Electric Field Due to Charge Distributions - Approach
Intensity of ...
Question
Intensity of electric field at a point at a perpendicular distance 'r' from an infinite line charge, having linear charge density 'λ' is given by :
A
E=(14πε0)λr
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B
E=(14πϵ0)2λr
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C
E=(14πε0)λr2
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D
E=(14πε0)2λr2
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Solution
The correct option is AE=(14πϵ0)2λr Consider a Gaussian surface (cylinder in this case) of radius r and length l.
Then according to the Gauss's Law of electrostatics:
∫→E.→dA=qenclosedϵ0
From the symmetry of the surface, we can easily say that the electric field would only be along the radial direction. And the amount of charge enclosed by this cylinder is qenclosed=λ×l and the surface area of the cylinder A=2πrl.