Electrostatics ofa shell question

[E92M3]

Homework Statement

A spherical shell with potential:
$$V(r=R)=V_0cos\theta$$
(a) Please solve the potential inside and outside the shell.
(b) if there is a point charge at the center and the potential of the shell keeps the same, solve the potential inside and outside.

Homework Equations

Laplace equation:
$$\nabla^2V=0$$

The Attempt at a Solution

Part a) I can do. Using the separation of variables in spherical coordinates I found that:
$$\frac{V_0 r}{R} cos\theta ; r\leq R$$
and
$$\frac{V_0 R^2}{r^2} cos\theta ; r\seq R$$

Part b) I'm having some trouble.
I think we should apply the uniqueness theorem since the potential is specified on the boundary (the sphere) but not sure how. Does that mean that the potential is the same as before inside the sphere? I was told that this is not the case by upperclassman who can't tell me why. They only know the answer is:
$$V(r,\theta)=\frac{V_0}{R} cos \theta +\frac{q}{4 \pi \epsilon_0}(\frac{1}{r}-\frac{1}{R})$$
I think that this is indeed the right answer since it satisfy the boundary condition, but I want to know why. It looks like that the first term accounts for the charge distribution on the sphere, but if that is the case shouldn't the second term be the point charge in ther center namely:
$$\frac{q}{4 \pi \epsilon_0}\frac{1}{r}$$
instead?
THe outside is even more tricky. Can I even apply the uniqueness theorem? The boundary where the potential is defined doesn't enclose the area outside the sphere. What can I do?

 

[gabbagabbahey]

I think we should apply the uniqueness theorem since the potential is specified on the boundary (the sphere) but not sure how. Does that mean that the potential is the same as before inside the sphere? I was told that this is not the case by upperclassman who can't tell me why.

No, the potential will be the same outside the sphere; it still satisfies Laplace's equation, and has the same values at its boundaries, $r=R,\infty$ , as before.

However, inside the sphere there is now a point charge and so the potential in that region doesn't satisfy Laplace's equation, but rather Poisson's equation corresponding to a point charge at the origin:

$$\nabla^2 V_{in}=-q\frac{\delta^3(\textbf{r})}{\epsilon_0}$$

They only know the answer is:
$$V(r,\theta)=\frac{V_0}{R} cos \theta +\frac{q}{4 \pi \epsilon_0}(\frac{1}{r}-\frac{1}{R})$$
I think that this is indeed the right answer since it satisfy the boundary condition, but I want to know why. It looks like that the first term accounts for the charge distribution on the sphere, but if that is the case shouldn't the second term be the point charge in ther center namely:
$$\frac{q}{4 \pi \epsilon_0}\frac{1}{r}$$
instead?

You can use the superposition principle here and say that the potential is a sum of two terms; one, $V_1$, that satisfies Laplace's equation, is finite at the origin, and takes on the value $V_0\cos\theta$ at $r=R$; and a second, $V_2$, which satisfies Poisson's equation corresponding to a point charge at the origin, and takes on the value of zero at $r=R$ (so that the total potential takes on the correct value there). That way, you have

$$\nabla^2 V_{in}=\nabla^2 V_1+\nabla^2 V_2=-q\frac{\delta^3(\textbf{r})}{\epsilon_0}\;\;\;\text{and}\;\;\; V(r=R)=V_0\cos\theta$$

as required.

The uniqueness theorem tells you that $V_1$ is exactly the same as the inside potential from part (a), while $V_2$ must be the potential due to a point charge at the origin, with reference point set at $r=R$ rather than infinity (so that $V_2(r=R)=0$ as required).

 

[E92M3]

Thanks I think I got it now. Just to make sure, so outside the sphere is the same as before because of the uniqueness theorem? I thought it only applies to the potential inside a volume bounded the a boundary where the boundary conditions are defined (inside the sphere). Here, are we treating infinity as the boundary too? If so then can I argue that no matter what I put inside the sphere won't affect the potential outside as long as the BC remain unchanged? But if I put some charge outside the sphere the external potential will be changed. Am I correct?

 

[gabbagabbahey]

Thanks I think I got it now. Just to make sure, so outside the sphere is the same as before because of the uniqueness theorem? I thought it only applies to the potential inside a volume bounded the a boundary where the boundary conditions are defined (inside the sphere). Here, are we treating infinity as the boundary too? If so then can I argue that no matter what I put inside the sphere won't affect the potential outside as long as the BC remain unchanged? But if I put some charge outside the sphere the external potential will be changed. Am I correct?

Yes, that's correct. Of course, if you allowed the potential at $r=R$ to change as you added charge inside the shell (as it normally would for most shell materials), then one of the boundary conditions would change and hence so would the potential outside.

 

[paranormal]

As a scientist, it is important to approach problems with a critical and analytical mindset. In this case, it seems like there may be some confusion or uncertainty in the solution provided by your upperclassman. It is always important to question and understand the reasoning behind a solution, rather than simply accepting it as correct.

In this situation, it may be helpful to revisit the uniqueness theorem and understand its implications for this problem. The uniqueness theorem states that if the potential is specified on the boundary, then there is only one unique solution to Laplace's equation within that boundary. This means that the solution provided by your upperclassman, which satisfies the boundary condition, is the only possible solution for the potential inside and outside the shell with a point charge at the center.

As for the specific form of the solution, it may be helpful to consider the physical interpretation of each term. The first term represents the potential due to the charge distribution on the shell, which is a function of the distance from the center and the angle of the point. The second term represents the potential due to the point charge at the center, which is a function of the distance from the center. Both terms are necessary to fully describe the potential at any point within or outside the shell.

Regarding the outside of the shell, the uniqueness theorem can still be applied as long as the potential is specified on the boundary. In this case, the boundary is the surface of the shell, so the potential at any point outside the shell can be determined using the same solution as inside the shell, as long as it satisfies the boundary condition.

In conclusion, it is important to question and understand the solutions provided, and to use the appropriate principles and theories to arrive at a logical and scientifically sound conclusion.