Why is there a greater degree of separation in a dipole-dipole interaction?

Question

My textbook says this:

Electrostatic forces between two ions decrease by the factor $\frac{1}{d^2}$ as their separation distance, $d$, increases. Dipole–dipole forces, however, vary as $\frac{1}{d^4}$. Because of the higher power of $d$ in the denominator, the attractive forces diminish with increasing $d$ much more rapidly than does $\frac{1}{d^2}$. As a result, dipole forces are effective only over very short distances.

Now mathematically I understand why this is true, but I don't understand why this is the case in reality. Can someone shed some light on this?

Answer 1

Like I answered my other question, this is in some ways very similar. Since electrostatic attractions are so strong, it is decreases less at a longer distances. The $+$ & $-$ are attracted together so strongly.

Dipole-dipole attractions aren't strong attractions, electrostatic attractions are stronger, since they are created by ionic bonding. Dipole-dipole occur in covalent bonding. Thus dipole-dipole attractions decrease with increasing distance faster.

This is just a piece of the whole picture.

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Check the refs: 1, 2, 3. They provide more in depth detail to the entire picture.

Answer 2

Imagine the dipole as a partial positive ($q_+$) and negative ($q_-)$ separated by a small distance $\delta$. The electrostatic potential due to the dipole is $\frac{q_+}{|d-\delta|}+\frac{q_-}{|d+\delta|}$ where $d$ is the distance to the midpoint between the two charges. (There is also an angle term which I left out for simplicity).

As $d$ increases $\frac{q_+}{|d-\delta|}\approx -\frac{q_-}{|d+\delta|}$, i.e. the electrostatic potential of the dipole (i.e. two very close charges of opposite charge) goes to zero faster than a single charge because at longer distances the two charges appear to be at the same position and "cancel each other out".

Feynman has given as a more thorough explanation.