the polarization surface charge density at ## r=b ##, the polarization charge per unit length at ## r=b ## must be equal and opposite that at ## r=a ##. (Recommend don't use ## \sigma_b ## at ## r=a ##. Better to call it ## \sigma_{pa} ##.(## p ## for polarization). ## \\ ## The free charge at ## r=b ## is the easiest. If the outside is grounded, by symmetry, the net charge enclosed must be zero to have ## E =0 ## outside of the coaxial metal layer, as well as inside this metallic layer. The inside surface of the metallic layer at ## r=b ## will thereby have some charge.## \\ ## Additional item: For charge per unit length and surface charge per unit length, suggest using ## \lambda ##. Do not use ## \sigma ## for that. ## \sigma ## is a surface charge density per unit area. e.g. For the core, you would have ## \lambda_{core}=\rho \, \pi \, a^2 ##. Also ## \lambda_{pa}=\sigma_{pa} \, 2 \pi \, a ##, etc.
 
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  • #3
Delta2
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I am a bit lost here, the central cylindrical rod is conducting/metal or non conducting? And the charge density given as ##\rho## is equal to ##\rho_{free}## or not?
 
  • #4
Charles Link
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@Delta² The central rod is ## \rho_{free} ## and non-conducting. It's free charge, (as opposed to polarization type charge), but is not free to move. It is embedded in the rod. ## \\ ## And it's not a polarization type charge that forms as the result of dipoles in the material.
 
  • #5
Delta2
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@Delta² The central rod is ## \rho_{free} ## and non-conducting. It's free charge, (as opposed to polarization type charge), but is not free to move. It is embedded in the rod. ## \\ ## And it's not a polarization type charge that forms as the result of dipoles in the material.
Ehm, I don't understand then why in your first equation you say ##\rho_{free}=0## at ##r=a##, ok its not free to move, but still it should be ##\rho_{free}=\rho## at ##r=a## , right or wrong?
 
  • #6
I think you omitted a step: ## \nabla \cdot D=\rho_{free}=0 ## at and around ## r=a ##, so that ## D_{above}=D_{below} ## which gives ## \epsilon_o \epsilon_r E_{above}=\epsilon_o E_{below} ##. (And then ## E_{below} ## is computed from knowing ## \rho ## which was given). ## \\ ## To get the polarization surface charge density at ## r=b ##, the polarization charge per unit length at ## r=b ## must be equal and opposite that at ## r=a ##. (Recommend don't use ## \sigma_b ## at ## r=a ##. Better to call it ## \sigma_{pa} ##.(## p ## for polarization). ## \\ ## The free charge at ## r=b ## is the easiest. If the outside is grounded, by symmetry, the net charge enclosed must be zero to have ## E =0 ## outside of the coaxial metal layer, as well as inside this metallic layer. The inside surface of the metallic layer at ## r=b ## will thereby have some charge.## \\ ## Additional item: For charge per unit length, suggest using ## \lambda ##. Do not use ## \sigma ## for that. ## \sigma ## is a surface charge density per unit area. e.g. For the core, you would have ## \lambda_{core}=\rho \, \pi \, a^2 ##. Also ## \lambda_{pa}=\sigma_{pa} \, 2 \pi \, a ##, etc.
By surface charge ##density## do you mean surface charge, because it's only the surface charge I want.

Is polarization charge same as bound charge? I failed to understand how to use the ##\lambda## concept.

However, I did a new calculation, I did a little wrong above.

See picture.

So now I think I have the total surface charge for both the cylinders. However, there's still bound charge present. How do I differentiate bound and free charge, so I eventually can extract only the free surface charge from the total surface charge?
IMG_0516.JPG
 

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  • #7
Charles Link
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Your calculations are leaving off an important step or two in the solution. I recommend you use the equation ## \nabla \cdot D=\rho_{free} ## where ## D=\epsilon_o \epsilon_r E ##. You can do the calculation just using ## E ## and deriving everything else from ## -\nabla \cdot P=\rho_p ##, but it's easier to use ## D ##. ## \\ ## With this equation, Gauss' law reads ## \int D \cdot dA=Q_{free} ##. You seem to get the correct answer for ## \sigma_pa ##, but you aren't showing the steps to get there. ## \\ ## And yes, your equation ## E_{above}-E_{below}=\frac{\sigma_{pa}}{\epsilon_o} ## is correct. ## \\ ## And, yes, it is bound polarization charge, usually called just simply polarization charge. ## \\ ## Using the ## D ## form of Gauss' law with a pillbox around ## r=a ## gives: ## \\ ## ## \epsilon_o \epsilon_r E_{above}-\epsilon_o E_{below}=0 ##, so that ## E_{above}=\frac{E_{below}}{\epsilon_r} ##. ## \\ ## This gives ## E_{below}(\frac{1}{\epsilon_r}-1)=\frac{\sigma_{pa}}{\epsilon_o} ##. ## \\ ## ## E_{below} ## is readily found: (Edit) ## E_{below} 2 \pi \, a L=\frac{\rho \pi a^2 \, L }{\epsilon_o} ##. ## \\ ## Now we can simply solve for ## \sigma_{pa} ##. (I think you got that part correct). ## \\ ## Because this is polarization of the dielectric between ## r=a ## and ## r=b ##, ## \lambda_{pa} L=-\lambda_{pb} L ##. ## \\ ## (It can be shown that ## -\nabla \cdot P=\rho_p=0 ## inside the dielectric, so that the only polarization charge is surface polarization charges. The net polarization charge must be zero. Alternatively, ## \int D \cdot dA =Q_{free} ## so that ##\epsilon_o \epsilon_r E(r) 2 \pi r=\rho \, \pi a^2 ##. We can compute ## \nabla \cdot E(r)=\frac{\rho_{total}}{\epsilon_o}=0 ## in the dielectric, (google "divergence in cylindrical coordinates"), so that ## \rho_p =0 ## in the dielectric. ) ## \\ ## This is why you need the surface charge per unit length.## \\ ## Now ## \lambda_{pa}=\sigma_{pa} 2 \pi a ## and ## \lambda_{pb}=\sigma_{pb} 2 \pi b ##. You need these last two relations to solve for ## \sigma_{pb} ## from ## \sigma_{pa} ##.
 
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