Question

What is Gauss theorem ? Derive Expression and Explain it .

Answer 1

Statement. Total electric flux through any closed surface, is equal to 1/ε times the total charge enclosed by the surface.

ФE=∫E.dS=q/ε

Where ε is the permittivity of the medium (for free space ε=ε0),

So                    ФE=∫E.dS=q/ε0

Where ∫E.dS is surface integral over the closed surface and q is the charge present in the closed surface S

The imaginary closed surface is called Gaussian Surface. If the surface encloses a continuous charge distribution then q is replaced by the intergal ∫ρdV,where ρ is the volume charge density.

Proof of gauss law of electrostatics:

Consider a source producing the electric field E is a point charge +q situated at a point O inside a volume enclosed by an arbitrary closed surface S. let us consider a small area element ds around a point P on the surface where the electric field produced by the charge +q is E. if E is along OP and area vector dS is along the outward drawn normal to the area element dS, (Try to make the diagram yourself),

Then the electric flux dФ through the area element dS is given by

dФ=E.dS=EdS cosθ

Where θ is the angle between E and dS and it is zero degree, therefore

dФ = EdS    (1)

Since the source producing E at dS is a point charge +q at O, therefore

E=1/4πε0 q/(OP)2=1/4πε0q/r2 (OP=r)

Substituting this value of E in equation (1),we get

dФ=E.dS=q/4πε0dS/r2 (2)

Hence , total electric flux Ф through the entire closed surface S would be

∫dФ=q/4πε0 r2∫dS                               (3)

But ∫dS     = 4πr2

Therefore, equation(3) becomes

Ф=∫E.dS= q4πε0 r2 /4πε0 r2

Or                                  Ф=∫E.dS=  q/ε0 (4)

Equation (4) represents Gauss law for electrostatics for a single point charge. If the source producing the electric field has more than one point charges such as +q1,+q2,+q3,-q4.-q5,-q6………..etc,then the total flux due to all of them would be the algebric sum of all the fluxes as,

Ф=∫E.dS=1/ε0(q1+q2+q3-q4-q5-q6…….)

Or                               Ф=∫E.dS=∑q/ε0

Hence proved a charge outside the Gaussian surface would contribute nothing to the electric flux

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