Question

Find the state and prove the gauss theorem in electrostatic

Answer 1

Statement of Gauss"s Theorem : The net-outward normal electric flux through any closed surface of any shape is equal to  

1/ε0 times the total charge contained within that surface , i.e.,

over the whole of the closed surface, q is the algebraic sum of all the charges (i.e., net charge in coulombs) enclosed by surface S.

Proof of Gauss"s Theorem :

Let a point charge +q coulomb be placed at O within the  closed surface. Let E be the electric field strength at P. Let

OP= r and the permittivity of free  space or vaccuum be ε0.

Answer 2

Answer:

Explanation:

According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

∮E⃗ .d⃗ s=1∈0q .

According to Gauss Law,

Φ = → E.d → A

Φ = Φcurved + Φtop + Φbottom

Φ = → E . d → A = ∫E . dA cos 0 + ∫E . dA cos 90° + ∫E . dA cos 90°

Φ = ∫E . dA × 1

Due to radial symmetry, the curved surface is equidistant from the line of charge and the electric field in the surface has a constant magnitude throughout.

Φ = ∫E . dA = E ∫dA = E . 2πrl

The net charge enclosed by the surface is:

qnet = λ.l

Using Gauss theorem,

Φ = E × 2πrl = qnet/ε0 = λl/ε0

E × 2πrl = λl/ε0

E = λ/2πrε0