QS Study

Field due to an Infinite Long Straight Charged Wire (Application of Gauss’s Law):

Consider a uniformly charged wire of infinite length having a constant linear charge density λ (charge per unit length). Let P be a point at a distance r from the wire (Figure) and E be the electric field at the point P. A cylinder of length l, radius r, closed at each end by plane caps normal to the axis is chosen as the Gaussian surface. Consider a very small area ds on the Gaussian surface.


Fig: Infinitely long straight charged wire

By symmetry, the magnitude of the electric field will be the same at all points on the curved surface of the cylinder and directed radially outward. E and ds are along the same direction.

The electric flux (φ) through curved surface = ϐ E ds cos θ

φ = ϐ E ds  [we know, θ = 0; cosθ = 1]

= E (2πrl)

(The surface area of the curved part is 2πrl)

Since E and dsare right angles to each other, the electric flux through the plane caps = 0

so, Total flux through the Gaussian surface, φ = E. (2πrl)

The net charge enclosed by Gaussian surface is, q = λl

By Gauss’s law, E (2πrl) = λl0

or, E = λ / 2πε0r

The direction of electric field E is radially outward if line charge is positive and inward if the line charge is negative.