# Free surface charges on concentric cylinders

## 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.

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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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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?

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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.

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?

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?

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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 $\sigma_{pa}$.