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Derive an expression for the electric field at a point due to an infinitely long thin charged straight wire using Gauss Law. |
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Answer» Solution :Statement : The total electric FLUX through a closed surface in free space is equal to `1/epsilon_0` times the NET charge enclosed by the surface. i.e., `phi=q/epsilon_0`  In the above fig. AB is the infinitely long wire E is the electric field P is a point at a distance R from the wire .r. is the radius of Gaussian cylinder .l. is the length of the Gaussian cylinder Let .q. be the charge enclosed by the Gaussian cylinder . Let `.lambda.` be the linear charge density on the wire . Flux through the end faces is zero because there are no components of electric field ALONG the NORMAL to the end faces. `phi`=flux through curved surface `phi`=E x area of curved surface (`phi`=E x area) `phi=E xx 2pirl to` (1) From Gauss.s theorem `phi=q/epsilon_0 to` (2) But `lambda=q/l rArr q=lambdal` (2)`rArr phi=(lambdal)/epsilon_0 to` (3) On comparing (1) and (3) , we get `Exx2pirl=(lambdal)/epsilon_0` `E=lambda/(2piepsilon_0)` `E=1/(2piepsilon_0) lambda/r` The direction of .E. is perpendicular to the wire and directed away from the wire . |
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