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


ayesha.fatima.kolkata 9th Nov, 2020
Answer (1)
apeksha150598 9th Nov, 2020
Hello,


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.

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