2.  Boundary Conditions at Surfaces


Following Jackson (we’ve got to section 1.5) we’ll look at interfaces between media. It’s a slight puzzle why Jackson does this now, he assumes you’ve seen the integral form of Maxwell’s equations in an undergraduate course, and we’ll go along.  First we’ll look at field discontinuities on going say from air into glass, then we’ll review the closely related problem of field discontinuity from a surface charge layer, finally we’ll examine a dipole layer, which we’ll need shortly.

Field Discontinuities at an Interface between Media

Imagine a locally flat interface between two uniform materials having different ε,μ.  (We don’t take the more general tensor form.)

In this section, we’ll assume there is no charge or current on the interface.

We’re interested in how the electric and magnetic field vectors change on crossing from one medium to the other. 

First we’ll examine the change in the normal to the surface components of the fields.

We find this by using the first two Maxwell equations (in integral form):

S D n da= V ρ d 3 x =0, S B n da=0 .  

The trick is to integrate the vector fields over the complete surface of a vanishingly small pillbox, very thin, having its circular flat surfaces one in each medium.

The area of the cylindrical sides goes to zero in the limit we’re interested in, and the pillbox is so small we can take the field constant over the range of the integral, so, (for no charge at the boundary surface between the media)

D 1 n = D 2 n , B 1 n = B 2 n .  

This is often written as  D 1 = D 2 , B 1 = B 2 .  

Note that since D 1 = ε 1 E 1 , D 2 = ε 2 E 2 ,  this is not the same as E 1 n = E 2 n !

The tangential components can be found using the other two Maxwell equations,

C H d l = S J + D t n da, C E d l = S B t n da .  

Recall the left-hand sides are integrals around loops, the right-hand sides integrals over surfaces spanning the loops.

We choose as loops of integration tiny rectangles with long sides parallel to the surface and very close but on opposite sides, and much shorter sides perpendicular to the surface. The contribution to the loop integral from these shorter sides vanishes in the thin limit, and the result is E 1|| = E 2|| , H 1|| = H 2|| .

It is straightforward to generalize to the case where the surface has nonzero charge density and/or current density:

n × H 2 H 1 = K ,

the vector K  being the surface current density.

Electric Field Discontinuity across a Charged Surface

This seems a good place to add a very similar scenario: a flat plane of charge in vacuum, charge density  σ  coul/m2.  Assuming no other charges are present, the electric fields on the two sides of the plane will be uniform, perpendicular to the plane and in opposite directions on the two sides, both pointing outwards if the plane is positively charged. The magnitude of the fields is easily found by using Gauss’ law for a pillbox shape, two round flat sides of area A  parallel to the sheet of charge and on opposite sides.  The cylindrical side is taken to go to zero in the limit. The enclosed charge is σA,  the flat sides contribute 2EA  to the Gaussian integral, so

E=±σ/2 ε 0 .

There is a discontinuity in the normal component (the only one present in this example) of the electric field.

Let’s think a little further concerning this discontinuity.  Imagine we’re looking in the neighborhood of a point P  which is a tiny distance δ  from the plane.  If we now move P  through the plane, it experiences this discontinuity in field. 

Now, mentally split the field in this small region into two contributions, that from charge within a small circle around where P  passed through the plane, the other contribution from the (further away) rest of the plane, that is, from a plane with a small hole in it, and we’re looking at the electric field on going through the middle of the hole, along a line perpendicular to the plane. The electric field from this rest of the plane has to be continuous along this line so the discontinuity is generated entirely by the very local charge distribution.

It follows that for any reasonably smooth surface, not necessarily plane, having in general varying charge density, there will be a discontinuity in the normal component of the electric field of magnitude depending on the local charge σ r / ε 0  on passing through it.

Electric Potential Change across a Dipole Surface

Dipole surfaces are certainly less common, although they do exist in biological systems. But they also come up in some important mathematics, as we’ll see in a while, so this is worth thinking through.

The Dipole

Before analyzing the dipole surface, let's remind ourselves about a single dipole. The dipole is defined as the limit of two equal and opposite charges ±q , separated by a small displacement δ ,  in the limit δ 0  under the condition that q δ = p  is held constant. 

(This drawing from here.)

The vector quantity p  is called the dipole moment. The vector δ  points from the negative charge to the positive charge.

The potential from a dipole at the origin is

φ r = lim δ 0 1 4π ε 0 q r δ q r = 1 4π ε 0 q δ 1 r = 1 4π ε 0 p r r 3 .

The electric field

E r = φ r = 1 4π ε 0 3 r ^ r ^ p p r 3 ,

where as usual r ^  means a vector of unit length.

Potential Discontinuity across a Dipole Sheet

We first consider a flat sheet of uniform strength dipole, meaning a sheet with uniform charge surface density σ  coul/m2, separated by a distance δ  from a sheet oppositely charged, in the limit δ0  holding δσ=D , the dipole surface density, constant.

We proved above that for a single uniform sheet of charge the electric fields above and below were E=±σ/2 ε 0 ,  taking that for the (upper) positive sheet of the dipole, the negative sheet will contribute fields E=σ/2 ε 0 ,  also extending throughout space, so the fields will exactly cancel except between the sheets, where E=σ/ ε 0 .  

The potential difference between the sheets φ 2 φ 1 =Eδ=σδ/ ε 0 =D/ ε 0

On crossing a uniform sheet of dipole there is a discontinuity in the potential equal to dipole strength / ε 0 .  

Electrostatic Potential from a Dipole Layer

Assume an arbitrary surface having uniform dipole strength D,  meaning that each increment of area has dipole strength Dd A .  

The electrostatic potential at the origin from the surface increment d A  is, from the dipole potential formula in the previous section, 

φ 0 = 1 4π ε 0 Dd A r r 3 = 1 4π ε 0 DdAcosθ r 2 = 1 4π ε 0 DdΩ,

since dAcosθ  is the projection of the area on to a plane perpendicular to r ,  equal to r 2 dΩ .

Therefore, at a point, the potential from the whole dipole surface equals the solid angle subtended.

Exercise:  Take a single dipole to be a tiny disc, and visualize the solid angle subtended at different points in space show this reproduces the dipole potential found above.

Closed Uniform Dipole Layer

Taking a point P outside the surface, since each increment of surface contributes to the potential at P proportional to the solid angle subtended (and remember the sign!) the total closed surface will contribute zero potential cancellation between the increments shown, which subtend the same solid angle at P, but clearly with opposite signs.  On the other hand, if P is inside the surface, the contributions all have the same sign, so the potential at all points inside is the same, equal to D/ ε 0 ,  in agreement with the potential discontinuity given above.