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

Reg_S
Messages
5
Reaction score
0
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)
 
Physics news on Phys.org
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.
 
  • Like
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.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top