## Gauss Law

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# Gauss Law

**Statement.**The total flux of the electric field E over any closed surface is equal to 1/ε0 times the total net charge enclosed by the surface.

φ = Φ E. dS = q/ ε

_{0}

**Explanation.**This law relates the flux through any closed surface and the net charge enclosed within the surface. Here q is the net charge inside the closed surface. This closed hypothetical surface is called Gaussian surface. Gauss’ law tells us the flux of E through a closed surface S depends only on the value of the net charge inside the surface and not on the location of the charges. Charges Outside the surface and not on the location of the charges. Charges outside the surface will not contribute to φ.

## Proof.

### i) For a charge inside the closed surface

Consider a single point charge +q located at a point O inside a closed surface S. Let dS be a small are element at a distance r from q.E = 1/4π ε

_{0}q/r

^{2}r

The flux through the area dS is given by

d φ = E. dS

= E dS cos θ

= (1/4πε

_{0}q/r

^{2})dS cos θ

= q/4πε

_{0}(dS cos θ/r

^{2})

But dS cos θ/r

^{2}= dΩ = Solid angle subtended by the area dS a O.

. : d φ = q/4πε

_{0}dΩ

The total flux through the entire closed surface S is given by

φ = Φ d φ = q/4πε

_{0}Φ dΩ = q/4πε

_{0}X 4π

(. : the total solid angle subtended by S at O is 4π, i.e.., Φ dΩ = 4π).

. : φ = q/ ε

_{0 }

Gauss’ law holds even if there are a number of charges q1, q2,……qn enclosed by a surface S. The principle of superposition that the electric field due to a number of charges is the vector sum of their individual fields.

. : φ = Φ E. dS = 1/ ε

_{0}∑

_{(i=1)}nqi = Q/ ε

_{0}

Here, Q is the total charge inside the surface.

### (ii) For a charge outside the closed surface

Consider a point charge +q situated at O outside the closed surface. Let an elementary cone from O with small solid angle dΩ cut the closed surface at two elements of area dS1 and dS2. Magnitude of flux through dS1 and dS2 are equal. Flux through dS1 is an inward flux. Flux through dS2 is an outward flux. Therefore,Total flux through dS1 and dS2 } = -q/4πε0 dΩ + q/4πε0 dΩ = 0

The entire closed surface can be considered to be made pairs of elements like dS1 and dS2. Thus the total flux, due to a charge outside, is zero.

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