Four point charges -q, +q, +q and -q are placed on y-axis at y = -2d, 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:

We have four point charges placed on the y-axis. Their positions and values are

$$q_1=-q\;{\rm at}\;(0,-2d),\qquad q_2=+q\;{\rm at}\;(0,-d),\qquad q_3=+q\;{\rm at}\;(0,+d),\qquad q_4=-q\;{\rm at}\;(0,+2d).$$

The point at which we want the electric field is $$P(D,0)\,,$$ i.e. a point on the x-axis with $$x=D$$ and $$y=0$$, where $$D\gg d.$$

For any point charge $$q_i$$ located at $$(0,y_i)$$ the position vector from the charge to $$P$$ is

$$\vec r_i=\bigl(D\,\hat i+(0-y_i)\,\hat j\bigr)=D\,\hat i-y_i\,\hat j.$$

The distance of the point from the charge is therefore

$$r_i=\sqrt{D^{\,2}+y_i^{\,2}}.$$

Statement of the formula: The electric field due to a point charge is

$$\vec E_i=\dfrac{k\,q_i}{r_i^{\,2}}\;\hat r_i =\dfrac{k\,q_i}{r_i^{\,3}}\;\vec r_i,$$ where $$k=\dfrac{1}{4\pi\varepsilon_0}$$ and $$\hat r_i=\dfrac{\vec r_i}{r_i}.$$

Substituting $$\vec r_i$$ we get the two Cartesian components for every charge:

$$E_{ix}= \dfrac{k\,q_i D}{\bigl(D^{\,2}+y_i^{\,2}\bigr)^{3/2}},\qquad E_{iy}= \dfrac{k\,q_i(-y_i)}{\bigl(D^{\,2}+y_i^{\,2}\bigr)^{3/2}}.$$

Now we first check the $$y$$-components. The charges at $$y=+d$$ and $$y=-d$$ have the same sign $$+q$$ but opposite $$y_i$$, so their $$E_{iy}$$ cancel. In the same way the charges at $$y=+2d$$ and $$y=-2d$$ have the same sign $$-q$$ and again opposite $$y_i$$, so those $$E_{iy}$$ also cancel. Hence,

$$\sum_{i=1}^{4}E_{iy}=0,$$ and the total electric field at $$P$$ is purely along the x-direction.

Therefore the magnitude of the net field equals the sum of the x-components:

$$E=\sum_{i=1}^{4}E_{ix}=kD\sum_{i=1}^{4} \dfrac{q_i}{\bigl(D^{\,2}+y_i^{\,2}\bigr)^{3/2}}.$$

Because $$D\gg d,$$ we expand each denominator with the binomial series. We start from

$$\bigl(D^{\,2}+y_i^{\,2}\bigr)^{-3/2}=D^{-3}\Bigl(1+\dfrac{y_i^{\,2}}{D^{\,2}}\Bigr)^{-3/2}.$$ Using $$(1+s)^{-3/2}=1-\dfrac{3}{2}s+\dfrac{15}{8}s^{2}-\ldots,$$ we substitute $$s=\dfrac{y_i^{\,2}}{D^{\,2}}$$ to obtain

$$\bigl(D^{\,2}+y_i^{\,2}\bigr)^{-3/2}=D^{-3}\Bigl[1-\dfrac{3}{2}\dfrac{y_i^{\,2}}{D^{\,2}} +\dfrac{15}{8}\dfrac{y_i^{\,4}}{D^{\,4}}-\ldots\Bigr].$$

Substituting this expansion back into the expression for $$E$$ gives

$$E=kD\sum_{i=1}^{4}q_i D^{-3}\Bigl[1-\dfrac{3}{2}\dfrac{y_i^{\,2}}{D^{\,2}} +\dfrac{15}{8}\dfrac{y_i^{\,4}}{D^{\,4}}-\ldots\Bigr].$$

Simplifying the common factor $$D$$ with $$D^{-3}$$ we have

$$E=k\sum_{i=1}^{4}q_i \Bigl[D^{-2}-\dfrac{3}{2}\dfrac{y_i^{\,2}}{D^{\,4}} +\dfrac{15}{8}\dfrac{y_i^{\,4}}{D^{\,6}}-\ldots\Bigr].$$

We now evaluate the sums term by term.

1. The monopole term (total charge):

$$\sum_{i=1}^{4}q_i=(-q)+(+q)+(+q)+(-q)=0.$$ So the entire $$D^{-2}$$ contribution vanishes.

2. The next (quadrupole) term involves $$\sum q_i y_i^{\,2}\!:$$

$$\begin{aligned} \sum_{i=1}^{4}q_i y_i^{\,2}&=(-q)(4d^{2})+(+q)(d^{2})+(+q)(d^{2})+(-q)(4d^{2})\\ &=-4qd^{2}+qd^{2}+qd^{2}-4qd^{2}\\ &=-6qd^{2}\;\;(\neq0). \end{aligned}$$

Hence the first non-vanishing contribution to $$E$$ is

$$E=-\dfrac{3}{2}k\bigl(\sum q_i y_i^{\,2}\bigr)D^{-4} +\text{higher-order terms}.$$ Since $$\sum q_i y_i^{\,2}$$ is merely a constant factor, this clearly shows

$$E\propto\dfrac{1}{D^{4}}.$$

All still-higher terms vary as $$1/D^{6},\,1/D^{8},\ldots$$ and are even smaller when $$D\gg d.$$ Thus the leading behaviour of the electric field magnitude is

$$E\;{\Large\sim}\;\dfrac{\text{constant}}{D^{4}}.$$

Hence, the correct answer is Option A.