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

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In summary, the potential inside the sphere is the same as the potential on the surface. The uniqueness theorem can be used to claim that the only solution of Laplace's equation inside the sphere is V=cos^2(θ).
  • #1
neelakash
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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.
 
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  • #2
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 [tex]\cos^2 \theta[/tex]. 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.
 
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  • #3
I see.The uniqueness theorem is for the "solution" of Laplace's equation.That should be determined first.
 
  • #4
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:
 
  • #5
Should we use Laplace's equation or Poisson's equation.How is Poisson's equation solved?
 
  • #6
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.

=)
 
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  • #7
in question there is no specification whether or not there is charge inside or not.So,we better neglect it.

Thank you for clarification.
 

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

What is the formula for finding potential in an interior sphere with f(θ)=cos^2(θ)?

The formula for finding potential in an interior sphere with f(θ)=cos^2(θ) is V(θ) = -GM/R + Ccos^2(θ), where G is the gravitational constant, M is the mass of the sphere, R is the distance from the center of the sphere, and C is a constant determined by the boundary conditions.

How do you determine the potential at a specific point inside the interior sphere?

To determine the potential at a specific point inside the interior sphere, you can use the formula V(θ) = -GM/R + Ccos^2(θ) and plug in the values for G, M, R, and C. Alternatively, you can use a calculator or computer program to solve the equation for you.

What is the significance of the function f(θ) in this formula?

The function f(θ) = cos^2(θ) represents the variation of potential with respect to the angle θ in the interior of the sphere. It is a way to take into account the changing distance from the center of the sphere as you move towards different points on the interior surface.

How does the potential change as you move closer to the center of the interior sphere?

The potential decreases as you move closer to the center of the interior sphere. This is because the gravitational force is stronger at points closer to the center, resulting in a lower potential energy.

Can this formula be used for any type of interior sphere?

Yes, this formula can be used for any type of interior sphere as long as the boundary conditions and constants are properly determined. It is a general formula for finding potential in an interior sphere with a given function f(θ).

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