Find the potential V(r, φ) inside and outside the cylinder

nickap34
Messages
4
Reaction score
0
Homework Statement
Consider two thin half-cylinder shells, made of a conducting material, that are the
right and left halves of a cylinder with radius R. They are separated from each other
at φ=π/2 and φ=3π/2 by small insulating gaps.
The left half, for which π/2<φ <3π/2, is held at potential –V0, and the right half,
which has 0<φ<π/2 and 3π/2<φ<2π, is held at +V0.
Find the potential V(r, φ) inside and outside the cylinder.
Relevant Equations
Unknown
Not even sure where to start.
 
Physics news on Phys.org
You want to start reading the PF guidelines. This is a second post from you with 'no idea'. I grant you it's not an easy exercise, but before we are allowed to help, you must simply post an effort.
And: 'Unknown' is a nono in PF.
What have you learned so far in your curriculum that might be relevant ?
 
I am thinking to find the inside potential, you take the double integral from 0 to L and 0 to 2π in cylindrical coordinates and do separation of variables
∫∫V(∅,z)sin(v∅)sin(knz)d∅dz
 
nickap34 said:
Find the potential V(r, φ) inside and outside the cylinder.
Is this the literal problem statement ? Because you bring in a ##z## and an ##L## that I don't see in there. Can the cylinder be considered infinitely long ?
 
nickap34 said:
I am thinking to find the inside potential, you take the double integral from 0 to L and 0 to 2π in cylindrical coordinates and do separation of variables
∫∫V(∅,z)sin(v∅)sin(knz)d∅dz

As @BvU says,please show your work. This problem can be solved in number of ways like solving Laplace equation,Using Green's Function,Poisson Integration with boundary conditions,etc. each being elegant though difficult.
 
As @Abhishek11235 stated you would likely want to use Green's functions.

This is a 2D problem in disguise so you want to use the 2D version of the Green's function integral

##\phi(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \int \sigma\left(\vec{r'} \right) G\left( \vec{r},\vec{r'} \right) \,da' - \frac{1}{4 \pi} \int \phi_s \frac{\partial G\left( \vec{r}, \vec{r'} \right)}{\partial n'} \, d\ell'##

You should only concern yourself with the second part of this integral since by definition G=0 on the surface.

The green's function normal derivative for a long cylinder should be easy enough to look up, and you know the potential on the surface. Have at it.

But i must say your lack of effort is disturbing.

Typically this problem (Jackson 2.13) is solved using the half-angle substitution but that can get real ugly real fast. When you find the greens function normal derivative, this page

https://math.stackexchange.com/ques...r21-2r-cos-theta-r2-12-sum-k-1-infty-rk-cos-k
will help you make sense of the integral.

Edit: hopefully I am not breaking forum rules by trying to help you.
 
  • Like
Likes Abhishek11235
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