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Question

# A very thin disc is uniformly charged with surface charge density σ>0. Then the electric field intensity on the axis at the point from which the disc is seen at an solid angle Ω is

A
σΩ2πϵ0
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B
σΩ4πϵ0
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C
σΩπϵ0
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D
σΩ8πϵ0
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Solution

## The correct option is A σΩ2πϵ0σ>0Electric field intensity on the axis at the point from which =Ω angle.Electric field near on infinite plane of uniform charge density.A much more important limit of the above result is actually for x much less than P. In this case, it is as through the disk were of infinite extent. So, the result correspond simply to the electric field near on infinite shell of these.Ex=2πKσ =σΩ2π4πϵ0K=Ω4πϵ0 =2ΩσΩ2πϵ0

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