Interesting problem appeared recently. It’s equivalent to solving the electrostatic potential above a plane given the surface potential on that plane.

Say the plane is at and we wish to find the electrostatic potential for given . We start with the divergence theorem

where is a well behaved vector field defined in the volume bounded by the closed surface . Let where and are arbitrary scalar fields. Then

and

where is the normal derivative at the surface (directed outwards from inside the volume ). Substituting the above equations back into the divergence equation, we find Green’s first identity:

Interchanging and in the above and subtracting the two equations (the term cancels) we obtain Green’s theorem (or Green’s second identity):

.

We now choose where is the observation point and is the integration variable and let , the electrostatic potential. Now use the facts that and and we find

If lies within the volume , we obtain:

If lies outside of then the left hand side of the above equation is 0.

This equation is not a solution of the boundary value problem but is simply an integral statement since the right hand side is an overspecification of the problem. We can define a Green’s function

where

and the function satisfies Laplace equation inside the volume :

We see if we use and the freedom that allows us, we can now use Green’s theorem to eliminate one or the other surface terms. If we take and plug this back into Green’s theorem we find

Since we are interested in the Dirichlet boundary condition (that is we are given the potential on ), we demand

for on . Then we have

If one is interested in the Neuman boundary condition (that is we have the normal component of the field on ), then we can define the corresponding Green’s function:

for on . Similarly as before, we can use Green’s theorem to find

where is the average value of the potential over the surface- typically an irrelevant constant.

For the Dirichlet case, it can be shown via Green’s theorem, that while for the Neuman case, can be forced to have the same property.

In our problem, consists of the upper half infinite volume . If we assume that falls to zero at large the surface integral only survives for . Since we are specifying the potential on the plane we need to find . By the method of images, we can immediately write it down

We see that when as required. We also note that the second term on the right hand side does satisfy Laplace’s equation in . Therefore by uniqueness we have the correct Green’s function. We next need to compute

Plugging this back into the equation for and assuming there are no charges in the volume () we find

This is the solution to our problem. We see that given we have an integral equation to find for . Furthermore, we see the Green’s function only depends on the differences of and , or equivalently, is simply a 2D convolution of and the kernel ( being viewed as simply a parameter for the surface integral to determine where in the potential should be evaluated at).

If one is more interested in the electrostatic field (vs. potential), we use and we can find

where again we can view these as 2D convolutions. We also note that can be shown for , as required.

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