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State Gauss' law in electrostatic.. A thin straight infinitely long conducting wire of linear charge densitylambdais enclosed by a cylinder surface of radius 'r' and length ''l' ,its axis coinciding with the length of the wire . Qbtain the expressionn for the electric field, indicating its direction at a point on the surface of the cylinder. |
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Answer» Solution :Consider an infinitely long straight charged wire of linear charge desnsity `lambda`. To find electric field at a POINT P situated at a distance r from the wire by using Guass. law consider a cylinder of length l and radius r as the Gaussian SURFACE. From symmetry consideration electric field at each point of its curved surface is `vecE` and is point outwards NORMALLY . Therefore, electric flux over the curved surface. ` int vecE hatn ds = E 2 PI r l`  On the side face 1 and 2 of the cylinder normal drawn on the surface is perpendicular to electric field E and hence these surfaces do not contribute towards the TOTAL electric flux. `therefore ` Net electric flux over the entire Gaussian surface `phi_E = E. 2pi r l` By Gauss law electric flux `phi_E = 1/(epsi_0)` (charge enclosed)`= (lambda l)/(epsi_0)` Comparing (i) and (ii) , we have `E. 2 pi rl= (lambdal)/(epsi_0)` `impliesE = (lambda)/(2 pi epsi_0 r)` As `E prop 1/r` , hence E -r shown in fig. 
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