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© GC Shiba
of cross-sectional area A through P is drawn as shown in figure.
The total electric flux passing through the two flat surfaces at the ends of PP' is
= 2 .
Since net charge enclosed by the area A is σ A, so by Gauss's theorem,
ℎ
=
∈ 0
, 2 . =
∈ 0
=
2 0
Clearly, E is independent of r, remains the same and does not change
with distance from the conductor. The field lines remain everywhere
straight, parallel and equally spaced.
6) Electric intensity due to charged plane conductor
Let us consider a positively charged
plane conductor having surface charge
density σ .( σ = charge/surface area). We
are interested to find the electric intensity at
point P outside the charged plane conductor
at a distance r. A cylindrical Gaussian
surface of cross-sectional area A through P
is drawn as shown in figure.
The total electric flux passing through the
Gaussian surface is = .
Since net charge enclosed by the area A is σ A, so by Gauss's theorem,
ℎ
=
∈ 0
, . =
∈ 0
=
0
This is the required expression for electric intensity due to the charged
plane conductor.
22 Electrostatics
