Question

Apply Gauss' theorem to calculate the electric field due to an infinitely long charged straight wire.

Answer 1

Amount of Electric flux passing through a closed surface or the surface integral of Electric field intensity is equal to the ratio of the Charge enclosed by the surface and the Permittivity of the free space.

ϕ=Qenclosedε0 ⇒∫E.dS=Qenclosedε0

Electric Field due to a Long Straight Charged Wire:

Consider a long straight wire charge by a linear charge density λ as shown in the figure below.



To find the electric field at a distance from the wire, consider a gaussian surface as shown in the figure.

Let the electric field at the surface is E. Then,

From the definition of gauss law,

∫E.dS=Qenclosedε0⇒E∫dS=Qε0⇒E(2πRL)=Qε0 ⇒ E=12πε0R(QL)⇒ E=λ2πε0R

✌️________✌️

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