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The electric field at a point near an infinite thin sheet of charged conductor is:-

The electric field at a point near an infinite thin sheet of charged conductor is:-

Grade:10

3 Answers

prav
11 Points
6 years ago
the electric flux through each cap isΦ1=E.dS=EdS cos 00=EdSAt the points on the curved surface,the field vector E and area vector dS make an angle of900 with each other.So, φ2=E.dS=EdS cos 900=0Therefore,cylindrical surface does not contribute to the flux.Hence, the total flux through the closed surface isΦ=φ1+φ1+ φ2 (there are two end caps)Or φ=EdS+EdS+0=2EdS (1)Now according to Gauss’s law for electrostaticsΦ=q/ε0 (2)Comparing equations (1) and (2),we get2EdS=q/ε0Or E=q/2ε0dS (3)The area of sheet enclosed in the Gaussian cylinder is also dS. Therefore,the charge contained in the cylinder,q=σdS (σ=q/dS)Substituting this value of q in equation (3),we getE=σdS/2ε0dSOr E=σ/2ε0
Adarsh
768 Points
6 years ago
Dear Rahul,
elctric field near the surface of thin infinite sheet is σ/2ε0.
we can derive it by using gauss law .
Thanks and Regards
Adarsh
SACHIN RAJPUT
15 Points
3 years ago
It's right answer is σ/2ε0.This is a derivation of gauss law.This is valid for electric field at a point near an infinite thin sheet.

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