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 $D \gg d$, will behave as
Answer: (A) $E \propto \dfrac{1}{D^4}$
The arrangement has total charge zero and, being symmetric about the origin, zero dipole moment. Its leading term is a quadrupole, whose field falls off as $1/D^4$.
Direct check: on the $x$-axis only the $x$-components survive. A pair $\pm a$ of charge $Q$ gives $E_x = \dfrac{2kQD}{(D^2 + a^2)^{3/2}} \approx \dfrac{2kQ}{D^2}\left(1 - \dfrac{3a^2}{2D^2}\right)$. Adding the $+q$ pair at $a = d$ and the $-q$ pair at $a = 2d$, the $1/D^2$ terms cancel:
$$E_x \approx \frac{2kq}{D^2}\cdot\frac{3(4d^2 - d^2)}{2D^2} = \frac{9kqd^2}{D^4} \propto \frac1{D^4}$$
Solution by Sreeraj P, M.Sc Physics