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

1. ghazi

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

2. ghazi

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3. littlecat

thank you! :)

4. ghazi

:)