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The method of image charges is a well-known and very useful tool for solving problems in electrostatics.

Unfortunately, when I was taught this method, it was presented simply as an algorithm. No real physical justification was given for its usage, and there was a complete absence of rigorous mathematics. The method has always been a little hazy to me since.

If someone could give a rigorous first-principles derivation of this method from Maxwell's equations (perhaps simply Coulomb's law?), it would be most appreciated.

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A derivation of the fact that the method of images works relies on the uniqueness theorem for solutions of Poisson's equation (up to a constant term). Are you looking for a derivation of that, or are you willing to have it used without explicit proof? (In the former case, questions about the derivation of the uniqueness theorem itself would be more appropriate on math.SE, I think) –  David Z Jan 24 '11 at 4:17
@David: Yeah; I can live with just taking the uniqueness theorem for granted, I think. The rest of the derivation is of interest though. :) –  Noldorin Jan 24 '11 at 14:40
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4 Answers

I think that Mark's answer is very good - and in fact, the Wikipedia article you quoted is not bad, either. But let me simplify it a bit:

The task where the method is useful is to find the electric field created by charges and planar (or straight) or spherical (or circular) conducting planes - or more general geometries where the method still works but that are unlikely to be encountered "naturally".

The electric field has to satisfy the usual conditions, $\mbox{div}(\vec D)=\rho$ and $\mbox{curl}(E)=0$. At the surface of the conductor, $\vec{E}$ has to be perpendicular to the surface.

What is the field induced by a configuration of charges and an infinite planar conducting plane? It's easy. You just put mirror charges on the other side from the surface of the planar conducting plane: they have the opposite signs of the charge and they're located just like the images in the mirror.

This is a solution - and because of its uniqueness, the only solution - to the original problem because $\mbox{div}(\vec D)=\rho$ in the physical part of the space where the charges are located. The divergence of the electric field also contains terms that are nonzero in the unphysical part of the space, behind the mirror, but we don't trust our solution over there, anyway. We replace it by $\vec{E}=0$ over there because it's a conductor.

The most important check is that the electric field constructed as the sum from the real charges and their images is perpendicular to the surface of the conducting plane. That's easily seen because the field lines from the configuration of the charges are symmetric with respect to the plane of the conductor's surface, by construction.

So one has the right divergence and the right curl of the electric field in the physical part of the space - and the right boundary conditions on its boundary, namely the perpendicularity of $\vec{E}$.

The only other simple generalization of this method involves spherical (or circular, in 2D problems) surfaces of the conductor. The mirror charges are located somewhere again. Their charge has a different magnitude, the location has to be calculated, but when you do it right, it will still be true that the electric field will be perpendicular to the spherical or circular surface of the conductor.

The easiest way to understand why the mathematics works is to learn some conformal symmetry. By conformal transformations, the plane from the simplest "flat mirror" problem may be mapped to a spherical plane. The locations and magnitude of the charges and the mirror charges also transforms in a calculable way, allowing us to solve a broader set of problem.

In principle, one may generalize the method to more complicated geometries of the surface of the conductor - more complicated configurations of charges may be needed and more complicated transformations (especially in 2D) may be helpful to find them; and the problem may also be generalized to problems with other boundary conditions on the surface - different components of the fields may be required to vanish. The set of all possible applications and generalizations can't be "quite classified": it's important to actually understand why and how the method works and then you may apply it whenever you want.

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Thanks @Lubos, this is helpful. I'll read it over more closely when I have a few more minutes. –  Noldorin Jan 24 '11 at 21:58
@Lubos: Could you please suggest a reference which explains the relevance of conformal symmetry in the method of images? –  Siva Dec 2 '12 at 7:03
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In electrostatics, we want to solve Poisson's equation for the electric potential $\Phi$.

$$\nabla^2 \Phi = 4 \pi \rho$$

Take, for example, free space, so $\rho = 0$. The resulting equation is Laplace's equation

$$\nabla^2 \Phi = 0$$

There are many solutions, for example $\Phi = 0$ or $\Phi = \Phi_0 e^{k(ix+y)}$ or others. Further, if $\Phi_1$ and $\Phi_2$ are both solutions, then $\alpha \Phi_1 + \beta \Phi_2$ is also a solution.

Poisson's equation has the same difficulty. If you have any solution to Poisson's equation, you can add any solution to Laplace's equation and get a new solution.

If someone said, "I just threw a ball; predict its location as a function of time," we would be in a similar predicament. We could determine a differential equation that describes the ball's motion, but there are many solutions, and we couldn't find the trajectory without knowing where the ball was thrown from and what its initial velocity was. These are boundary conditions. They take a differential equation with many solutions and narrow them down to one physically correct solution.

In the case of Poisson's equation, one sufficient set of boundary conditions is the potential everywhere on the boundary of the area we're studying. The "boundary" can be an infinity, if we want.

For example, if there is a grounded conductor, we set $\Phi = 0$ everywhere on that conductor, then set $\Phi = 0$ at infinity. If we can find any solution to Poisson's equation that also satisfies these two boundary conditions, it will be the only solution, and we will know the potential everywhere outside the conductor.

To prove this uniqueness theorem, imagine there are two solutions, $\Phi_1$ and $\Phi_2$, to Poisson's equation. Then $\Phi_3 = \Phi_1 - \Phi_2$ is a solution to Laplace's equation, and it has $\Phi_3 = 0$ on all the boundaries. Any solution to Laplace's equation has no local minima or maxima, so the extrema must occur at the boudaries. Since all boundaries are zero, $\Phi_3 = 0$ and $\Phi_1 = \Phi_2$.

Now consider the simplest application of the method of images. There is an infinite grounded conducting sheet and a point charge $q$ sitting above it at distance $d$. We need to find any solution to Poisson's equation such that $\Phi = 0$ at infinity and along the sheet.

Discard the sheet momentarily, and consider another point charge $-q$ a distance $2d$ from the first one, so that it looks like its mirror image, with the plane where the conductor was serving as mirror (the original point charge is above the plane; the new one is below the plane an equal distance).

Imagine a test charge sitting on the plane between the two mirror charges. Because the mirror charges have opposite sign, the net force on the test charge has no component in the plane. That means the plane is an equipotential surface, and the point charge can move out to infinity for free, where the potential is clearly zero. Hence, the potential is zero at infinity and all along the plane. The $\Phi$ you get in this scenario is easy to calculate because it's just the potential from two point charges, however, because it meets the boundary conditions, it is also the potential that arises from a point charge sitting above a conductor.

This method of images is not very general. It will work whenever you can think of a clever way to set up an image charge so that the potential with the original charge and the image charge satisfies the same boundary conditions as your original problem. You won't always be able to find a way to do that. In a general case, you'll need to use Green's functions or numerical solution methods.

Reference: This is mostly from Chapter 3 of David Griffiths' Introduction to Electrodynamics

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Thanks. This looks pretty thorough! I'll have a closer read later. –  Noldorin Jan 24 '11 at 21:57
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It takes advantage of the fact the solution is unique and fixed by the boundary conditions.

  • Any solutions is the solution.


  • Any solution that gets the boundary conditions right is a solution.
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Thanks, but I'm actually looking for a complete derivation from first principles. This hints at the mathematics involved, but I'm not so good at filling in the gaps! –  Noldorin Jan 24 '11 at 2:15
@Noldorin: Ah...I'd have to crack a book for that. I spent a long nigh convincing myself of this once, and have since taken it for granted. –  dmckee Jan 24 '11 at 2:24
@dmckee: Hehe fair enough. Maybe I could spend a similar long night understanding it too..? Any information you have to offer might ease that though! –  Noldorin Jan 24 '11 at 2:36
The link I included has a terse derivation though I think it glosses over the hard parts concerning the sufficiency boundary conditions. Jackson or Landau & Lifshitz are always reliable for these things. –  dmckee Jan 24 '11 at 2:40
Surprisingly, Jackson doesn't do a good job on this and L&L (Vol. 6, of course) gives a couple a good examples. –  Robert Smith Jan 24 '11 at 2:53
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The answers given so far are very good, but I want to add something that might help. The method of images begin by trying to derive Coulomb's law. Coulomb's law is the Green's function of the Laplace equation. It's general form is:

\begin{equation} G(\mathbf{x},\mathbf{x}^{\prime})=\frac{1}{R}+F(\mathbf{x},\mathbf{x}^{\prime}) \end{equation}

where $F$ is a function that satisfies the Laplace equation in your domain. In the case that the domain extends to infinity and the potential fall-off fast, $F=0$ and you have the usual Coulomb law. However, in the case of a domain limited by conductors (well, any boundary condition for which the solution exists and is unique will do, I'm just using the usual case), where you have $G=0$ on the boundary, $F$ will have to be adjusted so that this condition is satisfied.

Now comes the physical interpretation: since $F$ satisfies the Laplace equation inside your domain, but not outside, it can be interpreted as the potential due to a system of charges outside your boundaries. And those are exactly the image charges.

More details can be found in Jackson, Classical Electrodynamics. Section 1.10: Formal solution of electrostatic boundary-value problems with Green Function.

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