Electric dipole moment and charge density

Asked Oct 28 '21 at 13:52

Active Oct 28 '21 at 15:51

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I came across the following expression about the electric dipole moment:

$\vec p=\int_V \rho(\vec {r}')\vec r'$

But I don't understand it. What is $\vec r'$ supposed to represent and the density? Are we observing some region with a charge distribution and we take a small infinitesimal volume $dV$ and also the charge density there?

The expression I know for the dipole moment is: $\vec p= q\vec a$. Here $q$ is the charge value, and $\vec a$ a vector from the negative charge to the positive.

How does the above formula resembles to the last one?

edited Oct 28 '21 at 15:51

Qmechanic

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asked Oct 28 '21 at 13:52

imbAF

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Does this answer your question? Multipole Expansion: Electrostatics – Roger V. Oct 28 '21 at 13:56
That's what I am reading but that does explain as to why. It just presents it as a fact. – imbAF Oct 28 '21 at 14:00
Look up the posts (or book chapters) on multipole expansion. – Roger V. Oct 28 '21 at 14:06

Electric dipole moment and charge density

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