Question

find the electric field at a point p on the perpendicular bisector of uniformly charged rod.the length of rod is l,the charge on it is q and the distance of p from the centre of rod is a

Answer 1

from figure, it is clear that,

x = $atan\theta$ and $r=asec\theta$......(1)

differentiating, x = $atan\theta$

dx= $asec^2\theta.d\theta$ .....(2)

linear charge density of rod, $\lambda=\frac{q}{l}$

so, charge on element, $dq=\lambda dx$

now, electric field due to element at given point, $E_x = dE cos\theta=\frac{Kdq}{r^2}cos\theta$

= $\frac{K\lambda dx}{a^2sec^2\theta}cos\theta$ [ from equation (1),

= $K\lambda\frac{asec^2\theta.d\theta}{a^2sec^2\theta}cos\theta$

= $\frac{K\lambda}{a}\int\limits^{\theta}_{-\theta}{cos\theta}\,d\theta$

= $\frac{k\lambda}{a}\left[2sin\theta\right]$

now, $sin\theta=\frac{l/2}{\sqrt{(l/2)^2+a^2}}$

$sin\theta=\frac{l}{\sqrt{l^2+4a^2}}$

now, electric field due to rod at given point, $E_a=\frac{k\lambda}{a}\frac{2l}{\sqrt{l^2+4a^2}}$

putting , $\lambda=\frac{q}{l}$

so, $E_a=\frac{2kq}{a\sqrt{l^2+4a^2}}$