Electromagnetic Help: Finding Suitable Green Function

  • Thread starter Thread starter montser
  • Start date Start date
  • Tags Tags
    Electromagnetic
Click For Summary
SUMMARY

The discussion focuses on finding the suitable Green function for the potential described by the equation \(\nabla^2 \varphi= \frac{1}{l^2} \exp \left(\frac{-|x|}{l}\right)\). The solution provided indicates that under the boundary condition \(\phi \rightarrow 0\) as \(r \rightarrow \infty\), the Green function is expressed as \(G = -\frac{1}{4\pi|r-r'|}\). This solution is critical for solving differential equations in electromagnetic theory.

PREREQUISITES
  • Understanding of Green's functions in differential equations
  • Familiarity with boundary conditions in potential theory
  • Knowledge of the Laplace operator (\(\nabla^2\))
  • Basic concepts of electromagnetic theory
NEXT STEPS
  • Study the derivation of Green's functions for various boundary conditions
  • Explore applications of Green's functions in solving electromagnetic problems
  • Learn about the implications of the Laplace operator in potential theory
  • Investigate the role of exponential decay in physical systems
USEFUL FOR

Physicists, mathematicians, and engineers working in fields related to electromagnetic theory, particularly those involved in solving differential equations and applying boundary conditions.

montser
Messages
14
Reaction score
0
electromagnetic help!

I have a general question on finding the suitable green function for a given potential for example:
Find the suitable green function for \nabla^2 \varphi= \frac{1}{l^2} \exp (\frac{-|x|}{l})
 

Attachments

Physics news on Phys.org
I didn't get to see your Qus.pdf, but if the only BC is \phi-->0 as
r-->infinity, then G=-1/{4\pi|r-r'|}.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
6K
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K