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Obtain the formula for the electric field due to a along thin wire of uniform linear charge densitylambdawithout using Gauss's law. |
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Answer» Solution :Consider a long thin wire of length Land having uniform charge DENSITY `lambda.` Let P be a point situated at a normal distance r from the wire. Let there be asmall length elements situated at a distance y and having length DY as SHOWN in According to Coulomb.s law electric field due to this element at point P has a magnitude. `dE=( 1)/( 4pi in_0).(dq)/( (r^(2) +y^(2))) =(1)/( 4piin _0) .( lambda dy)/( (r^(2) +y^(2)) ` `oversetto (dE)` may be resolved into 2 components :(i)a COMPONENT`dE_x`in a DIRECTION normal to wire, and (ii)component `dE_y`in a direction parallwl to the wire. Obviously if we find electric field due to whole wire ,then `sum dE_y= 0 `because for every at +y there is corresponding element at -yand ` therefore ` Electric field due to whole conductor at point? ` E =int dE_x = int dE sin theta =(lambda )/( 4pi in _0) int (dy sin theta )/( (r^(2) +y^(2))` But` y= root theta , `hence dy ` =-r cosec ^(2)theta d theta` `therefore"" E= (lambda )/( 4pi in _0) underset( theta =pi ) oversetto ( theta =0 )int ((r-cosec ^(2) theta_x d theta ) sin theta)/( (r^(2) +r^(2)cot ^(2)theta) )=(lambda )/( 4pi in _0)underset ( pi ) overset (0) int-(1)/(r)sin theta d theta ` ` "" = (lambda )/( 4pi in _0)[cos theta ]underset pi overset 0=(lambda )/(4pi in _0)[cos 0^(@)-cos pi ] ` `=(lambda)/( 4 pi in _0)[1-(-1) ]= (lambda)/(2pi in _0r) ` ` (##U_LIK_SP_PHY_XII_C01_E02_006_S01.png" width="80%"> |
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