Page 21 - Electrostatics-11
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ℎ
=
∈ 0
=E × A
or, E × A =
0
=
0
=
4 2
0
Hence, charged spheres behave as whole charges are concentrated at its
center.
3) Electric intensity on the surface of a charged conducting
sphere
Consider a charge conducting sphere of radius R
having a charge +Q. We are interested to find the
electric intensity at point P on the surface of a charged
sphere. Let us draw a concentric sphere of radius r
through point P. This sphere is called a Gaussian sphere. r = R P
By symmetry, electric intensity would be the same at all
points on the surface of a Gaussian sphere. Electric flux
from the charged sphere is radially outward as shown in
the figure.
Now, electric flux passing through the gaussian surface
is
ℎ
=
∈ 0
=E × A
or, E × A =
0
=
0
=
4 2
0
Hence, the electric field just outside the surface.
20 Electrostatics
