## 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 *q*_{1} at a distance *r* = |** r**| of another charge

*q*

_{2}experiences a force

**given by**

*f**Coulomb's law*, using with

*ε*

_{0}the permittivity of vacuum,

(D.1)

The *field* *E* is the force a test charge of unit charge experiences from a charge *q* at position *r* so that we have

(D.2)

The energy increment d*w*_{12}(*r*) of a charge *q*_{1} in the field of a charge *q*_{2} is given by *dw*_{12}(*r*) = −*q*_{1}*E*_{2}·d** r** and accordingly the (potential or)

*Coulomb energy*reads

(D.3)

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 *q*_{1} and *q*_{2} = −*q*_{1}, separated by a (small) vector ** d** pointing from

*q*

_{2}to

*q*

_{1}, is called a

*dipole*, characterized by dipole moment

**=**

*μ**q*

_{1}

**. The potential of such a dipole fixed in space at large distance**

*d**r*from the center of the dipole is (

__Figure D.1__)

(D.4)

** Figure D.1** The potential of a dipole

*μ*=

*qd*at distance

*r*.

The field *E* is then given by

(D.5)

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 d*S* = ** n**d

*S*, located at position

*r*from the origin, where

**is the positive outward normal vector of the surface element and d**

*n**S*its area. Because the scalar product

*r*·d

*S*/

*r*is the projection of d

*S*on a sphere of radius

*r*, the product

*r*·d

*S*/

*r*

^{3}is the projection on a unit sphere centered at the origin. Hence, if

*S*encloses the origin, the closed surface integral

(D.6)

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*/

*ε*

_{0}if

*S*does contain

*q*and zero otherwise. For a general charge distribution

*ρ*within

*S*we obtain

*Gauss' law*

(D.7)

In order to go from this global description to a local one, we use the Gauss divergence theorem and find *Poisson's equation*

(D.8)

Using *E* = −∇*ϕ* we obtain

(D.9)

reducing to *Laplace's equation* ∇^{2}*ϕ* = 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

(D.10)

Molecules are polarizable and therefore a field induces in a molecule a dipole moment *μ*_{ind} = *α E*, 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**ε*

_{0}

*χ*with the

**E***susceptibility*

*χ*. Since

**represents a volumetric distribution of induced dipoles we have for the associated potential**

*P*(D.11)

where ** x** =

**−**

*r***and**

*ξ**x*= |

**| is used. The second line states that the potential is analogous to one caused by a surface charge**

*x***·**

*n***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**

*P**ρ*− ∇·

**, we have within the dielectric, since the surface charges do not enter Poisson's equation,**

*P*(D.12)

with the *dielectric displacement* ** D** ≡

*ε*

_{0}

**+**

*E***= (1 +**

*P**χ*)

*ε*

_{0}

**≡**

*E**ε*

_{r}

*ε*

_{0}

**≡**

*E**ε*, thus introducing the

**E***relative permittivity*(or

*dielectric constant*)

*ε*

_{r}= 1 +

*χ*, and

*permittivity*

*ε*≡

*ε*

_{r}

*ε*

_{0}. Outside the dielectric,

**= 0 and ∇·**

*P***=**

*D**ρ*reduces to

*ε*

_{0}∇·

**=**

*E**ρ*.

So, the dielectric displacement satisfies the condition ∇·** D** =

*ρ*, and

*ε*

_{0}∇·

*ε*

_{r}∇

*ϕ*= −

*ρ*is the governing equation for the resulting potential

*ϕ*. From this equation we can derive the requirement on the continuity of

**across the boundary between media by writing the integral equivalent of ∇·**

*D***=**

*D**ρ*to obtain

(D.13)

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

(D.14)

** Figure D.2** The continuity of

*D*at the boundary

*S*.

where *D***_{n}** is the component of

**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**

*D**V*goes to zero. Hence,

(D.15)

If a surface charge is present we obtain

(D.16)

In conclusion, the normal component *D***_{n}** of

**across a boundary between two dielectrics is continuous in the absence of surface charge, but if a surface charge is**

*D**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 *ε*_{2} embedded in a matrix with relative permittivity *ε*_{1} to which a field ** E** along the

*z*-axis is applied by a set of external charges that would be uniform if

*ε*

_{1}=

*ε*

_{2}. 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

(D.17)

In these coordinates the Laplace operator operating on a scalar reads

(D.18)

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

(D.19)

As the two parts involve different spatial variables, this expression can be only satisfied if each part equals a constant, say *λ*, and therefore

(D.20)

By substituting *x* = cos*θ* and *Θ*(*θ*) = *y*(*x*), the equation for *Θ* transforms to *Legendre's equation*, of which the generic solutions are *y*(*x*) = *A _{ν}P_{ν}*(

*x*) +

*B*(

_{ν}Q_{ν}*x*), where

*P*and

_{ν}*Q*are the Legendre functions of the first and second type. The label

_{ν}*ν*is determined by

*λ*=

*ν*(

*ν*+ 1), and

*A*and

_{ν}*B*are arbitrary constants. The full generic solution must be used when the

_{ν}*z*-axis (

*x*= cos

*θ*= ±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

*P*(

_{ν}*x*) for fractional

*ν*and the

*Q*(

_{ν}*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

*B*coefficients must be zero because all

_{ν}*Q*(

_{ν}*x*) functions lead to singularities. Moreover,

*ν*can only take integer values

*n*, so

*λ*=

*n*(

*n*+ 1) and

*Θ*(

*θ*) = Σ

*(cos*

_{n}A_{n}P_{n}*θ*). The first few Legendre polynomials

*P*(cos

_{n}*θ*) are

(D.21)

With *λ* = *n*(*n *+ 1), the solution for the equation for *R* becomes

(D.22)

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

__(D.23)__

To find the specific solution we must add the boundary conditions reading:

**I)** (*ϕ*_{1})_{r}_{→∞} = −*Er* cos*θ* because the field gradient far removed from the origin should approach the applied field ** E**.

**II)** (*ϕ*_{1})_{r}_{=a} = (*ϕ*_{2})_{r}_{=a} as *ϕ* should be continuous across the boundary.

**III)** *ε*_{1}(∂*ϕ*_{1}/∂*r*)_{r}_{=a} = *ε*_{2}(∂*ϕ*_{2}/∂*r*)_{r}_{=a} because at the boundary the normal dielectric displacement component *D***_{n}** should be continuous.

**IV)** *ϕ*_{2} 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 *D***_{n}** = 0 and condition I requires

*A*

_{1}= −

*E =*−|

**|, while all other**

*E**A*= 0. Conditions II and III say that, except for

_{n}*n*= 1,

(D.24)

From these equations it follows that *B _{n}* =

*C*= 0 for all

_{n}*n*≠ 1. For

*n*= 1 it is found that

(D.25)

Therefore

(D.26)

and *ϕ*_{1} and *ϕ*_{2} become

(D.27)

Because the potential due to the external charges is *ϕ* = −*Ez*, the contributions and due to the polarization become

(D.28)

The expression for is identical to that of an equivalent ideal dipole *μ*_{equ} 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

(D.29)

The field inside the sphere due to the polarization and the total field *E*_{2} inside the sphere are described by, respectively,

__(D.30)__

In case the sphere is a cavity so that *ε*_{1} = *ε* and *ε*_{2} = 1, the total field *E*_{2} becomes the *cavity field* *E*_{cav} = 3*ε E*/(2

*ε*+ 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

**. The reaction field will be proportional to**

*R***, having the same direction as**

*μ***so that**

*μ***=**

*R**f*. We choose the direction of the

**μ***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

*ε*

_{1}=

*ε*and

*ε*

_{2}= 1:

(D.31)

The *D*-terms in *ϕ*_{2} are due to the charges inside the cavity. Because the potential of an ideal dipole along the *z*-axis is , the contribution *D _{n}* = 0 for all

*n*except

*n*= 1, which reads . For

*ϕ*

_{2}, we thus have

__(D.32)__

From condition I, we conclude that *A _{n}* = 0 for all

*n*, so that

*ϕ*

_{1}becomes

__(D.33)__

From conditions II and III we obtain for all *n* ≠ 1,

(D.34)

which can be satisfied only if *B _{n}* =

*C*= 0 for

_{n}*n*≠ 1, while for

*n*= 1 the result is

(D.35)

Solving for *B*_{1} and *C*_{1} leads to

(D.36)

Substitution in __Eq. (D.32)__ and __(D.33)__ results in

__(D.37)__

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

(D.38)

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

__(D.39)__

Using again the potential due to the dipole given by , the contributions and due to the surface charges become

(D.40)

These potentials can be described by a dipole ** m** given by

(D.41)

The results in __Eq. (D.30)__ and __(D.39)__ are used in __Chapter 10__.

Further Reading

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.