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Grade: 10
        The electric field at a point near an infinite thin sheet of charged conductor is:-
8 months ago

Answers : (2)

prav
11 Points
							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
						
8 months ago
Adarsh
763 Points
							
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
8 months ago
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