# Three point charges inside a conducting spherical shell

1. Dec 21, 2015

### ShayanJ

1. The problem statement, all variables and given/known data
There are three point charges inside a conducting spherical shell of radius R. One of them of charge -2q is in the origin and the other two with charge q are in z=d and z=-d. Find the potential inside the sphere!

2. Relevant equations
$\nabla^2\phi=4\pi \rho$

3. The attempt at a solutioc
The reason I'm asking this is that this was a question in our midterm exam. One of my classmates solved this but the professor thinks he used a wrong method. But my classmate insists that he used the correct method. Today it was the second time they had a heated discussion.Its getting awkward. So I'm trying to help resolve this.

The professor says that the correct method is to divide the space inside the sphere to two regions: r<d and r>d. Now we can use Laplace's equation for two regions: r>d and r<d excluding the origin. For each region we have $\phi(r,\theta)=\sum_{l=0}^\infty (A_l r^l+\frac{B_l}{r^{l+1}})P_l(\cos\theta)$(should we add $-\frac{2q}{r}$ to it because of the charge at the origin or that comes somehow else?...Oh...I myself got confused because of that discussion!) . So we have four sets of coefficients to determine. The boundary conditions are: 1) The potential is zero on the surface of the shell. 2) The potential is continuous at r=d. 3) The electric field is discontinuous at r=d, $E_{r>d}|_{r=d}-E_{r<d}|_{r=d}=4\pi\sigma$, where $\sigma=\frac{q}{2\pi d^2 \sin\theta}\left[ \delta(\theta)+\delta(\theta-\pi)\right]$(not sure about the $d^2$ down there!!!).

My classmate says that he can write the potential as $\phi=\frac{q}{R_1}+\frac{q}{R_2}-\frac{2q}{R_3}+F(r,\theta)$ where the first three terms are the potentials for point particles and F is an unknown function we want to determine. Because we have $\nabla^2 \phi=4\pi q\left[\delta(R_1)+\delta(R_2)-2\delta(R_3)\right]$, we know that $\nabla^2 F=0$. So we can say that $F=\sum_{l=0}^\infty (A_l r^l+\frac{B_l}{r^{l+1}})P_l(\cos\theta)$.Actually he didn't give this line of reasoning so the professor said you can't do it and so the discussion got hot. Now that it seems to me both of them are acting excited and defensive, I was there to get into their discussion and suggested this line of reasoning to them but the professor said that you can't say $\nabla^2 F=0$ on the exact point of sources where there is singularity to which I don't know what to respond, I'm not even sure this is right or wrong. If someone can give a mathematically rigorous reason leading us one way or the other, the problem would be solved because either the above line of reasoning also leads us to partitioning of the space inside the sphere, or the one expansion we wrote is enough and we can fix the coefficients using only the fact that the potential is zero on the sphere(which seems weird to me!). Also my classmates needs to set $B_l=0$ but again there is argument on this.

To be honest it seems to me now this is going out of hand and most of the discussion is because of the awkward excitement on both sides so I really need strong reasons to persuade them.

Thanks

Last edited: Dec 21, 2015
2. Dec 21, 2015

### TSny

Uh oh, I'm not sure I want to get into the middle of this .

But your classmate's approach looks correct to me. I think $F(r, \theta)$ is just the potential inside the sphere due to induced charges on the inner surface of the sphere. So, $F$ will satisfy Laplace's equation everywhere inside the sphere and satisfies a boundary condition on the inner surface of the sphere such that the total potential at the surface is zero. This allows you to determine the solution for $F(r, \theta)$ everywhere inside the sphere in terms of a Legendre polynomial expansion.

Thus you can find the solution for the total potential inside the sphere in the form $\phi = q/R_1 + q/R_2 - 2q/R_3 + F(r, \theta)$ as your classmate argues. Of course you could expand the first two terms in Legendre series and these series would have different forms depending on whether $r > d$ or $r < d$.

The series expansion for $F(r, \theta)$ can be shown to correspond to the potentials of the image charges (located outside the sphere) corresponding to the two charges $q$. Thus, you can get $$\phi = q/R_1 + q/R_2 - 2q/R_3 + F(r, \theta) = q/R_1 + q/R_2 - 2q/R_3 + q'/R_5 +q'/R_6$$ where q' is an image charge. So, the solution can be written without any expansions in Legendre polynomials.

3. Dec 21, 2015

### ShayanJ

1) Will this give us the discontinuity of the electric field at r=d that we want? How can it be guaranteed when we don't impose it explicitly? Maybe the point charge parts do it for us with their singularity, is it right?
2) How is it that one approach needs three boundary conditions but another needs only one? This is really weird!

4. Dec 21, 2015

### TSny

The field is discontinuous at r = d only at the locations of the point charges q. The field is continuous at all other points where r = d.

Yes, by writing explicitly the contributions to the potential from the point charges, you are automatically taking care of boundary conditions at the point charges.

The answer is similar to the answer for your first question. The explicit expressions for the potentials of the point charges takes care of the boundary conditions at the point charges q (where r = d and $\theta = 0$ or $\theta = \pi$).