Obtain the formula for the electric field due to a along thin wire of uniform linear charge density `lambda ` without using Gauss's law.

Explanation

Consider a long thin wire of length L and having uniform charge density `lambda.` Let P be a point situated at a normal distance r from the wire. Let there be a small 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 -y and
` 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)/( 2 pi in _0r) `
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