Use Gauss' Law to calculate the electrostatic potential for this cylinder

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The discussion focuses on solving for the electrostatic potential of a cylinder using Gauss' Law and the Laplacian equation. The solution presented is V(r, phi) = a + b.ln(r) + summation terms involving r and sine functions. The boundary condition specified is V(R, phi) = V_0 sin(phi), leading to the suggestion of a solution form V(r, phi) = V_0 f(r) sin(phi) with f(R) = 1. Questions arise about the boundary conditions for both inner and outer potentials, particularly regarding continuity across the non-conducting cylinder. The potential must remain continuous, implying that the inner and outer potentials can be equated at the boundary.
Reg_S
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Homework Statement
An infinitely long hollow (non conducting) circular cylinder of radius R fixed at potential V =V•sin(phi) .
Relevant Equations
Using cylinder coordinates with z axis as a symmetric axis , argue V is independent of Z and V(r, -phi)= -V(r, phi)
b) Find electrostatic potential inside and outside of the cylinder
I solved laplacian equation. and got the solution of V(r, phi) = a. +b.lnr + (summation) an r^n sin(n phi +alpha n ) + (summation) bn r ^-n sin( n phi +beta n)
 
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Please help me how to use BCs and find the constant,
 
The boundary condition is V(R, \phi) = V_0\sin \phi (please don't use the same symbol for an unknown function and a given constant value). That immediately suggests trying a solution of the form V(r, \phi) = V_0f(r)\sin \phi with f(R) = 1.
 
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Likes topsquark and Reg_S
Thank you, Is it same BCs for inner and outer potential? just using relative term? Can we do (Phi)in = (phi)out at r=R?
 
The cylinder is non-conducting, so the potential is continuous across it.
 
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