Gauss's Law & its Applications

These slides describe Gauss's Law in electrostatics and its application to find out electric field due to point charge, uniformly charged spherical shell, and electric potential due to spherical shell.
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1. Electrostatics - II
Dr. Chhagan D. Mungmode
Associate Professor
M. G. College,Armori

2. Gauss’s Law:
The total electric flux passing normally through a closed surface of any shape in an electric field is equal to
time the total charge present within the surface.

If there is no charge present within the surface or charges lies outside the surface, then

Gaussian Surface:
The Gaussian surface is an imaginary closed surface in three-dimensional space draw around the
distribution of charges so that the flux of electric field is calculated.
Gaussian surface for a point charge and charged sphere is a sphere.
1
ϵ
0
ϕ=
∬
E⋅da=
1
ϵ
0
Σq
ϕ=
∬
E⋅da=0

3. Application of Gauss’s Law:
1. The electric field intensity due to point charge by Gauss's Law:
Let us consider, a source point charge particle of +q coulomb is
placed at point O in space. Let take a point P on the electric field of
source point charge particle. To find the electric field intensity at
point P, first put the test charge particle on the point P and draw
a gaussian surface which passes through the point P. After that take a
very small area around the point P. If the distance between the
source charge particle and small area is r then electric flux passing
through the small area

from the figure, the direction between and is parallel to each
other i.e. the angle will be 0. So the above equation can be written as
E
+q
0
dA
dA
dϕ=E⋅dA=EdAcosθ
E dA
dϕ=EdAcos0=EdA

4. The electric flux passing through the entire Gaussian surface and be find by closed integration of above
equation


According to Gauss's Law then above equation can be written as


The above expression is the electric field intensity due to point source charged particle.
ϕ=
∮
EdA=E
∮
dA
ϕ=
q
ε
0
ϕ=
q
ε
0
=E4πr
2

5. Electric Field Intensity due to uniformly charged spherical shell:
Consider a uniformly charged spherical shell of radius R and charge on it +q.
1. Point outside the spherical shell:
Consider a point P outside the shell at a distance r from the centre O of the sphere. Draw a Gaussian sphere of radius r enclosing the
spherical shell so that point P lie on the surface of the Gaussian sphere.

6. Hence it is clear that electric intensity at any point outside the spherical shell is such, as if the entire charge is concentrated at the
centre of the shell.
2. Field at the surface of the shell-

For this we have r = R


If is the charge density on the shell, then


3. Field inside the shell -

If the point P lies inside the spherical shell, then Gaussian surface is a surface of sphere of radius r

As there is no charge inside the spherical shell, Gaussian surface encloses no charge.

That is q=0

Hence the field inside the spherical shell is always zero.

E = 0

Hence the field inside the spherical shell is always zero.