Question

derive the expression forthe electric field intensity at a point r distance away from an infinitely thin long straight line of charge
​

Answer 1

Answer:

Consider an infinitely long thin straight wive with uniform linear charge q density λ. Let P be a point at ⊥r distance r from the wire. To calculate the E.F → E E→ at P, imagine a cylindrical Gaussian surface. ∴ The surface area of the curved part S = 2πrl Total charge enclosed by the Gaussian surface q = λl Electric flix through the end Surfaces of the cylinder is Φ = 0 Electric flux through the curved Surfaces of the cylinder is Φ2 = Ecosθ.s Φ2 = E x 1 x 2πrl The total electric flux Φ = Φ1 + Φ2 Φ = 0 + E2πrl, Φ2 = 2πrlE ………(1) A/C to Gauss law, from (1) and (2) Read more on Sarthaks.com - https://www.sarthaks.com/622084/derive-expression-electric-field-point-infinitely-thin-charged-straight-wire-using-gauss

Answer 2

Answer:

Here is your answer with the attachments!

Explanation:

Electric field due to an infinitely long charged wire:  Consider an infinitely long straight wire having uniform linear charge density λ. Let P be a point located at a perpendicular distance r from the wire. The electric field at the point P can be found using Gauss law. We choose two small charge elements A1 and A1 on the wire which are at equal distances from the point P. The resultant electric field due to these two charge elements points radially away from the charged wire and the magnitude of electric field is same at all points on the circle of radius r. From this property, we can infer that the charged wire possesses a cylindrical symmetry. (a)

see picture 1

Electric field due to infinite  long charged wire Let us choose a cylindrical Gaussian surface of radius r and length L. The total electric flux in this closed surface is

see picture 2

It is seen that for the curved surface, is parallel to → A A→and  → E E→ .d = EdA. For the top and  → E E→ bottom surface, is perpendicular to and  → A A→   → E E→.d → A A→ = 0 Substituting these values in the equation (2) and applying Gauss law

see picture 3

Since the magnitude of the electric field for the entire curved surface is constant, E is taken out of the integration and Qencl is given by Qencl = λL.

see picture 4