Page 22 - Electrostatics-11

© GC Shiba
4) Electric intensity due to linear distribution of charge (a

thin straight charged wire or rod)

Let us consider a thin long straight wire XY X
having linear charge density (λ = charge/length). Let

p be a point at a perpendicular distance r from the line
charge where electric intensity is to be determined. The
direction of electric flux is radially outward. A

cylindrical gaussian surface of radius r, length l and
coaxial with the line charge is drawn as shown in the
figure.
The total electric flux passing through Gaussian

cylinder (hypothetical) according to Gauss theorem is
ℎ
=
∈ 0 Y

=E × A (where E is electric intensity on the

surface of Gaussian cylinder and A
is surface area of Gaussian cylinder)

or, E × A =
0

=
2 0

=
2 0
Hence, the electric field intensity of a charged wire is inversely

1
proportional to the distance from the wire ( = ).

5) Electric intensity due to charged plane sheet

Let us consider a positively
charged plane sheet having surface

charge density σ .( σ =
charge/surface area). We are
interested to find the electric

intensity at point P outside the
charged plane sheet at a distance r.
A cylindrical Gaussian surface PP'

21 Electrostatics