With the help of gauss theorem, find out the electrical intensity at any nearby point due to uniformly charged thin and long wire.

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According to Gauss theorem, the total electric flux () through any closed surface (S) in free space is equal to1/0 times the total electric charge (q) enclosed by the surface, i.e,=SE.dS=q/0

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, a cylindrical Gaussian surface of radius r and length l coaxial with line charge is considered.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).
As is charge per unit length and length of cylinder is l, therefore, charge enclosed by assumed surface =(l). Thus, the electric field strength due to a line charge is inversely proportional to r.