# What does it mean that an expression satisfies Poisson's equation?

The criteria for an expression to determine the field and charge distribution when using the method of images is that:

1. the expression must satisfy Poisson's equation, which is, $$\nabla^2\phi=-\frac{\rho}{\mathcal{E}_0}$$.
2. the potential must approach 0 as r approaches infinity and
3. the potential from the image(s) and real charge must sum to 0 on the grounded sphere or a constant value on the Non-grounded sphere

I am using the method of images to solve for the field outside of and charge distribution on a sphere. I understand #2 and #3, but I don't quite understand #1. When I solve this for the field and the charge distribution I get:

$$\phi=\frac{1}{4\pi\mathcal{E}_0}\left[\frac{Q}{\sqrt{R^2+A^2-2RAcos{\theta}}}-\frac{q_i}{\sqrt{R^2+a^2-2Racos{\theta}}}\right]=0$$ on the surface

$$\ \sigma=-\mathcal{E}_0\frac{\partial\phi}{\partial r}=\fbox{\frac{-Q\left(A^2-R^2\right)}{4\pi R}\left[\frac{1}{\left(R^2+A^2-2RAcos{\theta}\right)^{3/2}}\right]}$$

I have verified by integrating to get the total charge on the outside of the sphere which should be $$-Q\frac{R}{A}$$ as in the Figure and it is.

I haven't put down more of the steps because that’s not the main point I am asking about.

I of course realize that the expression that I use has the form of Poisson's equation. I also realize that the problems are electrostatic and therefore have no extra energy from displacement currents and that they relate only to potential fields. What explicitly does #1 mean? Just that the form must match, and that it is a potential, and that there is no extra charge or energy from somewhere?

• I do not believe that this equation is true $\nabla^2\phi=-\frac{\sigma}{\epsilon_0}$, it certainly is not of Possion, with $\sigma$ representing surface charge density. Instead the $\sigma$ follows from the boundary conditions imposed on the metal: ${E}_n=\sigma/\epsilon_0$ where $\mathbf{E} = - \mathrm{grad} \phi$ Feb 4, 2020 at 0:40
• Thanks for the tip. Partial illumination has occured. I'm not sure the equation was wrong yet, but my sentence suggested something that is definitely not true. I think I have deleted the offending material. I don't think there are anymore errors, just a continuing mist upstairs.
– DMac
Feb 4, 2020 at 1:12
• #1 means that $\phi (\mathbf{x})= \frac{1}{4\pi\epsilon_0} \int_{all space} \frac{\rho}{|\mathbf{x}-\mathbf{x}'|} d^3\mathbf{x}'$ with the boundary condition $\frac{\partial \phi}{\partial s}|_{sphere} = 0$ (tangential derivative to the sphere) Feb 4, 2020 at 1:19
• Not sure I understand. Are you just restating #1 and #3? Is there redundancy in 1,2, and 3?
– DMac
Feb 4, 2020 at 1:26
• all I did was to answer your question "What explicitly does #1 mean?" Feb 4, 2020 at 1:39

First note that the potential of the real charges - the external charge and the surface charge on the sphere - will obey Poisson's equation, because real charges obey Maxwell's equations! When using the method of images you don't change the distribution of charge or the potential outside the sphere, so Poisson's equation is still obeyed in that region.

It is fairly obvious that you don't change the charge distribution outside the sphere -- you leave the single external charge alone -- but less obvious that you leave the potential unchanged. This relies on a uniqueness theorem, which is proven straightforwardly in Griffiths' Electrodynamics text book and probably others.

The uniqueness theorem says that if you specify the charge distribution in the region of interest, and you also specify the potential on the boundaries, then this fixes the potential everywhere in the region. (Slightly subtle point: Your region of interest is the space between the surface of the sphere and infinity. It does not contain its boundaries, but the boundary conditions still apply. Your boundary conditions are that $$V=V_0$$ on the surface of the sphere and $$V=0$$ at infinity.)

Once you have the potential outside the sphere, you can combine this with what you already know about the potential inside ($$V_0$$, because there is no electric field in a conductor) to determine the surface charge distribution.

• You have restated here that which I believe that I know. The thing that makes me think that I may be missing something is that I am using Poisson's equation here to solve for the distribution and the field. Isn't that enough to satisfy Poisson's equation? Or, are there other aspects to it that are implicit that I am not picking up on. I don't understand the term "satisfy Poisson's equation". Maybe another way to ask it is: Can you use Poisson's equation without satisfying Poisson's equation? It doesn't seem possible to me.
– DMac
Feb 3, 2020 at 23:43
• I could construct a potential that is equal to $V_0$ at the sphere's surface, $0$ at infinity, and $1$ everywhere else, and have just one static charge as specified in the problem. After all, I'm just constructing some arbitrary 2D function. But it wouldn't be physically realisable because it doesn't obey Poisson's equation. With the method of images you don't have that problem because you are using a physical distribution of charges to come up with your potential function, but in general it matters.
– Tony
Feb 4, 2020 at 0:52
• You could also think of some physical distribution of charges which replicates the original potential on the boundaries but alters the charge distribution in the region of interest (e.g. in your case puts some new charge outside the sphere). In this case the new physical distribution of charge would obey a Poisson's equation, but not the same equation as the real charge distribution obeys ($\rho$ will be different). So perhaps this requirement is to clarify that it has to be the original equation that is obeyed?
– Tony
Feb 4, 2020 at 1:14