PROOF OF GAUSS'S THEOREM

Let a charge of q coulombs be placed inside a hollow enclosure as shown in figure:

Let a small surface area dA of this enclosure subtend a solid angle d𝜔 at q and be at a distance d from the charge q.

The flux density D' (considered in the direction of the radius vector d) will be:

=q4πd2 Coulombs/ sq.meter  or  C/m2

and its component D, normal to the surface will be

=q4πd2cos 𝜃  C/m2

where 𝜃 is the angle between D and D'.

Now, the flux d𝜓 crossing normally the surface of area dA = D ⤬ area



or        dψ=q4πd2cos θ⋅dA    coulombs

By definition,

solid angle=surface arearadius

∴    dω=dA  cos θd2


∴    dψ=q4π⋅dω


or the total flux 𝜓 , crossing the total surface of the imagined enclosure is

∫dψ=q4π ∫dω

ψ=q4π×4π


or      ψ=q coulombs


Thus the total electric flux due to a charge at any point in an enclosure is equal to the charge enclosed in that enclosure.

If there are a number of charges +𝑞₁ , +𝑞₂ , -𝑞₃  and  +𝑞₄ placed in an enclosure, the total electric flux due to all the charges will be

𝜓 = 𝑞₁ + 𝑞₂ - 𝑞₃ + 𝑞₄

or     𝜓 = 𝛴𝑞

GAUSS'S THEOREM

Gauss's Theorem: 

The total flux normally traversing a surface completely surrounding a charge q is q coulomb.

If there are a number of charges inside the surface, the total flux is the algebric sum of the total charges.

𝜓 = 𝛴q

𝜓 = (q₁ + q₂ + q₃ + q₄ + .........  ) coulombs

All positive charges being taken together as positive and all negative charges as negative.

Thus, in the arrangement as shown in figure:

the value of the flux over the surface is given by

𝜓 = (q₁ - q₂ )  coulombs

since (q₁ - q₂ ) is the net charge inside.

MATHEMATICAL TREATMENT OF COULOMB'S LAW

Consider two charges of magnitude 𝑞₁ and 𝑞₂ , assumed to be concentrated as point sources at a distance d from each other.

According to the second law of electrostatics, the force F between them may be expressed mathematically as:

F∝q1×qkmfafaie2

F∝1d2

or        F∝q1q2d2

F=k⋅q1q2d2N

                                                      ..................................(1)

where, N is Newton and k is a Constant depending on the nature of the medium between the charges. 

The value of k has been found experimentally, 

k=14πϵ0ϵr

                                                     ....................................(2)

where ,

ϵr

ϵ0

ϵ0

ϵ0

ϵ0

𝜖ᵣ is the relative permittivity of the medium intervening the charges

𝜖₀ is the permittivity of the evacuated free space

The value of 𝜖₀ in rationalised  M.K.S. unit is 8.854×10⁻¹² farads per meter.

When d is measured in meters, the constant 4𝜋 , the proportionality factor for the rationalised M.K.S. units.

Substituting the value of k from Eq. (2) in Eq. (1) we get,

F=q1q24πϵ0ϵrd2N

where,

𝑞₁ and 𝑞₂ are measured in coulombs

d in meters

𝜖 in farad / meter

F in newtons

For air         𝜖ᵣ = 1

∴    F=q1q24πϵ0d2N

when     

𝑞₁ = 𝑞₂ = 1 coulomb

d = 1 meter

𝜖₀ =  8.854×10⁻¹² farad / meter

F=1×14π×8.854×10−12×12

F=8.997×109