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What is the electric field in between the plates of the capacitor, if sigma is the surface charge density? |
Answer» Solution :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. Considera 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. |
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