A very long straight uniformly charged thread carries a charge `lambda` per unit length. Find the magnitude and direaction of the electric field strength at point which is at a distance `y` from the thread and lies on the perpendicular passing through one of the thread's ends.

Explanation

The problem is reduced to finding `E_(x)` and `E_(y)` viz, the projections of `vec(E)` in fig. where is assumed that `lambda gt 0`.
Let us start with `E_(x)`. The contribution to `E_()` from the charge element of the segment `dx` is
`dE_(x) = (1)/(4pi epsilon_(0)) (lambda dx)/(r^(2)) sin alpha` (1)
Let us expression to the form convenlent for intergation. In our case, `dx = r d alpha//cos alpha, r = y//cos alpha`. Then
`dE_(x) = (lambda)/(4pi epsilon_(0) y) sin alpha d alpha`
Intergating this expressions over `alpha` between 0 and `pi//2`, we find
`E_(x) = lambda//4 pi epsilon_(0) y`.
In order to find the projection `E_(y)` is is sufficient to recall that `dE_(y)` differs from `dE_(x)` in that sin `alpha` in (1) in simply replaced by cos `aplha`.
This gives
`dE_(y) = (lambda cos alpha d alpha)//4pi epsilon_(0) y` and `E_(y) = lambda//4pi epsilon_(0) y`.
We have obtained an intersting result :
`E_(x) = E_(y)` independentaly of `y`.
`i.e., vec(E)` is oriented at the angle of `45^(@)` to the rod. The modulus of `vec(E)` is
`E = sqrt(E_(s)^(2) + E_(y)^(2)) = lambda sqrt(2)//4pi epsilon_(0) y`.