# Potential on conducting shell without image charge

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.

• Longest question I've ever written :p Commented Jun 5 at 4:10
• Electrostatic potential at the centre of a spherical shell is kQ/r even if charge is unevenly distributed on the surface Commented Jun 5 at 4:50
• The outer surface charge density would be uniform. Consider this: You have a charge q inside, an amount -q would get accumulated in the inner surface of the conductor, these two together would cancel each others field, and so the outer surface experiences no other forces. Thereby they arrange themselves in the most uniform way possible. Commented Jun 5 at 8:14
• With the title now changed (to exclude using images) I think the most elegant way is to use 3D conformal mapping to a spherical void where the point charge is then exactly in the center. By symmetry the charge distribution on the void surface is then uniform. (For the outer region we can invert the ball to a void. 3D conformal mappings are less versatile than their 2D counterparts, and need scaling of the potential with the Jacobian, but for spherical shapes and point charges they are well-suited!) Alternatively you could just solve for the charge distribution by "mathematical brute force". Commented Jun 15 at 13:22
• PS: and your assumption that the charge on the outer shell is $2Q$ is wrong. It is only $+Q$ if the total charge on the sphere is $0$ and it's only $\frac53 Q$ if the potential of the sphere is $0$. From your question it appears that the condition to be used is the former, but strictly speaking this information is missing. And the claim by @khaxan is also wrong. There a factor $\frac{11}{6}$ is missing! (But khaxan would be correct if the potential of the sphere was $0$, only that is inconsistent with "find the potential on the sphere". So I think total charge on the sphere should be $0$.) Commented Jun 15 at 14:38

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.

• Sorry, changing the post without leaving a comment here was kinda sneaky :( Commented Jun 16 at 15:28
• No big deal, it's clarified now. After all it is your question. (And an interesting one, are you still looking for other solution methods than mathematical brute force, or the method of images, or conformal mapping?) Commented Jun 16 at 18:04
• I think I'll stick with images, really, that seems most optimal to use and easy enough to understand at my level. There are a lot of problems with approximation that just won't go away without rigour. Thanks a lot! Commented Jun 16 at 18:30