Potential of Spherical Charges: Find A & B for Surface r<a

[kreil]

Homework Statement

Electric charges are distributed on a spherical surface of radius a so as to produce the potential

$$\Phi(\vec r)=A(x^2-y^2)+Bx$$

in the region r<a. Find the potential in the region r>a (hint: use the table of spherical harmonics).

Homework Equations

I am unsure, but I think I should start with the following general potential expression (solution to laplace's eqn in terms of the spherical harmonics Y).

$$\Phi(r, \theta, \phi)= \sum_{l=0}^{\infty} \sum_{m=-l}^l \left[ A_{lm}r^l+B_{lm}r^{-(l+1)} \right] Y_l^m(\theta, \phi)$$

The Attempt at a Solution

If I am to use the above potential eqn, I need to utilize boundary conditions to find the coefficients A and B, but the surface potential is not specified so I'm not sure where to start...can someone point me in the right direction?

Thanks for your comments.

 

[nickjer]

First write out the coefficients in that expansion for r<a. This can be done easily by expanding out the x and y in their angular components and finding combinations of spherical harmonics that match them.

Once you have that, you will need BC's. You know at the surface of the sphere the potential is continuous. You also know how the potential behaves at infinity. This should be enough information to solve for that expansion in the region r>a.

 

[kreil]

i'm having trouble finding the combination of spherical harmonics which describes:

$$\Phi(\vec r)=A(x^2-y^2)+Bx=Ar^2(cos^2 \theta -sin^2 \theta )+Brcos \theta=Ar^2cos 2 \theta + Brcos \theta$$

any hints/advice?

 

[nickjer]

This isn't 2 dimensions. So there should be a $$\phi$$ term in there. Remember:

$$x = r cos(\phi)sin(\theta)$$
$$y = r sin(\phi)sin(\theta)$$
$$z = r cos(\theta)$$

Also this might be useful:

$$sin(\phi) = \frac{e^{i\phi}-e^{-i\phi}}{2i}$$
$$cos(\phi) = \frac{e^{i\phi}+e^{-i\phi}}{2}$$

 

[paranormal]

First, let's start by defining some variables for clarity:
- r = distance from the center of the spherical surface
- a = radius of the spherical surface
- A and B = coefficients in the potential expression given
- x and y = coordinates on the surface of the sphere

To solve for the potential in the region r>a, we can use the general potential expression you provided, but we need to first determine the values of A and B. To do this, we can use the given potential expression at the surface of the sphere (r=a) and equate it to the general potential expression at r=a:

A(a^2-y^2)+Bx = \sum_{l=0}^{\infty} \sum_{m=-l}^l \left[ A_{lm}a^l+B_{lm}a^{-(l+1)} \right] Y_l^m(\theta, \phi)

Since the potential is only dependent on x and y, we can set y=0 and solve for A and B:

A(a^2)+Bx = \sum_{l=0}^{\infty} \sum_{m=-l}^l \left[ A_{lm}a^l+B_{lm}a^{-(l+1)} \right] Y_l^m(\theta, \phi)

Now, we can use the table of spherical harmonics to find the values of A and B for each value of l and m. Once we have these values, we can plug them back into the general potential expression and solve for the potential in the region r>a.

Hope this helps!