# Image charges, laplace equation and uniqueness theorem

Consider a well-known problem of the electric field generated by a system composed of a point charge in proximity of a large earthed conductor. It is said that the potential due to an image charge satifies all the boundary conditions - it this case, constant potential on the surface of a conductor - and therefore, by the uniqueness theorem, it is the only possible potential distribution.

However, we can of course imagine other non-trivial fields satisfying the boundary condition of constant potential on the surface, for example generated by three imaginary charges such that the whole system is again symmetric.

So my question is: what is the full set of boundary conditions needed to find the unique potential distribution?

This specific example is just an illustration of a more general question: how to use Laplace/Poisson equations to describe the field of point charges? The divergence of electric field has a singularity at the point charge. Another incarnation of this problem occurs when we want to use the formula for the energy stored in the electric field - we find the energy of the electric field generated by a point charge to be infinite. Does this mean that the concept of a point charge is too idealised?

• Where would these three imaginary charges be placed? – David Z Feb 15 '12 at 23:46
• Energy of a point charge is infinite. If you had to create a pair of point charges from a charge neutral area, you would have to (classically) supply infinite energy because of the $\frac{-kq^2}{r=0}$ infinite negative potential energy of two point charges making up a neutral point. – Manishearth Feb 16 '12 at 7:42
• The opposite also holds:If you destroy a pair of charges by bringing them together and making an overall charge-neutral point, then the infinite energy of the field will be radiated. Note that this is all by classical mechanics. Oh, and currently, we believe that electrons(leptons), W particles, and possibly quarks are true point charges.. – Manishearth Feb 16 '12 at 7:45

However, we can of course imagine other non-trivial fields satisfying the boundary condition of constant potential on the surface, for example generated by three imaginary charges such that the whole system is again symmetric.

I guess you mean that putting another set of image charges on both sides of the conductor plane so that the system is symmetric again. But point charges at specific places are also part of the boundary condition! Image charges cannot be put in the space where your electric field is to be calculated. In summary, there is only one way to put image charges, and there is only one solution of the electric field.

how to use Laplace/Poisson equations to describe the field of point charges? The divergence of electric field has a singularity at the point charge.

Point charges can be mathematically described by Dirac Delta function.

Does this mean that the concept of a point charge is too idealised?

Apparently yes, by classical electromagnetism. As electrons are recognized nowadays as true point charges, and they obviously don't have infinite energy, I guess there is some quantum effect when the radius goes down (vacuum polarization, blah blah blah).

• The infinite energy problem can also be solved by appealing to nonlinear electrodynamics (quantum is not necessary). – Willie Wong Feb 16 '12 at 15:11

There is no other solution of Laplace's equation satisfying the boundary condition of zero potential on the conductor and at infinity.

This can be proved as follows: supposing there are two solutions $\phi_1$ and $\phi_2$ on one side of the conducting plate, with a point charge q at a given position. Then $-\phi_2$ has a point charge of $-q$ at the same position, so that the difference $\phi_1-\phi_2$ satisfies the free Laplace equation on the domain, with no charges.

But the solution of Laplace's equation is the minimum of the field energy

$$\int |\nabla \phi|^2$$

So that the minimum of the free equation is achieved only when the gradient is everywhere zero, and this means that $\phi$ is constant.