¬eld energy due to the polarization of the environment, which is negative,

is

q2 1

UBorn = ’ 1’ . (13.60)

8πµ0 a µr

This polarization energy is often called the dielectric solvation energy or

Born energy after Born (1920), who ¬rst described this term. It is the (free)

344 Electromagnetism

energy change when a charge q is moved from vacuum into a cavity with

radius a in a dielectric continuum.

Another description of the Born energy is by using the alternative ¬eld

energy description of (13.41) in terms of charge-potential products. The

polarizable environment produces a potential φRP at the position of the

charge, called the reaction potential, and the energy in that potential is

UBorn = 1 qφRP . (13.61)

2

The reaction potential is the summed potential of all induced dipoles in the

medium:

∞

1 P (r)

φRP = ’ 4πr2 dr, (13.62)

2

4πµ0 a r

where

1 1 1 q

1’ 1’

P= D= . (13.63)

r2

µr 4π µr

Integration yields

q 1

φRP = ’ 1’ . (13.64)

4πµ0 a µr

This, of course, yields the same Born energy as was derived directly from

integrating the ¬eld energy in (13.60).

How does the solvation energy of a charge in a cavity behave in the case

that the medium is an electrolyte solution? Let us compute the reaction

potential, i.e., the excess potential at r = 0 due to the presence of the

medium beyond r = a. As before, the dielectric displacement is continuous

at r = a, and therefore

q

D(a) = (13.65)

4πa2

is also valid at the boundary in the medium, so that the boundary conditions

for φ are

dφ q

=’ , φ(r) ’ 0 for r ’ ∞. (13.66)

4πµa2

dr a

The di¬erential equation for φ in the range r ≥ a is

F zi F φ

∇2 φ = ’ c0 zi exp ’ (13.67)

i

µ RT

i

This equation can be solved numerically, given the ionic composition of

13.7 Quasi-stationary electrostatics 345

the medium. In the Debye“H¨ckel approximation, and using the radial

u

expression for the ∇ 2 operator, the equation simpli¬es to

1d dφ

r2 = κ2 φ, (13.68)

r2 dr dr

with the boundary conditions given above. The solution, valid for r ≥ a, is

q exp[’κ(r ’ a)]

φ(r) = . (13.69)

4πµ(1 + κa)r

The potential in the cavity (r ¤ a) is given by

dφ q

D(r) = ’µ0 = , (13.70)

4πr2

dr

and hence

q

φ(r) = + φRP , (13.71)

4πµ0 r

where φRP is a constant given by the boundary condition that φ is continuous

at r = a. This constant is also the excess potential (the reaction potential)

due to the mean-¬eld response of the medium. Applying that boundary

condition we ¬nd

q 1

φRP = ’ 1’ . (13.72)

4πµ0 a µr (1 + κa)

The excess energy of the charge “ due to interaction with the medium outside

the cavity in excess to the vacuum self-energy “ is, as in the previous case,

given by

Uexc = 1 qφRP . (13.73)

2

We see that for zero ionic strength the Born reaction potential (13.60) is

recovered, but that for large ionic strength the screening is more e¬ective,

as if the dielectric constant of the medium increases. The excess energy due

to κ:

1 q2 κ

Uexc (κ) ’ Uexc (κ = 0) = ’ , (13.74)

2 4πµ0 µr (1 + κa)

is the (free) energy resulting from transferring the cavity with charge from

an in¬nitely dilute solution to the electrolyte (assuming the dielectric con-

stant does not change). This term is due to the mean-¬eld distribution of

ions around a charge: the counterions are closer and produce a negative

energy. This term is responsible for the reduction of ionic chemical poten-

tials in electrolyte solutions, proportional to the square root of the ionic

concentration.

346 Electromagnetism

13.7.3 Dipole in a medium

The next example is the dielectric solvation energy for a dipole in the center

of a spherical cavity with radius a in a medium with homogeneous dielectric

constant µ. The problem is similar as the previous case of a charge, except

that we cannot make use of spherical symmetry. Let us choose the z-axis in

the direction of the dipole moment μ situated at r = 0, and use spherical

coordinates r, θ (polar angle with respect to the z-axis), and φ (azimuthal

angle of rotation around z, which must drop out of the problem for reasons

of symmetry).

For regions of space where there are no sources, i.e., for our problem

everywhere except for r = 0, the Poisson equation (13.47) reduces to the

Laplace equation

∇2 φ = 0, (13.75)

which has the general solution in spherical coordinates

∞ l

(Am rl + Blm r’l’1 )Plm (cos θ) exp(imφ),

φ(r, θ, φ) = (13.76)

l

l=0 m=’l

where Plm are Legendre functions and A and B are constants that must

follow from boundary conditions. In a bounded region of space (e.g., r ¤

a), the rl solutions are acceptable, but in an unbounded region with the

requirement that φ ’ 0 for r ’ ∞, only the r’l’1 solutions are acceptable.

The r’l’1 solutions are only acceptable for r ’ 0 if there is a singularity

due to a source at r = 0. The singularity determines which angular term

is acceptable. In the case of a dipole source, the singularity has a cosine

dependence on θ (l = 1), and only the l = 1 terms need be retained.

From these considerations, we can write the potentials in the cavity and

in the medium as

μ cos θ

(r ¤ a),

φcav (r) = + br cos θ (13.77)

4πµ0 r2

c cos θ

(r ≥ a),