There is no point charge here.Orodruin said:it is not correct. How did you arrive at these expressions?
Are you familiar with the point charge field in two dimensions?
It is the surface charge, not V, which is given. You have to find the discontinuity in ##\partial_r(V)## at the surface related to the surface charge density.Trollfaz said:Here's a detailed explanation
At the boundary r=R
$$V=\sum_{n=0}^{\infty}A_n R^{-n} cos(nθ)$$
$$\int_{0}^{π}\sum_{n=0}^{\infty}A_n R^{-n} cos(nθ)cos(mθ)dθ=A_m R^{-m} π$$
$$A_m=\frac{R^m}{π}\int_{0}^{π}V_Rcos(mθ)dθ$$
If we know V at boundary as a linear combination of cos(nθ), then we can filter out all the non zero terms of the infinite sum and find the full function of V.
Irrelevant to the point I was making. You have a potential monopole moment and the corresponding field of that is that of a point charge outside of the charge distribution.Meir Achuz said:There is no point charge here.
Which is … the field of a point charge in two dimensions. Just as the 1/r field outside of a spherically symmetric charge distribution in three dimensions.Meir Achuz said:There will be a ##ln( r)## term for outside if there is a constant surface charge.
"for a very long cylinder"vanhees71 said:The general expansion in terms of separation-ansatz solutions in cylindrical coordinates leads to the Bessel or Hankel functions. You seem to discuss the special case of ##z##-independent solutions, i.e., solutions of the Laplace equation in 2 dimensions in terms of polar coordinates, which is not the general case, of course.
Yes let's just assume the distribution is invariant with zvanhees71 said:Yes, then of course, you assume independence of ##z##. I've overlooked this assumption.
To derive the general solution to Laplacian in cylindrical coordinates, we start with the Laplacian operator in cylindrical coordinates and apply separation of variables technique. This involves assuming a solution of the form u(r, θ, z) = R(r)Θ(θ)Z(z) and substituting it into the Laplacian operator. By solving the resulting differential equations for R(r), Θ(θ), and Z(z), we can obtain the general solution.
When finding the general solution to Laplacian in cylindrical coordinates, boundary conditions such as the Dirichlet, Neumann, or mixed boundary conditions need to be considered. These conditions specify the values of the solution or its derivatives on the boundaries of the cylindrical domain and play a crucial role in determining the form of the general solution.
Sure! Let's consider the problem of finding the Laplacian of a function u(r, θ, z) = R(r)Θ(θ)Z(z) in cylindrical coordinates. By substituting this form into the Laplacian operator and solving the resulting differential equations for R(r), Θ(θ), and Z(z), we can obtain the general solution to Laplacian in cylindrical coordinates.
The general solution to Laplacian in cylindrical coordinates differs from the Cartesian coordinates case due to the cylindrical symmetry of the system. In cylindrical coordinates, the Laplacian operator involves derivatives with respect to the radial, azimuthal, and vertical directions, leading to a different form of the general solution compared to the Cartesian coordinates case.
The general solution to Laplacian in cylindrical coordinates has various applications in physics and engineering, such as in the study of heat conduction in cylindrical systems, electromagnetic fields around cylindrical conductors, and fluid flow in cylindrical geometries. By utilizing the general solution, we can analyze and solve a wide range of problems involving cylindrical symmetry.