Two this parallel threads carry a uniform charge with linear densities `lamnda` and `-lambda` . The distance between the threads is equal to `l`. Find the potential of the electric field and the magnitude of its strength vector at the distance `r gt gt l` at the angle `theta` to the vector 1 (fig).

Explanation

Let `p` be a point, at distance `r gt gt l` and at an angle to `theta` the vector `vec(l)` (Fig).
Thus `vec(E)` at `P = (lambda)/(2pi epsilon_(0)) (vec(r)+(vec(l))/(2))/(|vec(r)+(vec(l))/(2)|^(2))-(lambda)/(2pi epsilon_(0))(vec(r)-(vec(l))/(2))/(|vec(r)-(vec(l))/(2)|^(2))`
`= (lambda)/(2pi epsilon_(0)) [(vec(r) + vec(l)//2)/(r^(2) + (l^(2))/(4) + r l cos theta) - (vec(r) - vec(l)//2)/(r^(2) + (l^(2))/(4) - r l cos theta)]`
`= (lambda)/(2pi epsilon_(0)) ((vec(l))/(r^(2)) - (2l vec(r))/(r^(3)) cos theta)`
Hence `E = |vec(E)| = (lambda l)/(2pi epsilon_(0) r^(2)), r gt gt 1`
Also, `varphi = (lambda)/(2pi epsilon_(0))` In `|vec(r ) + vec(l)//2| - (lambda)/(2pi epsilon_(0))` In `|vec(r ) - vec(l)//2|`
`= (lambda)/(4pi epsilon_(0))` In `(r^(2) + r l cos theta + l^(2)//4)/(r^(2) - r l cos theta + l^(2)//4) = (lambda l cos theta)/(2pi epsilon_(0) r), r gt gt l`