What is the electric field in between the plates of the capacitor, if `sigma` is the surface charge density?

Explanation

In figure, a Gaussian surface, in the from of a pill box is shown. It is of area `triangleA` (on end faces) and of negligible thickness. Let `vecE_(1)` be the field below and `vecE_(2)`, the field above.
The flux through upper surface `=vecE_(2).triangle vecA`
`=(vecE_(2).hatn)triangleA`
Similarly flux through lower face `=vecE_(1). triangle vecA=(-vecE_(1).hatn) triangleA`
Hence by Gauss theorem
`(vecE_(2)-vecE_(1)). hatn. triangleA=(q)/(epsi_(0)) =(sigma triangleA)/(epsi_(0)) therefore (vecE_(2)-vecE_(1)).hatn=(sigma)/(epsi_(0))`
Inside a conducting surface, E=0 `therefore vecE_(2)=vecE=(sigma hatn)/(epsi_(0))`
b. Consider a closed path ABCD. If `E_(1)^(n) and E_(2)^(n)` are the tangential components, then work done is `E_(1)^(n) trianglel+E_(2)^(n) (-trianglel)=0 ("property of electric field")`
`E_(1)^(n)-E_(2)^(n)=0 or E_(1)^(n) =E_(2)^(n)`, showing that tangetial components are continuous.