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Question

Determine the direction of electric field intensity at a point near a uniformly charged infinite conducting plate. Obtain the relation for electric field intensity using Gauss's law. Draw the required diagram.

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Solution

Electric field intensity due to a uniformly charged infinite conducting plate.
The basic difference between then non-conducting sheer and conducting sheet is that the charge is non-conducting sheet remain at a place where the charge is given while the charge in a conducting sheet is spread out in a whole sheet and the electric field inside the conducting sheet is zero .
Value of electric field : in the figure conducting sheet is shown . when the some charge is given to conducting plate , it gets disturbed on the entire external surface of the conductor if the plane charge sheet is of uniform thickness and infinite size, then the surface. it is the same at the both surfaces.
Consider a cylindrical surface perpendicular to the sheet with the area of cross section (s) as shown in fig On the circular surface of the cylinder P is located.
The total charge enclosed Σq=σS
by the Gauss's law
ϕE=Σqϵ0=δϵ0.........(1)
By the defination of electric flux
ϕE=Eds=SEds1+sEds2+Eds3
ϕE=sEds1cos0+0+0
ϕE=Esds1=Es..............(2)
From (1) and (2)
Es=σsϵ0
E=σϵ0................(3)
On expressing in vector symbols
E=σϵ0^n
Where ^n is the unit vector perpendicular to the sheet from the equation (3) we see that the electric field at any point due to the sheet does not depends upon the distance of the point from the sheet i.e. the magnitude of electric field at all the points is equal .
Er0....(4)
The electric field intensity in any other medium .
E=σϵm=ϵϵrϵ0=σKϵ0.....(5)
Graph of electric field ; the electric field intensity does not depend upon the distance of point from the uniformly charged sheet i.e. it remains the same at all the points . the graph of it is obtained as shown in the figure.


1757137_1834150_ans_151dd296a3914f66a2cf2662fbf61bd2.png

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