Obtain the formula for the electric field due to a long thin wire of uniform linear charge density E without using Gauss's law. (Hint: Use Coulomb's law directly and evaluate the necessary integral.)

Explanation

Consider infinitely extended line of charge with constant linear charge density `lambda` . Let us determine its electric field at perpendicular distance r say at point P as shown in the figure. For this consider a charge element dq at distance y from central point O. Here, `dq = lambdady`
`(therefore lambda = (dq)/(dy))`
Magnitude of Coulombian force exerted by above charge element dq on test charge `q_(0)` placed at point P is,
`dF = k((dq)q_(0))/(CP)^(2) = (k(dq)q_(0))/(y^(2) + r^(2))`........(1)
Magnitude of electric field of charge element dq at point P is,
`dE = (dF)/(q_(0)) = (kdq)/(y^(2) + r^(2))`.........(2)
Similary magnitude of electric field of charge element dq (selected on opposite side of central point O) at point P is also, `dE = (kdq)/(y^(2) + r^(2))`
When we resolve above two electric fields into two mutually perpendicular components at point p we find that dEsinG components cancel each other (because of equal magnitudes and opposite directions). Hence, we get electric field at point P only due to `dEcostheta` components which are in same directions and so we can make their scalar addition to find resultant electric field E at point P Thus,
`E = int dEcostheta`
`int(kdq)/(y^(2) + r^(2)).r/sqrt(y^(2) + r^(2))`
From eqn. (2) and from `DeltaCOP`
`therefore E = intklambda dy xx r/(y^(2) + r^(2))^(3//2)`...........(3)
In right angled `triangleCOP`
`tantheta = y/r rArr y = r tan theta`.......(4)
For a given point P value of r is constant and so from above equation,
`(dy)/(d theta) = r(d)/(d theta) (tan theta)`
`therefore (dy)/(d theta) = r sec^(2)theta`
`therefore dy = rsec^(2)theta d theta`........(5)
Now, `(y^(2) + r^(2))^(3//2) = (r^(2)tan^(2)theta + r^(2))^(3//2)` (From equation (4))
`=(r^(2) sec^(2)theta)^(3//2)`
`=r^(2)sec^(3)theta`.........(6)
From equation (4)
`tan theta = y/r`
When `y =-infty, tan theta = -infty rArr theta = -pi/2`
`y=+infty, tan theta =+ infty rArr theta = + pi/2` ............(7)
From eqn (3), (5),(6) and (7),
`E = int_(-pi//2)^(+pi//2) klambda(r sec^(2)theta d theta) xx r/(r^(3) sec^(3)theta)`
`therefore E=(klambda)/r {sin(pi/2)- sin(-pi/2)}`
`therefore E =(2klambda)/r`
`therefore E = (lambda)/(2pi epsilon_(0)r) (therefore k=1/(4pi epsilon_(0)))`