What is the surface charge density on the surface of a spherical cavity, scooped from a conductor, with an electric quadrupole at the center?

With a little bit of intuition about spherical tensors, we can “guess” the correct answer up to a constant.

Let the electric quadrupole be oriented purely in the z direction and have magnitude [math]Q[/math].

We know that induced charge can lie only on the inner surface of the conductor; in the interior of the conductor, and outside its outer surface, there is no charge and no electric field. Therefore all multipole moments of the configuration as a whole (the lone quadrupole at the centre plus the induced charge on the inner surface) must vanish. Therefore the induced charge on the inner surface must form a pure quadrupole, oriented purely in the z direction, with quadrupole moment [math]-Q[/math].

The [math]2^\ell[/math]-pole moment is a spherical tensor of rank [math]\ell[/math] and each component is just the inner product of the charge distribution with the corresponding spherical harmonic, with the spherical surface of radius [math]r[/math] weighted by [math]r^\ell[/math]. Since the spherical harmonics form a orthonormal basis for (sufficiently smooth) functions on the sphere, in order to get the induced charge to be a pure quadrupole oriented purely in the z direction, it must have the distribution of the corresponding spherical harmonic, [math]Y^0_2[/math]. Therefore

[math]\begin{equation} \sigma(\theta) = \sigma_0 (3 \cos^2 \theta - 1) \end{equation}[/math]

Performing the integration and setting the quadrupole moment for the induced charge to [math]-Q[/math], we obtain

[math]\begin{equation} \sigma_0 = -\frac{5Q}{16\pi r^4} \end{equation}[/math]

where [math]r[/math] is the radius of the inner surface of the conductor.

Without any additional work, we can also see how to get the induced charge for any other quadrupole moment: just take the equally weighted linear combination of the corresponding spherical harmonics.