The electric field at a point near an infinite thin sheet of charged conductor is
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The electric field at a point near an infinite thin sheet of charged conductor is sigma/2ۥ
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Let us today discuss another application of gauss law of
electrostatics that is Electric Field Due To An Infinite Plane
Sheet Of Charge:-
Consider a portion of a thin, non-conducting, infinite plane
sheet of charge with constant surface charge density σ.
Suppose we want to find the intensity of electric field E at
a point p near the sheet, distant r in front of the sheet. To
evaluate the field at p we choose another point p on the
other side of sheet such that p and p are equidistant from
the infinite sheet of charge(try to make the figure
yourself). Now we draw a small closed Gaussian cylinder
with its circular ends parallel to the sheet and passes
through the points p and p .suppose the flat ends of
p and P have equal area dS.The cylinder together with flat
ends from a closed surface such that the gauss’s law can
be applied.
By symmetry,the magnitude of electric field E at all the
points of infinite plane sheet of charge on either sides end
caps is same and along the outward drawn normal,for
positively charged sheet.
Therefore, the electric flux through each cap is
Φ =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
90 with each other.
So, φ =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=φ +φ + φ (there are two end caps)
Or φ=EdS+EdS
+0=2EdS
(1)
Now according to Gauss’s law for electrostatics
Φ=q/ε (2)
Comparing equations (1) and (2),we get
2EdS=q/ε
Or
E=q/2ε dS
(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/2ε dS
Or E=σ/2ε
This is the relation for electric filed due to an infinite plane
sheet of charge.
Thus, the field is uniform and does not depend on the
distance from the plane sheet of charge.u can see the figure at https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcTzS8N3CldKoEqcO7pEBrHWXPl2bNYLZwdmqwwfO6AKicRmlqMnwMCbcg
electrostatics that is Electric Field Due To An Infinite Plane
Sheet Of Charge:-
Consider a portion of a thin, non-conducting, infinite plane
sheet of charge with constant surface charge density σ.
Suppose we want to find the intensity of electric field E at
a point p near the sheet, distant r in front of the sheet. To
evaluate the field at p we choose another point p on the
other side of sheet such that p and p are equidistant from
the infinite sheet of charge(try to make the figure
yourself). Now we draw a small closed Gaussian cylinder
with its circular ends parallel to the sheet and passes
through the points p and p .suppose the flat ends of
p and P have equal area dS.The cylinder together with flat
ends from a closed surface such that the gauss’s law can
be applied.
By symmetry,the magnitude of electric field E at all the
points of infinite plane sheet of charge on either sides end
caps is same and along the outward drawn normal,for
positively charged sheet.
Therefore, the electric flux through each cap is
Φ =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
90 with each other.
So, φ =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=φ +φ + φ (there are two end caps)
Or φ=EdS+EdS
+0=2EdS
(1)
Now according to Gauss’s law for electrostatics
Φ=q/ε (2)
Comparing equations (1) and (2),we get
2EdS=q/ε
Or
E=q/2ε dS
(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/2ε dS
Or E=σ/2ε
This is the relation for electric filed due to an infinite plane
sheet of charge.
Thus, the field is uniform and does not depend on the
distance from the plane sheet of charge.u can see the figure at https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcTzS8N3CldKoEqcO7pEBrHWXPl2bNYLZwdmqwwfO6AKicRmlqMnwMCbcg
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