Find the electric field at a point P on the perpendicular bisector of a uniformly charged rod. The length of the rod is L, the charge on it is Q and the distance of P from the centre of the rod is a.

Explanation

Let us take an element of length dx at a distance x from
the centre of the rod . The charge on this
element is
` dQ=(Q)/(L) dx`.
The electric field at P due to this element is
`dE=(dQ)/(4piepsilon_0 (AP)^2)`
. By symmetry, the resultant field at P will be along OP
(if the charge is positive). The component of dE along
OP is
`dE cos theta = (dQ)/(4piepsilon_0 (AP)^2) . (OP)/(AP) = (aQdx)/(4piesilon_0 L(a^2+x^2 )^(3/2)`
Thus, the resultant field at P is
`E =int dE cos theta`
`= (aQ)/(4piesilon_0L) int_(-L/2)^(L/2) (dx) /(a^2+x^2 (3/2)).`
We have `x = a tan theta or dx = a sec^2theta dtheta.`
theus,`int(dx)/(a^2+x^2)^(3/2) = int (alpha sec^2theta dthetatheta)/(alpah^3 sec ^3 theta)
`= (1)/(a^2) int cos theta d theta = (1)/(a^2) sin theta = (1)/(alpha^2) (x)/(x^2 + alpha^2 )^(1/2)`
. From(i)
`E=(alphaQ)/(4piepsilon_0 La^2) [(x)/(x^2 + alpha^2 )^(1/2) ]_(-L/2)^(L/2)`
`= (aQ)/(4piepsilon_0 La^2) [(2L)/(L^2+4alpah^2)^(1/2)]`
`= (Q)/(2piepsilon_0 alpha sqrt(L^2+4a^2))`.