# Why does the electric dipole moment of the electron tell us about its sphericity? [duplicate]

There are a bunch of experiments that claim to show that the electron is highly spherical by measuring the electron electric dipole moment. See e.g.:

However, according to the Cartesian expansion of a charge distribution (see http://en.wikipedia.org/wiki/Multipole_expansion), if we assume the electron is made up of two small balls of negative charge $q$ separated by a small distance $2\mathrm{d}z$, and calculate the dipole moment in the centre of the two balls, the dipole moment is zero.

$$(-q)(-\mathrm{d}z)+(-q)(\mathrm{d}z)=0$$

which means even a highly non-spherical shape produces a dipole moment of zero.

So if the dipole moment is measured to be tiny, why does that imply the electron is highly spherical?

Some similar questions on this site:

• What is the mass density distribution of an electron? is a different question because I am not asking about the mass distribution, I am asking about how the dipole moment in a Cartesian expansion tells us about the sphericity of the electron.
• Do electrons have shape? is also not what I am asking; I am asking why it is that the dipole moment can be used to tell you that the electron is spherical when in the example I give of a non-spherical object, the dipole moment is zero and is therefore not a measure of the degree to which the electron is spherical.
• An electric dipole consists of two charges of different sign, not of the same sign! – Photon Mar 6 '15 at 13:00
• @Photon The dipole moment of an arbitrary charge distribution does not require charges to be of opposite sign. An "electric dipole" is not the same thing as the dipole moment term in a cartesian expansion. – kotozna Mar 6 '15 at 13:49
• Relarted and possibly a duplicate: Do electrons have shape? – John Rennie Mar 6 '15 at 14:49
• More on electron dipole moment. – Qmechanic Mar 9 '15 at 13:14