# Free surface charges on concentric cylinders

• Philip Land
In summary, we are considering an infinitely long cylindrical rod with radius a carrying a uniform charge density ##\rho##. It 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##. By using the discontinuity in E and applying Gauss's law, we can calculate the free and bound surface charges at r=a and r=b. The bound charge per unit area can be found by integrating, while the free surface charge can be determined by considering the symmetry of the problem. It is important to use the correct symbols when representing charge per unit length and surface charge per unit area, using ##\lambda## forf

## 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

## The Attempt at a Solution

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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.

• Delta2
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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• Philip Land and Delta2
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?

@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.

• Delta2
@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?

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? #### Attachments

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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• Delta2 and Philip Land