# Why does charge build up at the boundary surface of two media?

On a homework problem, we are asked to to use the first two Maxwell equations,

$$\nabla\cdot \mathbf{B} = 0$$

$$\nabla \cdot \mathbf{D} = \rho$$

to show that along the boundary surface of two different media (different permittivity constants), the magnetic field component that is normal to that boundary is continuous, and also that the electric displacement field $\mathbf{D}$ has a discontinuity on the surface because of the surface charge density on that boundary.

How do I show this? Am I supposed to show, through Maxwell's equations (the ones provided) that an electromagnetic four-potential causes surface charge density to accumulate on the boundary? Why does this happen?

• The polarization changes from one dielectric to the other - this means there will be some net charge on the interface. Think about the implications of $D=\epsilon_0 E + P$. – Floris Feb 11 '15 at 3:21
• Isn't the polarization change only for the component of the electric field tangential to the boundary surface? – Arturo don Juan Feb 11 '15 at 3:46
• If one of the "materials" was a vacuum, there would be no polarization in that material - but there would be in the other material. So no - I am pretty sure that there is a component of polarization normal to the surface (parallel to the field). – Floris Feb 11 '15 at 3:54
• Wouldn't there be no phase change then if you were talking about incident light incident right on (nearly completely perpendicular to) the boundary of a medium? (This doesn't happen) In that case, you only have your $\mathbf{D}$ component in the direction parallel/tangential to the boundary surface. – Arturo don Juan Feb 11 '15 at 4:16
• I think I was just confusing 'parallel' to the surface and 'parallel' to the normal to the surface. – Arturo don Juan Feb 26 '15 at 1:01