Q. Four point charges $-q, +q, +q$ and $-q$ are placed on y-axis at $y = -2d, y = -d, y = +d$ and $y = +2d$, respectively. The magnitude of the electric field $E$ at a point on the x-axis at $x = D$, with $D >> d$, will behave as :
Solution:
Electric field at $p = 2E_{1} \cos\theta_{1} -2E_{1}\cos\theta_{2} $
$ = \frac{2Kq}{\left(d^{2}+D^{2}\right)} \times\frac{D}{\left(d^{2}+D^{2}\right)^{1/2}} - \frac{2Kq}{\left[\left(2d\right)^{2}+D^{2}\right]} \times\frac{D}{\left[\left(2d\right)^{2}+D^{2}\right]^{1/2}} $
$ =2KqD \left[\left(d^{2}+D^{2}\right)^{-\frac{3}{2}} - \left(4d^{2} +D^{2}\right)^{-\frac{3}{2}} \right] $
$ = \frac{2KaD}{D^{3}} \left[\left(1- \frac{d^{2}}{D^{2}}\right)^{-3/2} -\left(1+ \frac{4d^{2}}{D^{2}}\right)^{-3/2}\right] $
Applying binomial approximation $ \because d < < D $
$ = \frac{2KqD}{D^{3}} \left[ 1- \frac{3}{2} \frac{d^{2}}{D^{2}} - \left(1- \frac{3\times4d^{2}}{2D^{2}}\right)\right] $
$ = \frac{2KqD}{D^{3}} \left[\frac{12}{2} \frac{d^{3}}{D^{2}} - \frac{3}{2} \frac{d^{2}}{D^{2}}\right] $
$ = \frac{9kqd^{2}}{D^{4}} $
