Question

Using Gauss's law, calculate the electric field at a point distance s from a long wire bearing uniform charge density. i need help please.​

Answer 1

Answer:

E = 2k  $\frac{\lambda}{ r}$

Explanation:

Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.

We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.

            Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀

tells us that the linear charge density is

            λ = q_ {int} /l

            q_ {int} = l λ

we substitute

            E A = l λ /ε₀

is area of ​​cylinder is

           A = 2π r l

we substitute

            E = $\frac{ l \ \lambda}{ \epsilon_o \ 2\pi \ r \ l }$

             E = $\frac{\lambda}{ 2\pi \epsilon_o \ r}$

the amount

            k = 1 / 4πε₀

            E = 2k  $\frac{\lambda}{ r}$