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Given the expression for electric field intensity at a point due to a thin infinitely long straight wire. Give the meaning the of symbols used. |
Answer» Solution :ELECTRIC field intensity due to a thin long wire : A line charge is in the form of a thin charged rod with unform linear charge density `lambda` (charge PER unit length), To determine the electric field intensity `vecE` at any point P at a perpendicular distance r from the rod let a right CIRCULAR closed cylinder of radius r and lenghtl with the infinitely long line of charge as its axis (as SHOWN in figure). The magnitudeof `vecE` at every point on the curved surface of the cylinser is the same as all such points are at the same distance from the line charge . Also `vecE` and unit vector `hatn` normal to curved surface arein the same direction so `theta=0^(@)`.`THEREFORE`Contribution of curved surface of cylinder towards electric flux , `oint_(s)vecE*vecds=ointvecE*hatnds` `=Eoint_(s)ds=E(2pirl)`where`(2ppiel)` is area of the curved surface of the cylinder. On the ends of cylinder , the angle between electric field `veceandhatn` is `90^(@)` . So it will not contribute to electric flux on cylinder. `rArrointvecE*vecds=Exx2pirl` Charge enclosed in the cylinder i.e= linear charge density `xx` lenght `q=lambdal` According to Gauss.s theorem `ointvecE*vecds=(q)/(epsilon_(0))` `rArr(2pirl)xxE=(lambdal)/(epsilon_(0))` `E=(lambdal)/(2piepsilon_(0)rl)` `thereforeEprop(l)/(r)` (As `(lambda)/(2piepsilon_(0)r)=`constant) |
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