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

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SUMMARY

The discussion focuses on applying Gauss' Law to calculate the electrostatic potential for a non-conducting cylinder. The solution to the Laplacian equation is presented as V(r, φ) = a + b ln(r) + Σ an r^n sin(n φ + αn) + Σ bn r^-n sin(n φ + βn). The boundary condition specified is V(R, φ) = V0 sin(φ), leading to the suggestion of a solution form V(r, φ) = V0 f(r) sin(φ) with f(R) = 1. The continuity of potential across the cylinder is emphasized, and questions regarding the boundary conditions for inner and outer potentials are raised.

PREREQUISITES
  • Understanding of Gauss' Law in electrostatics
  • Familiarity with the Laplacian equation and its solutions
  • Knowledge of boundary conditions in electrostatic problems
  • Basic concepts of cylindrical coordinates in physics
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  • Study the application of Gauss' Law in cylindrical symmetry
  • Learn about solving Laplace's equation in cylindrical coordinates
  • Research boundary condition techniques in electrostatics
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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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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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