Find Potential in Interior Sphere w/ f(θ)=cos^2(θ)

  • Thread starter Thread starter neelakash
  • Start date Start date
  • Tags Tags
    Potential
Click For Summary

Homework Help Overview

The problem involves finding the electric potential inside a unit sphere given a potential function on the surface, specifically f(θ) = cos²(θ). The context is rooted in electrostatics and the application of Laplace's and Poisson's equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the uniqueness theorem in relation to Laplace's equation and the conditions necessary for its application. Questions arise regarding the appropriateness of using Laplace's versus Poisson's equation, particularly in the absence of charge within the sphere.

Discussion Status

The discussion is ongoing, with participants clarifying the conditions under which the uniqueness theorem applies and exploring the implications of the boundary conditions. There is a recognition that a solution must satisfy both Laplace's equation and the given boundary conditions before applying the uniqueness theorem.

Contextual Notes

There is uncertainty regarding whether to assume the presence of charge within the sphere, which affects the choice between Laplace's and Poisson's equations. Participants note the lack of explicit information about charge in the problem statement.

neelakash
Messages
491
Reaction score
1

Homework Statement



Find the potential in the interior of a sphere of unit radius when the potential on the surface f(θ)=cos^2(θ).

Homework Equations





The Attempt at a Solution



I think the correct procedure is to apply uniqueness theorem.We know when the potential at every point of the surface is given,and the potential in that region obeys Laplace's (Here, Poisson'sequation),the potential function is unique.

So,I think it would be the same inside the sphere.

Please check if it is correct.
 
Physics news on Phys.org
neelakash said:

Homework Statement



Find the potential in the interior of a sphere of unit radius when the potential on the surface f(θ)=cos^2(θ).

Homework Equations


The Attempt at a Solution



I think the correct procedure is to apply uniqueness theorem.We know when the potential at every point of the surface is given,and the potential in that region obeys Laplace's (Here, Poisson'sequation),the potential function is unique.

So,I think it would be the same inside the sphere.

Please check if it is correct.

That's not right.
What the uniqueness theorem says is that, if you find a function V so that
(i) It satisfies Laplace's (or Poisson's) equation inside the region
(ii) It satisfies the given boundary conditions
then, V is the unique solution inside the region.

In your case, you took V as \cos^2 \theta. Does it satisfy Laplace's equation inside the region?

You can find the potential insde the sphere, by solving laplace's equation in spherical coordinates by the separation of variables technique, and then fitting the answer to your boundary conditions.
 
Last edited:
I see.The uniqueness theorem is for the "solution" of Laplace's equation.That should be determined first.
 
neelakash said:
I see.The uniqueness theorem is for the "solution" of Laplace's equation.That should be determined first.

Yes. After you have find a solution satisfying the Laplace equation and the boundary condition(s), then you can use the Uniqueness theorem to claim that solution is the only solution allowed. :smile:
 
Should we use Laplace's equation or Poisson's equation.How is Poisson's equation solved?
 
neelakash said:
Should we use Laplace's equation or Poisson's equation.How is Poisson's equation solved?


In your case, there is no net charge in the sphere, right?
so, Laplace is enough.

For Poisson's equation, you have to do the homogeneous part (Laplace part, solved by separating variables) and inhomogeneous part (the additonal term on the right, use Green function usually) separatly.

=)
 
Last edited:
in question there is no specification whether or not there is charge inside or not.So,we better neglect it.

Thank you for clarification.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
4K
Replies
6
Views
3K
  • · Replies 37 ·
2
Replies
37
Views
13K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K