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How do you derive E = σ/2ε as the electric field around a charged plane?
 one year ago
 one year ago
How do you derive E = σ/2ε as the electric field around a charged plane?
 one year ago
 one year ago

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ghaziBest ResponseYou've already chosen the best response.1
gauss law states that \[\int\limits_{0}^{A} E.ds= \frac{ q }{ \epsilon }\] now \[\sigma\] is surface charge density which is equal to = charge per unit area=\[\frac{ q }{ A }\] therefore \[E.2A= \frac{ q }{ \epsilon } and E=\frac{ q }{ 2A*\epsilon }= \frac{ { \sigma } }{ 2 epsilon }\]...here i have taken 2 A because side surface area of a slab contribute to the electric field where as the upper surface area makes 90 degrees with electric field vector giving dot product of E and A =0 ..since cos 90= 0
 one year ago
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