Question

A thin straight rod of length 2a carrying a uniformly distributed charge q is located in vacuum. Find the magnitude of the electric field strength as a function of the distance r from the rod'* centre along the straight line

(a) perpendicular to the rod and passing through its centre;

(b) coinciding with the rod'* direction (at the points lying outside the rod).

Investigate the obtained expressions at r >> a.

Answer 1

(a) It is clear from symmetry considerations that vector E must be directed as shown in the figure. This shows the way of solving this problem: we must find the component dEr of the field created by the element dl of the rod, having the charge dq and then integrate the result over all the elements of the rod. In this case

where λ = q/2a is the linear charge density. Let us reduce this equation of the form convenient for integration.

(b) Let, us consider the element of length dl at a distance l from the centre of the rod, as shown in the figure.

The field at P, due to this element.

if the element **** on the side, shown in the diagram, and