Question

E14M.10 Consider a boundary that contains a very thin layer of surface charge. Assume that the boundary is "smooth,” meaning that we can model any sufficiently small region as being flat. Consider a gaussian surface that is a tiny rect- angular solid that straddles the boundary and is so small that the enclosed boundary is essentially flat. We can there- fore squeeze the gaussian surface's sides perpendicular to the boundary until they have essentially zero area. In this limit, show that Gauss's law requires a certain component of the electric field to change by o/E, as we go from below the boundary to above the boundary (where o is the local charge per area), and describe which component changes.

Answer 1

r = unit normal vector take a rectangular Gaussiah pill bon which is essentially flat & the Width (w) of the rectangular bon is infinitesimal lie wyo), it dă,= da ¢ is the infinitesimal area rector of the top side then daž=-da ħ is the area rector of the bottom side. let EA o be the Electric field just above the surface & ĒB be the euctric field tust below the surface, Now, as wayo, the flun through the Gaussian Pillbon is, E. dão Encãi tēnei when - EañHA + Gēgin) da Ā A: a

Now, by Gauss's law $. cn = document Charge here, the enclosed Qene = o da charge density there r surface is the so (Eniñ - En f) 2A = od now, ñ is the to the surface, of É takes out component of of Ê which is unit normal vector so the dot product the perpendicular Foie component perpendicular to the surface. of So, the perpendicular component Ē to the surface suffers a discontinuity by an amount E. where o is the surface Charge density.

let loop the above below ABCD be the hectangular of length b & bredth a. line segment Ap is tust the surface & BC is just the surface. Let the breathayo electro- Now since, ax = ő for -static condition, hence, ABCO since, ano, side AB & Co so are contribution of essentially zesso, let to AD D from Î be the unit vector parallel & BC, & directed towards A.

so, TED - Ea. f) 6-6.)b.co ABCO sina, bzo » Ēa. f - ēm. f =0 Now, Ēn. F is the parallel Component of E r to the surface fust above it & Eb F is the Purcelles componet of Ēk to the surface fust below it. So, the Electric remains component of some surface parallel field to continious. so, The normal component of Electric field to surface with surface Change density (o suffers a Change by a s the parallel Component remains unchanged.