Liquid-State Physical Chemistry: Fundamentals, Modeling, and Applications (2013)
Appendix D. Elements of Electrostatics
In this appendix we review briefly some aspects of electrostatics that are relevant to our discussion on polar liquids. Free use has been made of the literature indicated in the bibliography.
D.1 Coulomb, Gauss, Poisson, and Laplace
In vacuum a charge q1 at a distance r = |r| of another charge q2 experiences a force f given by Coulomb's law, using with ε0 the permittivity of vacuum,
The field E is the force a test charge of unit charge experiences from a charge q at position r so that we have
The energy increment dw12(r) of a charge q1 in the field of a charge q2 is given by dw12(r) = −q1E2·dr and accordingly the (potential or) Coulomb energy reads
where the potential , as defined by E ≡ −∇ϕ, is introduced. As conventional, the force is thus f = −∇w.
A pair of opposite and equal charges q1 and q2 = −q1, separated by a (small) vector d pointing from q2 to q1, is called a dipole, characterized by dipole moment μ = q1d. The potential of such a dipole fixed in space at large distance r from the center of the dipole is (Figure D.1)
Figure D.1 The potential of a dipole μ = qd at distance r.
The field E is then given by
The Coulomb energy of a dipole in a field with potential ϕ can be calculated to be w = μ·∇ϕ = −μ·E (see Chapter 3).
Consider now an arbitrary closed surface S with surface elements dS = ndS, located at position r from the origin, where n is the positive outward normal vector of the surface element and dS its area. Because the scalar product r·dS/r is the projection of dS on a sphere of radius r, the product r·dS/r3 is the projection on a unit sphere centered at the origin. Hence, if S encloses the origin, the closed surface integral
If S does not enclose the origin, r cuts S at two points, the contributions from these two intersections cancel and the integral is zero. It follows that the integral is q/ε0 if S does contain q and zero otherwise. For a general charge distribution ρ within S we obtain Gauss' law
In order to go from this global description to a local one, we use the Gauss divergence theorem and find Poisson's equation
Using E = −∇ϕ we obtain
reducing to Laplace's equation ∇2ϕ = 0 if ρ = 0, which is the usual case for a dielectric material. For a distribution of volume charge ρ and surface charge σ, the potential becomes
Molecules are polarizable and therefore a field induces in a molecule a dipole moment μind = αE, where the proportionality factor α is a material-dependent parameter called the polarizability. For macroscopic matter this occurs also, and we have the polarization P = ε0χE with the susceptibility χ. Since P represents a volumetric distribution of induced dipoles we have for the associated potential
where x = r − ξ and x = |x| is used. The second line states that the potential is analogous to one caused by a surface charge n·P and a volume charge −∇·P. Using the same reasoning as before, and going again from a global to a local description via the Gauss divergence theorem using the volume charge density ρ − ∇·P, we have within the dielectric, since the surface charges do not enter Poisson's equation,
with the dielectric displacement D ≡ ε0E + P = (1 + χ)ε0E ≡ εrε0E ≡ εE, thus introducing the relative permittivity (or dielectric constant) εr = 1 + χ, and permittivity ε ≡ εrε0. Outside the dielectric, P = 0 and ∇·D = ρ reduces to ε0∇·E = ρ.
So, the dielectric displacement satisfies the condition ∇·D = ρ, and ε0∇·εr∇ϕ = −ρ is the governing equation for the resulting potential ϕ. From this equation we can derive the requirement on the continuity of D across the boundary between media by writing the integral equivalent of ∇·D = ρ to obtain
Applying this relation to an infinitesimally small pill-box-shaped volume with a bottom and top surface with area A parallel to the boundary on each side of the boundary (Figure D.2), we have
Figure D.2 The continuity of D at the boundary S.
where Dn is the component of D normal to the boundary (with the direction from medium 1 to 2 as positive). If the charge density at the boundary remains finite, the integral tends to zero when volume of the pill-box V goes to zero. Hence,
If a surface charge is present we obtain
In conclusion, the normal component Dn of D across a boundary between two dielectrics is continuous in the absence of surface charge, but if a surface charge is σ is present this component jumps by σ.
D.2 A Dielectric Sphere in a Dielectric Matrix
To illustrate the use of the above theory, we now calculate the potential field and some associated relations outside (domain 1) and inside (domain 2) a sphere with radius a and relative permittivity ε2 embedded in a matrix with relative permittivity ε1 to which a field E along the z-axis is applied by a set of external charges that would be uniform if ε1 = ε2. Generally, it is most convenient to use a coordinate system associated with the boundary conditions and for spherical entities these are the spherical coordinates r, θ, φ defined in terms of the Cartesian coordinates x, y, z by
In these coordinates the Laplace operator operating on a scalar reads
Since we interested in an axially symmetric system, we have ϕ = ϕ(r,θ) and use a trial solution in the form ϕ(r,θ) = R(r)Θ(θ) from which we obtain
As the two parts involve different spatial variables, this expression can be only satisfied if each part equals a constant, say λ, and therefore
By substituting x = cosθ and Θ(θ) = y(x), the equation for Θ transforms to Legendre's equation, of which the generic solutions are y(x) = AνPν(x) + BνQν(x), where Pν and Qν are the Legendre functions of the first and second type. The label ν is determined by λ = ν(ν + 1), and Aν and Bν are arbitrary constants. The full generic solution must be used when the z-axis (x = cosθ = ±1) is excluded from the region where the potential has to be calculated. In our case, however, the axis has to be included. It appears that the Pν(x) for fractional ν and the Qν(x) for all values of ν have a singularity at x = ±1, that is, at θ = 0 and θ = π. The consequence is that all Bν coefficients must be zero because all Qν(x) functions lead to singularities. Moreover, ν can only take integer values n, so λ = n(n + 1) and Θ(θ) = ΣnAnPn(cosθ). The first few Legendre polynomials Pn(cosθ) are
With λ = n(n + 1), the solution for the equation for R becomes
Hence, the generic solution for the Laplace operator for axial symmetry, recalling that outside the sphere is domain (1) and inside the sphere is domain (2), becomes
To find the specific solution we must add the boundary conditions reading:
I) (ϕ1)r→∞ = −Er cosθ because the field gradient far removed from the origin should approach the applied field E.
II) (ϕ1)r=a = (ϕ2)r=a as ϕ should be continuous across the boundary.
III) ε1(∂ϕ1/∂r)r=a = ε2(∂ϕ2/∂r)r=a because at the boundary the normal dielectric displacement component Dn should be continuous.
IV) ϕ2 is finite at r = 0 since the potential may not have a singularity at r = 0.
Satisfying boundary condition IV can only be achieved if Dn = 0 and condition I requires A1 = −E = −|E|, while all other An = 0. Conditions II and III say that, except for n = 1,
From these equations it follows that Bn = Cn = 0 for all n ≠ 1. For n = 1 it is found that
and ϕ1 and ϕ2 become
Because the potential due to the external charges is ϕ = −Ez, the contributions and due to the polarization become
The expression for is identical to that of an equivalent ideal dipole μequ at the center of the sphere, surrounded by vacuum with the dipole vector directed along the external field E (the z-axis), and given by
The field inside the sphere due to the polarization and the total field E2 inside the sphere are described by, respectively,
In case the sphere is a cavity so that ε1 = ε and ε2 = 1, the total field E2 becomes the cavity field Ecav = 3εE/(2ε + 1).
D.3 A Dipole in a Spherical Cavity
If in a spherical cavity in an infinite matrix a dipole μ is placed at the center, this dipole will polarize the surrounding matter and the resulting polarization will give rise to a field at the dipole, called the reaction field R. The reaction field will be proportional to μ, having the same direction as μ so that R = fμ. We choose the direction of the z-axis along the dipole vector so that we can employ again the axi-symmetric solutions [Eq. (D.23)] to the Laplace equation. In this case, the boundary conditions are, using ε1 = ε and ε2 = 1:
The D-terms in ϕ2 are due to the charges inside the cavity. Because the potential of an ideal dipole along the z-axis is , the contribution Dn = 0 for all n except n = 1, which reads . For ϕ2, we thus have
From condition I, we conclude that An = 0 for all n, so that ϕ1 becomes
From conditions II and III we obtain for all n ≠ 1,
which can be satisfied only if Bn = Cn = 0 for n ≠ 1, while for n = 1 the result is
Solving for B1 and C1 leads to
Substitution in Eq. (D.32) and (D.33) results in
Because the potential due to the dipole is given by , we see from the first part of Eq. (D.37) that formally the potential in the dielectric matrix can be described as the field of an effective (or virtual) dipole at the center of the cavity given by
From the second part of Eq. (D.37) we see that the field in the cavity is the field of the dipole in vacuum plus the uniform reaction field R given by
Using again the potential due to the dipole given by , the contributions and due to the surface charges become
These potentials can be described by a dipole m given by
The results in Eq. (D.30) and (D.39) are used in Chapter 10.
Böttcher, C.J.F. (1973) Theory of Electric Polarization, vol. 1, 2nd edn, Elsevier, Amsterdam.
Jackson, J.D. (1999) Classical Electrodynamics, 3rd edn, John Wiley & Sons, Inc., New York.
Robinson, F.N.H. (1973) Electromagnetism, Oxford Physics Series, Clarendon Press, Oxford.
Stratton, J.A. (1941) Electromagnetic Theory, McGraw-Hill.