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State Gauss theorem in electrostatics. Apply this theorem to obtain the expression for the electric field at a point due to an infinitely long, thin, uniformly charged straight wire of linear charge density λ C m–1 .

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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 \(\vec{dS_1}\) is directed along the direction of electric field (i.e., angle between \(\vec E\) and \(\vec{dS_1}\) is zero); the \(\vec{dS_1}\) and elements \(\vec{dS_2}\) and \(\vec{dS_3}\) are directed perpendicular to field vector \(\vec E\) (i.e., angle between \(\vec{dS_2}\) and \(\vec E\) is 90° and so also angle between \(\vec{dS_3}\) and \(\vec E\)).

Electric Flux through the cylindrical surface

= E∮dS1 (since electric field E is the same at each point of curved surface)

= E 2π rl (since area of curved surface  = 2π rl)

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

\(\therefore\) By Guass's theorem

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

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