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littlecat

  • 2 years ago

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

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  1. ghazi
    • 2 years ago
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    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
    • 2 years ago
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    |dw:1345975850526:dw|

  3. littlecat
    • 2 years ago
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    thank you! :)

  4. ghazi
    • 2 years ago
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    :)

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