Potential on conducting shell without image charge

Question

Edit 2: I know how image charges work. But I'm still trying to rationalize without it.

Edit 1: scrap this, realized what I needed was image charges. This entire question is sort of a jury-rig rationalization.

In a system where we have a thick conducting spherical shell (thickness comparable to radius), some charges within the shell, and some charges exterior to the shell; I've been asked to find electrostatic potential of the conducting shell.

The diagram attached shows a simple example of the setup described:

Here we have a $Q$ charge on the external surface of the shell, shell having radius $2R$ and thickness $R$. There's a $Q$ charge in the cavity of the shell, at a distance of $R/2$ from the centre. There's also a $Q$ charge outside the shell, at a distance $3R$ from the centre.

I think I can safely say that the shell's outer charge becomes $2Q$ and the inner surface of the shell acquires a non-uniform charge density of $-Q$. As for the influence of the external $Q$, I believe the outer surface $2Q$ arranges to accommodate that.

Here's my reasoning for the electric potential of the shell. The shell is a conductor, so potential all throughout the material is constant, yes? For a while, let me remove all the charges inside the shell ($Q$ and density of $-Q$) and pretend they don't exist.

(I think this step is valid because potential is a scalar and is associated with space.)

Now we essentially have just a shell with outer charge density $2Q$ and constant potential throughout the interior, and the external charge. The potential at the centre in the modified system is equivalent to the potential of the $2Q$ charge density + the potential at the centre due to external charge $Q$. (This is one of the steps of logical difficulty.)

I reason that the potential at the centre of this modified system is the same as the potential on the surface of the shell (still in the modified system).

Now when we return to the original system and resume the presence of internal charges: I think that the potential on the surface of the conductor stays the same. I presume this is valid because while approaching the outer surface of the shell, the inner charges contribute nothing to the potential (flux does not leave the cavity at the centre of the shell).

So to find out the potential on the outer surface of the shell, I will merely have to sum the potentials at the centre of the shell due to the outer surface itself and due to the external charge.

Is this correct? (I mean, the answer checks out)

Ah, I've got one more question. If I was asked to find the potential at the centre of the shell due to all the charges, I think I would go about doing that by adding the potentials due to: external $Q$, outer surface $2Q$, inner surface $-Q$, and inner point charge $Q$. This corresponds to the answer I've been given.

The validity of this seems to be fuzzy to me at best. Aren't the surface charge densities of $2Q$ and $-Q$ non-uniform? How do we calculate the potential then? (In the first section, too, I assumed we could calculate the potential due to $2Q$). Is it because when bringing a charge from very far away to the surface, at least, we can roughly approximate it to a point charge?

I would be grateful if you could address why this works (at least in the few questions I have done) and whether it is correct at all. Thank you.

Answer 1

PS: The question's title was changed in an edit by OP, it previously didn't contain "without images", that's why this answer was valid. (And of course it gives the right numbers.)

Since we are asked "find the potential of the shell", we must assume that it is not grounded (so it is not kept at $V=0$) and that means the total charge on the shell remains $0$. If not, the net charge on the shell would be an extra free parameter, $Q_s$, which would have to be specified before we can answer the questions.

On the outside we don't see what happens inside, but only that the total charge contained is $+Q$. We can now:

1. Distribute the internal charge $+Q$ equally over the surface, without the external $Q$ present. That gives $V({\bf x})=Q/(4\pi\varepsilon_0|{\bf x}|)$.
2. Subsequently put the external $Q$ at its position ${\bf x}_{\rm ext}=(3R,0,0)$ and give it the correct image charge $Q_{\rm ext\_im}=-2Q/3$ inside the sphere at location ${\bf x}_{\rm ext\_im}=(4R/3,0,0)$
3. Compensate for the image charge because the sphere is not grounded and has to remain chargeless, by adding $-Q_{\rm ext\_im}$ uniformly distributed at the surface.

So the external potential is now known: $$ V_\text{ext}({\bf x}) = \frac{5Q/3}{4\pi\varepsilon_0|{\bf x}|} +\frac{-2Q/3}{4\pi\varepsilon_0|{\bf x}-{\bf x}_{\rm ext\_im}|} +\frac{Q}{4\pi\varepsilon_0|{\bf x}-{\bf x}_{\rm ext}|} $$ which gives the potential at (any point of) the surface of the shell as: $$ V_\text{shell} = \frac{5Q}{24\,\pi\varepsilon_0\,R} \tag{1} $$

On the inside we don't see what happens at the outside, only that the shell is at potential $V_\text{shell}$. We also have the charge $Q$ at position ${\bf x}_{\rm int}=(-R/2,0,0)$ and we give it an image charge $-2Q$ at position ${\bf x}_{\rm int\_im}=(-2R,0,0)$ so the potential in the cavity is: $$ V_\text{int}({\bf x}) = V_\text{shell}+\frac{Q}{4\pi\varepsilon_0|{\bf x}-{\bf x}_{\rm int}|} +\frac{-2Q}{4\pi\varepsilon_0|{\bf x}-{\bf x}_{\rm int\_im}|} $$ which gives the potential in the center as: $$ V_\text{cen} = \frac{11Q}{24\,\pi\varepsilon_0\,R} \tag{2} $$ NB: In the above we assume that the reader is accustomed with using [image charges] for a sphere, which always scale with the relative distance to the spherical image surface, and of course have opposite sign.