Question

Using Gauss’s law derive an expression for the electric field intensity at any point near a uniformly charged thin wire of charge/length λC/m.

Answer 1

Gauss Theorem : The net outward electric flux through a closed surface is equal to 1/ε₀  times the net charge enclosed within the surface i.e.,

Electric field due to infinitely long, thin and uniformly charged straight wire: Consider an infinitely long line charge having linear charge density l coulomb meter^-1 (linear charge density means charge per unit length). To find the electric field strength at a distance r, we consider a cylindrical Gaussian surface of radius r and length l coaxial with line charge. The cylindrical Gaussian surface may be divided into three parts: 

(i) Curved surface S₁  (ii) Flat surface S₂  and (iii) Flat surface S₃. By symmetry the electric field has the same magnitude E at each point of curved surface S₁  and is directed radially outward. We consider small elements of surfaces S₁S₂ and S₃. The surface element vector d vector S₁ is directed along the direction of electric field (i. e., angle between vector E and d vector S₁ is zero); the elements d vector S₂ and d vector S₃ are directed perpendicular to field vector E  (i. e., angle between d vector S₂ and vector E is 90° and so also angle between d vector S₃ and vector E). 

Electric Flux through the cylindrical surface

As λ is charge per unit length and length of cylinder is l, therefore, charge enclosed by assumed surface =(λl)

Thus, the electric field strength due to a line charge is inversely proportional to r.