the electic field at a point near an infinite thin sheet of charged conductor
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Φ1=E.dS=EdS cos (0) = EdS
At the points on the curved surface,the field vector E and area vector dS make an angle of
900 with each other.
So, φ2=E.dS=EdS cos 90=0
Therefore,cylindrical surface does not contribute to the flux.
Hence, the total flux through the closed surface is
Φ=φ1+φ1+ φ2 (there are two end caps)
Or φ=EdS+EdS+0=2EdS (1)
Now according to Gauss’s law for electrostatics
Φ=q/ε0 (2)
Comparing equations (1) and (2),we get
2EdS=q/ε0
Or 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 get
E=σdS/2ε0dS
Or E=σ/2ε0
Hope it helps u
Plz mark as brainliest
At the points on the curved surface,the field vector E and area vector dS make an angle of
900 with each other.
So, φ2=E.dS=EdS cos 90=0
Therefore,cylindrical surface does not contribute to the flux.
Hence, the total flux through the closed surface is
Φ=φ1+φ1+ φ2 (there are two end caps)
Or φ=EdS+EdS+0=2EdS (1)
Now according to Gauss’s law for electrostatics
Φ=q/ε0 (2)
Comparing equations (1) and (2),we get
2EdS=q/ε0
Or 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 get
E=σdS/2ε0dS
Or E=σ/2ε0
Hope it helps u
Plz mark as brainliest
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