A rod of length l, has a uniform positive charge per unit length and a total charge Q. calculate the electric foeld at a point P located along the long axis of the rod and at a distance a from one end

Explanation

The field `dvec(E )` at P due to each segment of charge on the rod is in the negative x direction because every segment carries a positive charge.
Because the rod is continous we are evaluating the field due to a continuous charge distribution rather than a group of individual charges. Because every segment of the negative x direction, the sum of their contributions can be handled without the need to add vectors.
Let us assume the rod is lying along the x-axis , dx is the length of one small segment and dq is the charge on that segment. Beacuse the rod has a charge per unt length 1, the charge dq on the small segment is `dq=lambda dx`. The magnitude of the electric field at P due to one segment of the rod having a charge dq is
`dE=k_(e)(dq)/(x^(2))=k_(e)(lambda dx)/(x^(2))`
The total field at P is
`E=int_(a)^(l+a)K_(e )lambda(dx)/(x^(2))`
If `K_(e )` and `lambda = Q//l` are constants and can be removed from the integral, them
`E=int_(a)^(l+a) k_(e)lambda (dx)/(x^(2))[-(1)/(x)]_(a)^(l+a)`
`=k_(e) Q/l(1/a-1/(l+a))=(k_(e)Q)/(a(l+a))`
If a to b, which corresponds to sliding the bar to the left until its left end is at the origin, then `E to oo` that represents the condition in wich the observation point P is at zero distance from the charge at the end of the rod, so the field become infinite.