 # The electric dipole moment for a charge distribution is: P=int_V rho(x)xd^3 x. If we imagine that rho=const on a spherical volume then it's easy to see that P depends by the origin of the frame of reference, it can be zero or something totally different. For this reason it's very difficult for me to give a physical meaning at this quantity waldo7852p 2022-09-20 Answered
The electric dipole moment for a charge distribution is:
$\stackrel{\to }{P}={\int }_{V}\rho \left(\stackrel{\to }{x}\right)\stackrel{\to }{x}{d}^{3}x$
If we imagine that $\rho =const$ on a spherical volume then it's easy to see that $\stackrel{\to }{P}$ depends by the origin of the frame of reference, it can be zero or something totally different. For this reason it's very difficult for me to give a physical meaning at this quantity (because it's strange that depends by the origin).
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The dipole moment as you have written does indeed change if you change the origin of your coordinates. The reason for this is that the dipole moment on this form is part of the multipole expansion, of which the dipole is is one of many multipoles.
You can think of the multipole expansion as somewhat related to the Fourier series expansion: we know what the field is for each type of multipole, so we can break down the field from a complicated charge distribution in terms of a sum of multipole fields. Similarly with the Fourier expansion, we can break down a complicated function's behavior in terms of simple harmonic (sine and cosine) functions.
In the analogy with the Fourier series, the coordinate dependence of the dipole moment is like the change of phase picked up in the Fourier series if you change the coordinates. The individual terms in he series will change depending on the coordinate system, but the sum of the whole series will still converge to the same function/field.
With the multipole expansion the electrostatic field is expressed in its spherical coordinate components ${E}_{r}$, ${E}_{\varphi }$ and ${E}_{\theta }$. In the example of a single point charge. If you place it at the origin, then the multipole expansion becomes very easy, since only the monopole term survives. You get
$\stackrel{\to }{E}=\frac{q\stackrel{\to }{r}}{4\pi {ϵ}_{0}|\stackrel{\to }{r}{|}^{3}},$
or simply put, the components of the field are ${E}_{r}=\frac{q}{4\pi {ϵ}_{0}|\stackrel{\to }{r}{|}^{2}}$, ${E}_{\varphi }=0$ and ${E}_{\theta }=0$. If you on the other hand place your point charge/origin somewhere else, we are get the familiar
$\stackrel{\to }{E}=\frac{q\left(\stackrel{\to }{r}-{\stackrel{\to }{r}}_{0}\right)}{4\pi {ϵ}_{0}|\stackrel{\to }{r}-{\stackrel{\to }{r}}_{0}{|}^{3}},$
which is much more complicated to express in terms of spherical components, but the multipole expansion is what helps you find a series expression for those field components.

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