Gauss’s Law

Gauss law may stated as

“The flux of electric field E throw any closed surface is 1/ℰ₀ times of total charge inclosed by the surface & if the surface does not inclose the charge the flux is equal to zero. i.e.

⨜E.dS=q/ℰ₀ (If q is inside the surface)

⨜E.dS=0        (If q is outside the surface)

Proof:-

Consider a closed surface as shown in the figure.

Consider the point charge q at O drawing a small conical solid d𝞈. We find that it will cut the surface an odd number of times.

Consider in this case the surfaces dS₁,dS₂ & dS₃ obviously the direction of normal shows that the contribution of surface dS₂ is opposite to that of dS₁ & dS₃.Therefore total flux

𝞥_E=∲E.dS

      =∲E₁.dS₁+∲E₂.dS₂+∲E₃.dS₃

Where E₁,E₂ & E₃ are the respective field on the surfaces dS₁,dS₂ & dS₃ at a distance r₁,r₂ & r₃ respectively.Hence

𝞥=∲1/4𝛑ℰ₀ q/r₁² dS₁ cos𝛉₁-∲1/4𝛑ℰ₀ q/r₂² dS₂

        cos𝛉₂+∲1/4𝛑ℰ₀ q/r₃² dS₃ cos𝛉₃

   =q/4𝛑ℰ₀[∲dS₁ cos𝛉₁/r₁-∲dS₂ cos𝛉₂/r₂+∲dS₃ cos𝛉₃/r₃]

The quantity ds cos𝛉/r² is called the solid angle d𝞈 which is same for the all surfaces.Therefore

                             dS cos𝛉/r=d𝞈

𝞥_E=q/4𝛑ℰ₀[ ∲d𝞈-∲d𝞈+∲d𝞈]

𝞥_E=q/4𝛑ℰ₀⨯4𝛑

𝞥_E=q/ℰ₀

In the second case when the charge lies outside the surface the small solid angle d𝞈 will intersect the surface to an even number of times then proceeding as above we have

∲E.dS=q/4𝛑ℰ₀[∲d𝞈-∲d𝞈+∲d𝞈-∲d𝞈]

=0

i.e.              ∲E.dS=0

This proved the Gauss’s law.