# Electric Field Due to An Infinite Line Of Charge derivation

We have already derived and discussed the gauss law of electrostatics. Let  us discuss its applications one by one. Toady I will derive and discuss the electric field due to an Infinite line Of Charge.

Electric Field Due to An Infinite Line Of Charge Or Uniformity Charged Long Wire or Thin Wire:-

An infinite line of charge may be a uniformly charged wire of infinite length or a rod of negligible radius.

Consider an infinite line of charge with a uniform linear charge density λ that is charge per unit length. To find the intensity of electric field at a distance r at point P from the charged line, draw a Gaussian surface around the line in the form of a circular cylinder of radius r and length l, closed at each ends by plane parallel circular caps normal to the axis

All the points on the curved surface of this Gaussian cylinder are at the same normal distance where on the curved surface of the Gaussian cylinder is equal in magnitude and radially outward in direction (Try to make the figure yourself as assignment).

Therefore, for any area element dS taken on the cylindrical surface,the field vector E and area vector dS both are along the same direction.

Thus,

1=E.dS =EdS cos 00=EdS

Hence, the total electric flux through the entire curved cylindrical surface of the Gaussian Cylinder is

Φ1=E(2πrl)

The electric flux through the cylinder flat caps of the Gaussian Cylinder is zero because the electric field at any point on either of the caps is perpendicular to the line charge or to the area vectors on these caps.

Hence, E and dS are at an angle 900with each other.

So that                             φ2=∫E.dS=∫EdS cos 900=0

On both the caps.

Thus, the total flux through the closed Gaussian Cylinder is,

Φ=φ123

Φ=E(2πrl)                                                                (1)

Also the total charge enclosed by the Gaussian surface ,q=λl

Now according to Gauss’s law of electrostatics

Φ=∫E.dS=q/ε0

Or                                  φ=λl/ε0 (λ=q/l)  (2)

By comparing (1) and (2),we get

E2πrl=λl/ε0

E=λ/2πε0r

Therefore ,the field is inversely proportional to r.

It is radially outward in direction if the charge is positive and radially inward if the charge is negative.