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

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Homework Help Overview

The discussion revolves around applying Gauss' Law to determine the electrostatic potential for a non-conducting cylinder. Participants are exploring the implications of boundary conditions on the potential function derived from the Laplacian equation.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the general form of the potential solution and the application of boundary conditions to find constants. Questions arise regarding the consistency of boundary conditions for inner and outer potentials.

Discussion Status

The conversation is ongoing, with some participants providing insights into the boundary conditions and their implications for the potential function. There is no explicit consensus yet, but guidance on the form of the potential and the nature of the boundary conditions has been shared.

Contextual Notes

Participants are considering the implications of the cylinder being non-conducting, which affects the continuity of the potential across its surface. There are also concerns about the notation used for constants and functions in the context of boundary conditions.

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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