Question

state Gauss law applying this law obtain electric field due to an infinitely long thin uniform charged straight wire

Answer 1

Use this law to derive an expression for the electric field due to an infinitely long straight wire of linear charge density A cm–1. Gauss's law in electrostatics: It states that total electric flux over the closed surface S is times the total charge (q) contained in side S.

Answer 2

Hey,

Here is the answer

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.