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  • A system of two equal and opposite charges separated by a certain distance is called an electric dipole.
  • Electric dipole moment ($p$) is defined as the product of either charge ($q$) and the length ($2a$)of the electric dipole.
  • Magnitude of electric field ($E$) due to an electric dipole at a distance $r$ from its centre in a direction making an angle $\theta$ with the dipole is given by the equation,$$E=\frac{1}{4\pi\epsilon}.\frac{p\sqrt{3\cos^2\theta+1}}{r^3}$$
    where, $p=2aq$ ($2a$ is the distance of separation of the charges $q$).

From the above equation, I thought that, electric field around a neutral body won't be zero. Electric field around a neutral body would be zero if and only if the distance of separation between the dipole charges is zero (from the equation, we can notice that $E$ tends to zero as $p$ tends to zero or $2a$ tends to zero). So, even if a body is neutral, I thought that electric field need not be zero.

I got a doubt here. We know that, in a neutral atom, electron and proton have equal and opposite charges, and even they are separated by a certain distance. So, I assumed a pair of electron and proton to behave as a dipole, so that I could fit them to the above dipole equation. I thought, as the distance between electron and proton is not zero, there must exist electric field around them. As a result, we can expect electric field around all the atoms, thus producing a vector added field around the substance. But, we don't feel electric field around all the substance. Is it that electric field around the substances negligible? I don't know whether I am correct or wrong, or whether I have misunderstood the concept. If any is the case, please explain.

[My book mentions one related example: "The molecules of water, ammonia, etc behave as electric dipoles. It is because, the centres of positive and negative charges in these molecules lie at a small distance from each other" It is to be noted that, molecules are considered here and atoms are neglected.]

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So, I assumed a pair of electron and proton to behave as a dipole

Classically, this is correct. It would behave as a dipole.

But in quantum mechanics we don't talk about a localized electron, but about orbitals. This is because particles have an associated wavefunction $\psi(t)$, which tells you the probability of finding your particle at a particular place.

In the case of an electron and a proton (hydrogen atom), the orbital has spherical symmetry. This means that the will be no net charge.

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@BrandonEnright Just as a note, "localised" is perfectly valid. Although I think "localized" looks better. –  jinawee Dec 16 '13 at 22:04
    
Yeah sorry, I blame my ignorant American spell checker :-p I actually edited to fix the linkebreak in the quote and the comma without space. –  Brandon Enright Dec 16 '13 at 22:07
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