Tensors for Physics

10.4 Further Applications in Electrodynamics 173

The center of the sphere is put atr=0. The electric field, far from the sphere,
i.e. forr→∞is the imposed fieldE∞. The electric potentialφobeys the Laplace
equationΔφ =0. The boundary condition isφ =const., on the surface of a
conductor. Hereφ=0 is chosen forr=R. The electrostatic potential is a scalar,
and it should be linear in the imposed vectorial fieldE∞.Forr>R, the ansatz

φ(r)=arμEμ∞+br−^3 rμEμ∞

is made. It obeys the required symmetry and it is a solution of the Laplace equation
because it involves the ascending and descending vectorial multipole potential func-
tions. The coefficientsaandbare determined by the boundary conditions. First,
notice that the electric field is given by

Eμ=−∇μφ(r)=−aEμ∞+ 3 br−^5 rμrνEν∞.

Now, the conditionE→E∞,forr→∞, impliesa=−1. Then,φ=0forr=R,
leads tob=R^3. Thus the solution for the present potential problem is

φ(r)=−rμE∞μ +

R^3

r^3

rμEμ∞=−rμEμ∞+

1

4 πε 0

r−^3 rμpindμ. (10.38)

The second term of the computed potential has the form typical for a dipole potential.
The induced dipole moment is introduced by

pindμ = 4 πR^3 ε 0 E∞μ =αε 0 E∞μ. (10.39)

The coefficient

α= 4 πR^3 = 3 Vsph, (10.40)

is the polarizability, cf. Sect.5.3.3. It is proportional to the volumeVsph=( 4 π/ 3 )R^3
of the sphere.

10.4.2 Electric Polarization as Dipole Density

The total charge density of a material is the sum of the densityρof the free charges
and the densityρiof the bound internal charges, i.e. of the atomic nuclei and the
electrons bound in atoms and molecules. When all charges are counted, the Gauss
law is

ε 0 ∇μEμ=ρ+ρi,