Electric potential of a charged dielectric-coated conducting cylinder

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Homework Statement



We have an infinitely-long perfect electrical conductor (PEC) cylindrical core (Region 3) of radius a surrounded by a dielectric layer (Region 2) of permittivity ε2 and radius b. The combined conductor/dielectric cylinder is embedded in an infinite space (Region 1) of permittivity ε1. The core has a constant static potential of K. What is the electric potential at every point within the two regions surrounding the core?


Homework Equations



Polar Laplace equation:
[tex]\nabla^2V = \frac{1}{\rho}\frac{\partial}{\partial\rho}\left( \rho\frac{\partial V}{\partial\rho}\right)+\frac{1}{\rho^2} \frac{\partial^2V}{\partial\phi^2}=0[/tex]

The Attempt at a Solution



To solve this, I first used separation of variables to define V as a product of functions of variables ρ and ϕ:
[tex]V(\rho,\phi)=P(\rho)\Phi(\phi) [/tex]
When this is divided from Laplace's equation and the result multiplied by ρ2, we get:
[tex]\frac{\rho}{P} \frac{d}{d\rho} \left(\rho\frac{dP}{d\rho}\right)=-\frac{1}{\Phi} \frac{d^2\Phi}{d\phi^2}[/tex]
This equation can only be true if both sides are both equal to a nonzero constant m2. This means that:
[tex]\frac{d}{d\rho}\left(\rho\frac{dP}{d\rho}\right)-\frac{m^2}{\rho}P=0[/tex]
[tex]\frac{d^2\Phi}{d\phi^2}+m^2\Phi=0[/tex]
It can be shown that:
[tex]V=\sum_{m=1}^{\infty}\left(C_{1m}\rho^{m}+\frac{C_{2m}}{\rho^{m}}\right)\left[C_3\cos(m\phi)+C_4\sin(m\phi)\right][/tex],
however, only the m = 1 term contributes to the solution. Also, because of the circular symmetry of the regions, the sine and cosine terms will add to one at any angle and the constants C3 and C4 are equal to each other and can be absorbed by the constants of the P term. Therefore the general solution becomes:
[tex]V=C_1\rho+\frac{C_2}{\rho}[/tex].
As ρ goes to infinity V goes to 0, so the C1 for Region 1 must equal 0. The final solution becomes:
[tex]V_1=\frac{A}{\rho}[/tex]
[tex]V_2=B\rho+\frac{C}{\rho}[/tex]
[tex]V_3=K[/tex]

I believe I started the problem off correctly, but I'm not entirely convinced this is the correct solution. Does anyone see any problems with my derivation or solution?
 

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