understanding polarity of water ?

Oxygen is electronegative, and "steals" (up to) two electrons, to effect an electron configuration similar to neon. In such a state, the full valence shell is spherically symmetric... and spherically symmetric charge distributions, perceived from afar, are those of point charges, of the same total charge, located at the centers of those charge distributions. So, the potential, of O^-- would be

[math]\approx \frac{-2 q_e}{4 \pi \epsilon_0 r}[/math]

But then you must re-remove two electrons, from two 2P orbitals... given that Oxygen forms bonds about its equator, into P_x + P_y (hybrid) orbitals, the equator seems the appropriate place to re-remove two units of charge from... the P₊ + P_- orbitals' charge distributions, are "donuts" centered on the Oxygen nucleus, evidently w/ radii of roughly 100 pm, which is the size of the oxygen atom, namely the level 2 orbitals of the oxygen.

Now, at least in the equatorial plane, the potential from a ring of charge, of radius R, out at a distance D > R from the origin, is:

[math]= \int \frac{dq}{4 \pi \epsilon_0 r} = \int_{\phi=0}^{2 pi} \frac{\mu R d\phi}{4 \pi \epsilon_0 r(\phi)}[/math]

[math]= \frac{\mu R}{4 \pi \epsilon_0} \int d\phi \frac{1}{D}\sum_{l=0}^{\infty} \left( \frac{R}{D} \right)^l P_l(cos \theta)[/math]

The odd-numbered Legendre Polynomials all integrate to zero. But, the first few even LP's give rise to:

[math]= \frac{\mu R}{4 \pi \epsilon_0 D} \int d\phi \left( 1 + \left(\frac{R}{D}\right)^2\frac{1}{2}\left( 3 cos^2\phi - 1\right) + ...\right)[/math]

[math]= \frac{\mu R}{4 \pi \epsilon_0} \left( 2 \pi + \frac{1}{2} \left( 3 \frac{1}{2} 2 \pi - 2 \pi \right)\left(\frac{R}{D}\right)^2 + ...\right)[/math]

[math]= \frac{2 \pi \mu R}{4 \pi \epsilon_0D} \left(1 + \frac{1}{4}\left(\frac{R}{D}\right)^2 + ...\right)[/math]

[math]\approx \frac{Q}{4 \pi \epsilon_0 D} \left( 1 + \frac{1}{4}\left(\frac{R}{D}\right)^2 \right)[/math]

Now, if the ring of charge, mathematically modeling the absence (presence) of two electrons (holes) from the Oxygen's other-wise-spherically-symmetric charge distribution, contains two units worth of charge (the missing two electrons), then the total potential, from a normal neutral Oxygen atom, would be:

[math]V_O(equator) \approx \frac{-2 q_e}{4 \pi \epsilon_0 D} + \frac{+2 q_e}{4\pi\epsilon_0D}\left(1 + \frac{1}{4}\left(\frac{R}{D}\right)^2\right)[/math]

[math]= \frac{\frac{1}{2} q_e}{4 \pi \epsilon_0 D} \left(\frac{R}{D}\right)^2[/math]

That equation implies, that (near the equatorial plane) the missing electrons, generate a positive potential, which falls off as 1/r³... such a potential could account for the electro-negativity ("electro-positivity"?) of oxygen, which "wants" to grab (up to) two other electrons, into equatorial P_x,y orbitals...

does that equation seem potentially qualitatively accurate (if not quantitatively precise) ?