Find the electric flux crossing the wire frame ABCD of length l, width b, and center at a distance OP = d from an infinite line of charge with linear charge density `lambda`. Consider that the plane of the frame is perpendicular to the line OP

Answer

`(lambda l)/( pi epsilon_(0)) tan^(-1) ((b)/( 2 d))`

Explanation

The flux passing through the strip of area dA is
`dphi = EdA cos theta`
`= lambda/(2piepsilon_0sqrt(d^2+x^2)) (ldx) d/(sqrt(d^2+x^2))`
`= (lambdadl)/(2piepsilon_0) (dx)/(d^2+x^2)`
`:. Phi = int dphi = (lambdadl)/(2piepsilon_0) int_(-b//2)^(b//2) (dx)/(d^2+x^2)`
`= (lambdadl)/(2piepsilon_0) 1/d [ tan^(-1)x/d]_(-b//2)^(b//2)`
`= (lambdal)/(2piepsilon_0) xx 2 tan^(-1) (b/(2d))`
` = (lambdal)/(piepsilon_0) tan^(-1) (b/(2d))`
.
Alternative method
The number of lines passing through a surface is proportional to the
flux passing through the surface. The number of lines passing through
the plane surface ABCD is equal to the number of lines passing
through the curved surface. The amount of flux passing through the
plane surface is equal to the line flux passing through the curved
surface.
From figure , `tan (theta/2) = (b/(2d)) or theta = 2tan^(-1) (b/(2d))`
`2pi` angle corresponds to flux `lambdal//epsilon_0`.
Hence, `theta` angle corresponds to the flux, i.e.,
`phi = theta/(2pi) ((lambdal)/(epsilon_0)) = (lambdal)/(piepsilon_0) tan^(-1) (b/(2d))`
.