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