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Consider an infinitely long thin wire with uniform linear charge density λ. That is, the charge per unit length is λ and this assumed to be the same for all points on the wire. From considerations of symmetry, we can say that the electric field is radially directed. The magnitude of the electric field is same at all those points which are equidistant from the wire. Let us find the electric field intensity at any point P at a perpendicular distance r from the axis of the wire. A cylinder of radius r and arbitrary length l coaxial with the axis of the wire is chosen as the Gaussian surface.Since the Gaussian surface is a closed surface, this cylinder is closed at each end by plane caps normal to the axis. The magnitude of is constant at all points on the surface of the cylinder, as those points are equidistant for the wire and is normal to the surface (outward).

The total electric flux through the Gaussian surface is the algebraic sum of the flux due to two end caps and the cylindrical part of the Gaussian surface.
For all points on the end caps, the electric field
is perpendicular to the area element vector . Therefore, the electric flux through the caps is zero.
The electric flux through the curved surface of the cylinder is given by .
As already discussed, angle between and is 0°, i.e., directed radically outward from the axis (∵ +vely charged)
∴
As is same at all points on the curved surface,

Therefore, the total electric flux, the algebraic sum of the flux through the two end caps and curved surface is
φ E = 0 + 0 + E(2πrl)
Now according to Gauss's law,
the total electric flux through the Gaussian surface is equal to times the net charge enclosed in the surface.
In this case, net charge Q = λl, for cylindrical surface
So, the electric field of the charged wire is inversely proportional to the distance from the charged wire.
The direction of electric field is radially outward if the line charge is positive and radially inward if it is negative.