The problem is reduced to finding Ex and Ey viz. the projections of vector E in Fig, where it is assumed that λ > 0.
Let us start Ex. The contribution to Ex from the charge element of the segment dx is
Let us reduce this expression to the form convenient for integration. In our case, dx = rdα/cos α, r = y/cos α. Then
Integrating this expression over α between 0 and π/2, we find
In order to find the projection E it is sufficient to recall that differs from dE in that sin α in (1) is simply replaced by cos α.
This gives
We have obtained an interesting result :
Ex = Ey in dependently of y,
i.e. vector E oriented at the angle of 45° to the rod. The modulus of is