How to Find Bound Charges in a Polarized Cylindrical Dielectric Shell?

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To find bound charges in a polarized cylindrical dielectric shell, the polarization is given by P=k/s^2, affecting both surface and volume charge densities. The surface charge density can be calculated using the formula sigma=P·n, applying it at the inner radius R1 and outer radius R2. For the volume charge density, the divergence of P leads to ro=-divergence of P, resulting in ro=k/s^2. It is confirmed that using different limits for s at the inner and outer surfaces is appropriate for calculating the respective surface charges. Understanding these calculations is crucial for accurately determining the bound charges in the dielectric shell.
azone
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Homework Statement


We have a long cylindrical, dielectric shell in the z-axis with inner radius R1 and outer radius R2. The polarization is given by P=k/s^2 (in cylindrical coordinates, it is only in the shat direction, i.e. no zhat or phihat)


Homework Equations


Find the bound surface charge density and volume charge density.


The Attempt at a Solution


I know the general formula for surface charge is sigma=P.normal and for volume charge it is ro=-divergence of P. But I'm wondering if things are slightly altered because it is a shell and not a solid cylinder. Or can I just use the formula and then have a limit for s?
For the volume charge I'm thinking I can just use the formula which will give me ro=k/s^2. But for the surface charge I was thinking of doing two different ones for s=R1 and s=R2 to find inner surface charge and outer surface charge.
Any help would be appreciated.
 
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azone said:
Or can I just use the formula and then have a limit for s?

What do you mean by a limit for s? Since the polarization is given as a function of s, you can find the volume charge density, which would also depend on s, from the divergence.

But for the surface charge I was thinking of doing two different ones for s=R1 and s=R2 to find inner surface charge and outer surface charge.
Any help would be appreciated.

Yes, that seems correct.
 
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