Saved Bookmarks
| 1. |
An infinitely thin straight wire has uniform linear charge density lamda . Obtain the expression. For the electric field (E) at a point lying at a distance x from the wire, using Gauss' law. (b) Show graphically the variation of this electric field E as a function of distance x from thewire . |
|
Answer» Solution :Consider an infinitely long straight charged wire of linear charge DENSITY `lamda`. To find electric field at a point P situated at a DISTANCE r from the wire by using Gauss. LAW consider a cylinder of length l and radiusr as the Gaussian surface. From symmetry consideration electric field at eac point of its curved surface is `vecE` and is POINTED outwards normally. Therefore , electric flux over the curved surface. `=intvecE.hatnds = E 2pirl` On the side faces 1 and 2 of the cylinder normal drawn on the surface is perpendicular to electric field E and hence these surface do not contribute towards the total electric flux. `:.` Net electric flux over the entire Gaussian surface `phi_E=E.2pirl ""....(i)` By Gauss law electric flux `phi_E=1/in_0` (charge enclosed ) `(lamdal)/in_0 ""...(ii)` Comparing (i) and (ii) , we have `E.2pirl=(lamdal)/in_0` `implies E=lamda/(2piin_0r) and vecE=lamda/(2piin_0r).hatr` (B) The E - r graph is as shown here.  
|
|