Application of Gauss theorem

There are following applications of Gauss theorem :-
1.Electric field due to a straight long conducting wire :-
Let a straight long conducting wire of length 'L' is present and 'q' charge present on it. A positive point charge +q0 is present at point 'P' at 'r' distance from center of wire.
Imagine a cylindrical Gaussian surface of radius 'r' around it. This surface is divided into three parts - one surface area of cylinder and two circles.
Assume an unit area ds is present at point 'P'

By the definition of electric flux
ɸ  =  ∮Eds Cosθ
ɸ = [∫Eds Cosθ]S1 + [∫Eds Cosθ]S2 + [∫Eds Cosθ]S3
here S1 = upper circle , S2 = lower circle , S3 = surface area of cylinder
for S1
E⊥ds , θ = 90 , Cos 90 = 0
for S2
E⊥ds , θ = 90 , Cos 90 = 0
for S3
E∥ ds , θ = 0 , Cos 0 = 1
so ɸ = [∫Eds ] S3
ɸ = E∫ds
since ∫ds = s
so ɸ = E.s    .....(1)

Now according to Gauss law
ɸ = q / ε0
since in straight wire their is linear charge distribution
λ = q / L
q = λL
ɸ = λL/ε0    ......(2)
equation (1) = (2)
so E.s = λL / ε0
E = λL /sε0
here s = surface area of cylinder = 2πrL
so E = λL / 2πrLε0
E = λ / 2πrε0
E = 2λ / 4πrε0
since 1/4πε0 = k  (coulomb's constant)
E = 2kλ /r    .....(3)
equation (3) is the electric field due to straight long conducting wire.2. Electric field due to non-conducting charged plate :-
Let a non-conducting plate and 'q' charge present on it. A point 'P' is present at 'r' distance from centre of plate.
Now imagine a cylindrical Gaussian surface through plate.
( Here cylindrical surface is present at same distance from plate. If we imagine spherical surface around it then it is not present at equal distance from centre of plate.

This cylindrical surface also divided into three parts - surface area of cylinder(s1) and two circles(s2,s3).
According to definition of electric flux
ɸ = ∮Eds Cosθ
ɸ = [∫Eds Cosθ] for s1 + [∫Eds Cosθ] for s2 + [∫Eds Cosθ] for s3
for s1
E⊥ds , θ = 90 , Cos 90 = 0
for s2
E ∥ ds , θ = 0 , Cos 0 = 1
for s3
E ∥ ds , θ = 0 , Cos 0 = 1
so ɸ = [∫Eds] for s2 + [∫Eds] for s3
since s2 = s3 = s = area of circular surface
ɸ = 2∫E.ds
since ∫ds = s
ɸ = 2Es    .....(1)
from Gauss law
ɸ = q/ε0
here charge surface charge distribution on plate then
σ = q / s
q = σs
ɸ = σs/ε0   .....(2)
equation (1) = (2)
so 2Es = σs/ε0
E = σ/2ε0    .....(3)
It is the electric field intensity due to a non-conducting charged plate.
3. Electric field intensity due to conducting charged plate :-
Let a conducting charged plate and 'q' charge plate on it. A point 'P' is present at 'r' distance from its centre. So imagine a cylindrical Gaussian surface from its centre to point 'P'. It also divided into three parts - one surface area of cylinder (s1) and two cicles (s2, s3).

We know that in a conductor charge present at surface so according to definition of flux
ɸ  = ∮ Eds Cosθ
ɸ = [∫Eds Cosθ] for s1+ [∫Eds Cosθ] for s2 +[∫Eds Cosθ] for s3
for s1
E⊥ds , θ = 90 , Cos 90 = 0
for s2
E ∥ ds , θ = 0 , Cos 0 = 1
for s3
E = 0 inside a conductor electric field is zero.
so ɸ = ∫Eds
ɸ = Es     .....(1)
here s is area of circle
from Gauss law
ɸ = q/ε0
here charge surface charge distribution on plate then
σ = q / s
q = σs
ɸ = σs/ε0   .....(2)
equation (1) = (2)
so Es = σs/ε0
E = σ/ε0    .....(3)
It is the electric field intensity due to a conducting charged plate.

4. Electric field intensity due to charged sphere :-
Let a charged sphere of radius 'R' and 'q' charge present on it. We know that in a conductor charge flow at its surface.

 A point 'P' is present at three different points -
(A) Point present outside the sphere :-
When point 'P' present outside the sphere at 'r' distance. Imagine a spherical gaussian surface of radius 'r' around sphere.

Assume an unit positive charge ds is present at P then from definition of electric flux
ɸ = ∫Eds Cosθ
since E∥ ds ,θ = 0 , Cos0 = 1
ɸ = ∫Eds
ɸ = Es     .....(1)
According to Gauss law
ɸ = q/ε0  .....(2)
equation (1) = (2)
Es = q/ε0
E = q/ε0s
here s = 4πr^2
so E = q/4πε0r^2   .....(3)
It is the electric field intensity due to charged conducting sphere at outer point.
Here an alternate method after equation (1)
According to Gauss law
ɸ = q/ε0
by surface charge distribution
σ = q/s
q = σs  
ɸ  = σs/ε0    ......(4)
again equation (1) = (4)
E.s = σs/εo
E = σ/ε0     .....(5)
It is also the electric field intensity due to charged conducting sphere at outer point but equation (3) is more usable.

(B) Point present at surface of sphere :-
 When point 'P' is present at surface of sphere then r =R

so from equation (3)
E = q/4πε0R^2     .....(6)
It is the electric field intensity due to charged conducting sphere at surface point.

(C) Point present inside the sphere :-
We know that inside a charged conductor electric field is zero. Because charge present at surface of conductor .So inside a conductor charge is absent(q = 0).

since E = q/4πε0r^2
when q = 0
E = 0
So inside a charged sphere electric field intensity is zero means no any electric field present.