Question

the electric field at a point near an infinite thin sheet of charged conductor is

Answer 1

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 p1near the sheet, distant r in front of the sheet. To evaluate the field at p1 we choose another point p2 on the other side of sheet such that p1and p2are 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 p1and p2.suppose the flat ends of p1and P2have 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.

Answer 2

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

90° 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
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