Electric field $E$ due to a dipole moment

Question

Why is the formula of $$E = kp/z^3~ ?$$ Why is it $z^3$ ? I want a physics answer not math please.

Answer 1

A dipole may be represented by a pair of nearby charges of the opposite sign (but the same absolute values), $+Q$ and $-Q$ at a distance $\vec r$ from each other.

At the distance $R$ from the dipole (and I will only consider the case when we measure the field on the axis that includes the dipole itself), one adds $E$ going like $+1/R^2$ from $+Q$, and $-1/(R+r)^2$ from $-Q$, where $r\ll R$. But in this limit (we must know that the electric field from charges goes like $1/R^2$, from the Coulomb's law), $$ \frac{1}{R^2} - \frac{1}{(R+r)^2} = \frac{ (R+r)^2 - R^2 }{ R^2 (R+r)^2 }=\dots $$ and we expand (without mathematics, use physical platinum models of rectangles to perform the multiplication) $$ \dots = \frac{R^2+2Rr+r^2-R^2}{R^2(R+r)^2}\sim \frac{2Rr}{R^4} = \frac{2r}{R^3}$$ For $r\ll R$, I could approximate the denominator and neglect $r^2$ relatively to the larger $2rR$ in the numerator. In the result, $2r/R^3$, the factor of $r$ combines with the (universally omitted) factor of $Q$ to give us the dipole $p$, because $p=Qr$, so we get $kp/R^3$. $k=1/(4\pi\epsilon_0)$ in the SI units is the universal coefficient one always inserts to the formula for the electric field from charges, and you used $z$ instead of $R$ (I should have used $z$ as well because I studied the electric field on the $z$ axis only, anyway).

I did these weird expansions to avoid the method of the "derivative" (calculus) but I can't avoid mathematics entirely. After all, the very question contains things like $1/R^3$, so it is clearly impossible to address the question without using mathematical symbols of the same kind or their basic meaning or properties.

The operation with the fraction was included to replace what normal adult physicists would do by the derivative of $1/R^2$. If even the fractions are too advanced mathematical one, one could translate the operations with the fractions into plain English and discuss the various powers of $R$, why only integral powers appear, and why all of them cancel and the first surviving term goes like $1/R^3$. But any way to prove these things without the mathematical expressions is just a waste of time. Mathematics is very effective and even if some satisfactory "verbal" proof exists, it is trivially included in the mathematical calculation.

Answer 2

E=k*p/z^3. This is because usually, the Electric field is calculated at a point whose distance from the midpoint of the dipole(z) is very large as compared to distance between the two opposite charges(say a). That is z>>a. Thus, the original value of dipole at a point on the equatorial line=kp/(z^2+a^2)^3/2. as z^2+a^2=z^2.(z>>a). Thus, E=kp/z^3.