How Do I Solve Poisson's Equation for V?

  • Thread starter Thread starter leoflindall
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on solving Poisson's equation for the potential function V, specifically in one dimension, where the equation is represented as \(\nabla^{2}V = \frac{\rho}{\epsilon_{0}}\). The solution involves integrating the equation twice, leading to the general form of V based on the charge density function \(\rho(x)\). For example, if \(\rho(x) = \epsilon_0 x^2\), the resulting potential function is \(V(x) = \frac{1}{12} x^4 + cx + k\). The complexity increases significantly when extending the problem to two dimensions, requiring additional techniques such as symmetry and polar coordinates.

PREREQUISITES
  • Understanding of differential equations, specifically second-order equations.
  • Familiarity with Poisson's equation and its physical significance.
  • Knowledge of integration techniques in calculus.
  • Basic concepts of charge density and electric potential in electrostatics.
NEXT STEPS
  • Study the method of solving second-order differential equations.
  • Learn about the applications of Poisson's equation in electrostatics.
  • Explore techniques for solving differential equations in two dimensions, including the use of polar coordinates.
  • Investigate specific charge density functions and their impact on potential solutions.
USEFUL FOR

Students and professionals in physics, particularly those focusing on electrostatics, mathematical physics, and applied mathematics, will benefit from this discussion.

leoflindall
Messages
38
Reaction score
0

Homework Statement



[tex]\nabla[/tex][tex]^{2}[/tex]V=([tex]\rho[/tex])/([tex]\epsilon[/tex][tex]_{0}[/tex])


Homework Equations



-

The Attempt at a Solution



I have a function of x which i can supstitute into Charge density and boundary conditions, however my problem with this is very simple. How to i manipulate the equation to get a funvtion for V. Obviously i need to get rid of the div of the div of v and i assume this has to be done through some form of intergration?

I assume this isn't as simple as intergrating both sides twice?

I would appreciate any guidance that can be given on this.

Regards.

Leo
 
Physics news on Phys.org
In general, differential equations in physics are hard to solve. From what you posted, I understand that the function [itex]\rho = \rho(x)[/itex] is given, and you are working in one dimension only.

In that case,
[tex]\nabla^2 V = \frac{d^2 V}{dx^2}[/tex]
so you can solve your equation (I'm wiping the constant into rho, for notational convenience)
[tex]\frac{d^2 V}{dx^2} = \rho(x)[/tex]
by integrating twice.

For example, when [itex]\rho(x) = \epsilon_0 x^2[/itex] you would simply get
[tex]V''(x) = x^2[/tex]
so
[tex]V'(x) = \frac13 x^3 + c[/tex]
and
[tex]V(x) = \frac{1}{12} x^4 + c x + k[/tex]

The only snag might be that the integration is very hard to do: just pick a nasty function like [tex]\rho / \epsilon_0 = \sqrt{1 + x e^x}[/tex] and your scr*wed :)

Note that once you go to two dimensions, things already get far less trivial, you would have to solve
[tex]\frac{\partial^2V}{\partial x^2} + \frac{\partial^2 V}{\partial y^2} = \rho(x, y)[/tex]
which takes a little more than two integrations (usually, you already need things like symmetry and polar coordinates here to make anything of it).
 
That makes makes perfect sense! Thank you for your help!
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
11
Views
3K
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K