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.

Gauss theorem

Gauss theorem :- According to it the electric flux passing through a close surface is 1/ε0 multiplier of charge present on that surface.
Let Q charge present on a close surface then it's electric flux
ɸ = Q/ε0
It is only applicable for close surface.

Verification of Gauss theorem :- Let an unit positive charge +q is present at any point and we will find electric field at 'r' distance from it. So imagine a spherical close surface (Gaussian surface) of radius 'r' around that charge

Assume an unit area dA at surface of Gaussian surface
so electric field due to +q at dA
E = kQ/r^2 ....(1)
according to definition of electric flux
dɸ  = EdA Cosθ
here E ∥ dA (∥ means parallel)
so θ = 0 , Cos0 = 1
dɸ = E.dA    ....(2)
putting value of E from equation (1) into (2)
dɸ = (kQ/r^2)dA   ....(3)
Integrating both sides of equation (3)
∫dɸ = ∫(kQ/r^2)dA
since ∫dɸ = ɸ
ɸ  = kQ/r^2∫dA
since ∫dA = A
here kQ / r^2 is constant with respect to A.
so ɸ = kQ/r^2.A
here A is surface area of spherical surface = 4πr^2
& k = 1/(4πε0)
so ɸ = 1/(4πε0)[Q/r^2]4πr^2
ɸ  = Q/ε0   ....(4) Hence Proved.
It is the statement of Gauss Law.

Charge distribution

Charge distribution :- When charge is equally distributed in a conductor then it is called charge distribution.
It has following types :-

1. Linear charge distribution :-  When charge distributed along the length of conductor. In wire, straight rod, long conductor charge distributed along its length.
If is denoted by λ.

It total charge of conductor of  'L' length  is 'Q' then
λ = Q / L
here λ is also known as linear charge density.
Linear charge density :- It means charge per unit length.

2. Surface ( Area ) charge distribution :- When charge distributed along the surface of conductor. In conducting plate charge distributed along its surface or area.
It is denoted by σ .

If total charge of plate of area 'S' is 'Q' then
σ = Q / S = Q / A
here σ is also known as surface charge density.
Surface charge density :- It means charge per unit area.

3.Volume charge distribution :- When charge distributed along the volume of conductor. In solid sphere, solid cylinder, any solid conductor charge distributed along its volume.
It is denoted by ρ.

If total charge of solid sphere of volume 'V' is  'Q' then
ρ = Q / V
here ρ is also known as volume charge density.
Volume charge density :-  It means charge per unit volume.

Electric field lines and flux

Electric field lines :- These are the imaginary lines which shows the electric field in a conductor.
It has following properties :-

1. These are imaginary open curves.
2. They travel from positive (+) to negative (-).
3. These are never intersect each other.
4. A tangent line on these line at a point shows the direction of electric field at that point.
5. Where quantity of electric field lines is high there intensity of electric field also high. Like in a dipole electric field intensity is high at both poles of dipole.
   

Electric flux :- It is the total numbers of electric field lines passing through any close surface.
It is the scalar product of electric field (E) and surface area of close surface(A).
It is denoted by ɸ.
ɸ = EA Cosθ

It surface area is small then electric flux is given by
dɸ = EdA Cosθ
ɸ = ∫dɸ = ∫ EdA Cosθ
here θ is angle formed between electric field and surface area.

Torque and Workdone by a dipole

Torque on a dipole :- When a dipole placed in electric field then torque will generate. It is generate when dipole present at any angle θ.
Let a dipole AB is placed at angle θ in an uniform electric field E.

we know that torque = force*perpendicular distance between line of action
so τ = F* AC     ....(1)
here AC = perpendicular distance
since E = F / q
so F =qE           ....(2)
and from triangle ABC
Sinθ = AC/AB
Sinθ = AC/2a
AC = 2a Sinθ    ....(3)

so from equation (1)
  τ = qE * 2a Sinθ
since 2aq = p
so τ = pE Sinθ   ....(4)
It is the torque due to dipole.

Work done by rotating a dipole in electric field :-
Let a dipole AB is placed at angle θ1 in uniform electric field E. It is rotate from θ1 to θ2 by torque and angular displacement dθ.
we know that work done in rotational motion = torque * angular displacement
dW = τ * dθ
since τ = pE Sinθ
so dW =  pE Sinθ dθ
integrating both sides
∫dW = ∫pE Sinθ dθ
since ∫dW = W
W = pE ∫ Sinθ dθ (here limit of integral is θ1 to θ2)
since ∫ Sin dθ  = -Cosθ
W =  -pE [Cosθ] (limit θ1 to θ2]
W = - pE[Cos θ2 - Cosθ1]
W = pE [ Cos θ1 - Cosθ2]
It is the work done by rotating a dipole in electric field.

if θ1 = 0 & θ2 = θ
W = pE [Cos0 - Cosθ]
since Cos0 = 1
so W = pE [1 - Cosθ]
this work is stored as Potential energy of dipole.
U = W = pE[1 - Cosθ]

Electric field 2

Dipole :- When two equal and opposite charge placed at very small distance then this formation is called dipole and distance between them is 2a.
here 2a = 1Å (One Angstrom ) = 10^-10 meter

Let two charges +q and -q are formed a dipole so

Dipole momentum :- It is the product of one charge of dipole and distance between them. It is denoted by 'p'.
p = 2*q = 2aq
It's unit is coulomb*meter (c*m).

Electric field due to a dipole :- There are two conditions to find electric field intensity due to a dipole -
1. Electric field intensity at axis of dipole :-
Let a dipole AB. A positive charge +q0 is present at point p at r distance from center of dipole O.

distance of p from +q is (r-a) and from -q (r+a) so
electric field intensity at p due to +q
E1 = kq / (r-a)^2          ....(1)
electric field intensity at p due to -q
E2 = kq / (r+a)^2         ....(2)
so total electric field intensity at p due to dipole
E = E1 - E2
E = [kq / (r-a)^2] - [kq / (r+a)^2]
E = kq [1 / (r-a)^2 - 1 / (r+a)^2]
E = kq [{(r+a)^2-(r-a)^2} / {(r-a)^2(r+a)^2}]
E = kq [{(r^2+a^2+2ar) - (r^2+a^2-2ar)} /  {(r-a)(r+a)}^2]
E = kq [(r^2+a^2+2ar - r^2-a^2+2ar) / (r^2 - a^2)^2}
E = kq [4ar / (r^2-a^2)^2}
since r >> a then r^2 >>>> a^2 so neglecting a^2
E = k [2*2aq*r / (r^2)^2]
since 2aq = p (dipole momentum )
E = 2kpr / r^4
E = 2kp / r^3                ....(3)
It is the electric field intensity at axis of dipole.
Here k = 1 / 4πε0  = 9*10^9 Nm^2/c^2

2. Electric field intensity at equator of dipole :-
 Let a dipole AB. A positive charge +q0 is present at r distance from mid point 'o' of dipole at point p.

now connect point A&B to P so AP & BP are hypotenuse so AP = BP = {(r^2+a^2)}^1/2so electric field intensity at p due to +q
E1 = kq / {(r^2+a^2)^1/2}^2
E1 = kq / (r^2+a^2)       ....(1)
electric field intensity at p due to -q
E2 = kq / (r^2+a^2)       ....(2)
here from equations (1) &  (2)
E1 = E2
so they cancel each other means E = 0 which is not possible.
so E1 & E2 are divide into components
E1 Sinθ & E2 Sinθ are equal and opposite so they cancel each other and E1 Cosθ & E2 Cosθ are in same direction so they add.
so resultant electric field intensity is given by
E = E1 Cosθ + E2 Cosθ
E = 2E1 Cosθ               ....(3)  
From triangle PBO
Cosθ = BO / BP
Cosθ = a / (r^2+a^2)^1/2  ....(4)
so again from (3)
E = 2[ {kq / (r^2+a^2)}{a / (r^2+a^2)^1/2}]
E = 2[kqa / (r^2+a^2)^3/2]
since 2aq = p (dipole momentum)
so E = kp / (r^2+a^2)^3/2
since r >>a so r^2 >>>> a^2 neglecting a^2
E = kp / (r^2)^3/2
E = kp / r^3        ....(5)
It is the electric field intensity at equator of dipole.
Here k = 1 / 4πε0  = 9*10^9 Nm^2/c^2

Electric Field 1

Electric field :- It is the area around any charge. When any charge enters in it then that charge feel force ( attraction or repulsion ).

Electric Field Intensity :- The force acting on a unit positive charge is known as electric field intensity.
It is denoted by 'E'.
E = F/q0
here q0 is unit positive charge.
or we can say force per unit charge is known as electric field intensity.
It's unit is newton per coulomb(N/c).

Electric field intensity due to a point charge :-
Let a positive charge +q is present at point 'o' and another charge +q0 present at 'p', r distance from o.

Due to same charge a repulsive force acting
F = kqq0/r^2
when this charge enters into electric field then electric field intensity on it
E = F/q0  ( according to the definition of electric field intensity)
so E = (kqq0/r^2)/q0
    E = kqq0/r^2q0
    E = kq/r^2
here k = 1/4πε0 = 9*10^9 Nm^2/c^2
so E = 1/4πε0(qq0/r^2)

Electric force

To find force between two static charges a law given by scientist Charles-Augustin de Coulomb in 1784. Which is known as Coulomb's law or Coulomb's inverse square law.

Coulomb's law :-
When two point charge present at a distance then force between them is -

1. Directly proportional to product of charge.
2. Inversely proportional to square of distance between them.

Let two point charge q1 and q2 present at a distance r then
F α q1.q2   ........(1)
F α r²         ........(2)
 now from equation (1) and (2)
F α (q1.q2)/r^{2  ......(3)
F = k(q1.q2) /}r2 N
here k is coulomb's constant.
k = 1/4πε₀  = 9*10^9 Nm^2/c^2
here N = Newton ( unit of force)
ε₀  is permittivity of vacuum = 8.85*10^-12 c^2/N.m^2
Permittivity of vacuum:- It is the capability of vacuum to flow the charges.

Note:-

• Coulomb's force is applicable only for static charge.
• It is applicable for point charges.
• It is low range force.

Point charge:- Charges whose size are very smaller then distance between them called point charge.

Electric charge

Electric current:- It is the flow of charge.
Charge :- We know that every material is formed by molecules, molecules by atoms. It every atom neutron,proton,electrons are present. Electron is a moving particle.Which is responsible for charge.

• It is denoted by 'q','Q'.
• It's unit is coulomb (c).

It has two types-
Positive charge- It is due to deficiency of electrons.

Negative charge- It is due to excess of electrons.

NOTE-  In every atom number of electron = number of proton
means atom is neutral (Charge is zero).
When electron emits from it then number of proton > number of electron. Means it is positively charged.
When electron receive by atom then number of proton < number of electron. Means it is negatively charged.
Since electron is negatively charged and proton is positively charged.

Electric current is discovered by German scientist Benjamin Franklin.
According to them when a glass rod is rubbed on silk cloth then electron from rod transfer to silk.due to it rod become positive and silk cloth negative. Like that when ebonite rod is rubbed on fur then electrons from fur transfer to rod. Due to it fur become positive and rod negative.

Force between charges:-
When positive charged rod is freely suspended and another same positive charged rod taken near to it then suspended rod moves upward. Means same charge repel each other.
When positive charged rod is freely suspended and negative charged ebonite rod taken near to it then suspended rod moves downward. Means opposite charge attract each other.

+ and +, - and -repel
+ and - attract

Properties of charge:-
1. Quantization of charge :- The total charge of any conductor is intiger multiplier of number of electron present on it.
q = ne
here q charge of conductor
n = intiger numbers 0,1,2,3,4............n
e = charge of electron 

1.60217662 × 10^-19 coulombs

1.6 × 10^-19 coulombs (approx).

2. Additive :- When multiple charge present on any conductor then total charge of that conductor is algebric sum of all present charges.

q = q1 + q2 + q3 +........ qn

3. Conservation of charge :- Like energy, charge is neither be created nor be destroyed. It only transfer from one conductor to another. Means total charge charge of conductor is conserved.

Energy

Energy :- It is the capacity or ability of work done by a body.

• It is a scalar quantity.
• Its unit is joule and dimension is [M^1L^2T^-2].
• It has following types-

1. Internal Energy - It is the energy of system due to its atom or molecules.
2. Mechanical Energy - It is the sum of kinetic and mechanical energy of system.
   
   
   
   
   
   
3. Sound Energy - It generate the sensitivity in ear drums. It is required to listen voice.
4. Light Energy - It generate the sensitivity in eyes. It is required to watch object.
5. Electrical Energy - It is due to flow of electrons in any conductor.
6. Solar Energy - It is obtain from Sun light.
7. Chemical Energy - It is generate due to chemical reactions.
8. Nuclear Energy - It is due to nuclear reactions like Nuclear fission (Atom Bomb), Nuclear fusion (Hydrogen Bomb).
9. Thermal Energy - It is due to heat (temperature)  of any object.
10. Gravitational Energy - It is the work done by earth to attract any object.

Kinetic Energy - It is due to motion of body (object).
example - Wind mills work on the kinetic energy of air.

Let an object of mass m is in rest(initial velocity u=0). When F force applied on it then it accelerated by a with velocity v and displacement s.
by using third equation of motion
v^2-u^2 = 2as
since u = 0
so v^2 = 2as
a = v^2/2s
since F = ma
so F = mv^2/2s
and work W = F.s
W =  (mv^2/2s).s
W = mv^2/2
This work is stored as kinetic energy.
so K.E. of body = mv^2/2

Potential Energy - It is due to position of object.
example - Chemical energy from a bettery.
                Energy of a body placed at a height from earth surface.

Let an object of mass m is placed at h height from earth.
force F = mg ( here a = g gravitational acceleration )
work W = F.h
so W = mgh
This work is stored as potential energy.
P.E. of body = mgh