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# Appendix D: Exact Solution of the Polarization of a Dumbbell

It does not seem to have been noticed previously that all the mathematical apparatus necessary for computing the components of αe for the spherical particle dumbbell has been in place since the publication of a paper by Davis [144]. The problem is the following: Take two highly conducting spheres of the same radius that are connected by a very thin conducting wire so that both spheres in this dumbbell are at the same potential. Apply an electric field that is uniform at infinity and solve for the potential. Once this is done, calculate the normal component of the electric field right at the surface of the dumbbell, thus giving the charge density (charge/unit area) on the dumbell surface. The dipole moment of the dumbbell is then calculated from the average of r over the surface and divided by the dumbbell volume.

Davis [144] gives the solution for the potential when the spheres are both held at zero potential and this solution is all that is required for solving our dumbbell problem. In calculating the polarization there are two difficult integrals that arise, which fortunately have been tabulated by Apelblat [145]. The required integrals are given by,

The polarizability of the dumbbell then equals,

where rp is the ratio of the center-center separation to the sphere diameter and the constants Z2 and Sj are defined by,

where cosh(µ1) = rp . These infinite sums are rapidly converging so that numerical computation is straightforward. In the limit rp → ∞ only the first term of So is important and we find the simple limiting behavior,

The electric polarizability for two spheres without the wire [46] is quite different in general. For rp → ∞ the intrinsic conductivity [σ] of the untethered spheres approaches 3 (the single sphere result) and for nearly touching spheres (rp → 1), [σ] approaches 7ζ(3)/2 [see eqn. (27)].

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