Question

State Guass’s law in electrostatics. Use this law to derive an expression for the electric field due to an infinitely long straight wire of linear charge density λCm^-1.

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 λ coulomb metre^-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 S₁, S₂ 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.