Chapter 1: Fundamentals Free

We start with Gauss's law and Maxwell's formulation of Faraday's law:

When charges are at rest (electrostatics) there is no magnetic field due to their motion; we also assume that any other magnetic field present is static. Hence the curl of E is zero, and the electric field can be written as the gradient of a potential,

It then follows from Gauss's law that the potential V satisfies Poisson's equation,

An electric field causes charges to move. In insulators the electrons and nuclei adjust locally; in conductors some of the electrons can move macroscopically. Hence in electrostatics the electric field inside a conductor is zero: the charges have moved to make it so. The potential is constant inside and on the conductor. It follows that any excess charge resides on the surface. (A metal contains nuclei and electrons, some of the latter being free to move. The bulk is neutral, with proton charges being cancelled by electron charges. In Gauss's law $∇⋅E=4πρ$ the charge density ρ is zero inside the conductor. It is understood that field and charge density have been averaged over a volume large enough to contain many nuclei and electrons, but still microscopic.)

Note that the electrostatic field is normal to a conducting surface: if there were a tangential component, charges would move. The surface charge density is proportional to the magnitude of the field, as shown in the following. Gauss's law in its integral form, namely that the integral of the scalar product of the field with an (outward-facing) surface element over a closed surface is equal to 4π times the charge enclosed by the surface, $∮dS⋅E=4πQ$⁠, applied to a cylindrical volume element with one end-face inside the conductor and the other just outside, gives the local surface charge density σ:

We recall some basic formulas. A single charge Q at the origin has the electric potential

It generates a radial (or central) electric field

A dipole p has the potential and field

A conducting sphere in an external field experiences charge separation: it becomes “polarized.” We shall take the external field to be E₀ = [0, 0, E₀] (pointing along the positive z direction). Then positive charge accumulates at the north pole of the sphere, negative charge at the south. Far from the sphere the resulting field due to this charge redistribution is that of a dipole, of magnitude p. The polarizability α is defined by p = αE₀. We shall see that the polarizability of an isolated conducting sphere of radius a is α = a³.

We wish to find the electric potential V(r) everywhere outside a sphere of radius a in the specified external field; it is constant on and inside the conducting sphere. The potential must be V = −E₀z far from the sphere, which we take to be centered on the origin. There are no charges outside the sphere, apart from those at infinity, which give rise to the specified external field. Hence the Poisson equation to be satisfied by the potential V(r) becomes the Laplace equation. In spherical polar coordinates (r, θ, ϕ) this reads

Here we need only solutions independent of the azimuthal angle ϕ; these are of the form V(r, θ) = rⁿP_n(cos θ), r⁻ⁿ⁻¹P_n(cos θ) where P_n(cos θ) are the Legendre polynomials, of which the first four are

The external potential is V_ext = −E₀z = −E₀r cos θ; its polar-angle dependence is purely in P₁ = cos θ. Hence we look for a solution in the form V = −E₀r cos θ + Ar⁻² cos θ + V₀. We can choose the constant potential on the sphere to be zero, in which case, for a conducting sphere of radius a,

The potential is zero on and inside the sphere, and correctly tends to the external potential at large distances. The term E₀a³r⁻² cos θ corresponds to that of a dipole at the center of the sphere, V(r) = p · r/r³, with p pointing in the direction of the external field, upwards in Fig. 1.1.

The normal component of the field at the surface of the sphere is found by taking the radial derivative of the potential function (1.11). It is

The electric field is three times larger than the applied external field E₀ at each pole. The electric field magnitude at the surface of the sphere is maximum at the poles, zero at the equator. There is an accumulation of positive charge at the north pole, and negative charge at the south pole, so at a distance the polarized sphere acts as a dipole.

Comparison of (1.12) with (1.8) shows that the magnitude of the dipole is p = E₀a³. The polarizability α is defined by p = αE₀, so the polarizability of an isolated conducting sphere of radius a is a³.

Capacitance is a measure of the ability to store charge. Usually it is applied to two or more conductors; the general formulation is due to Maxwell and will be discussed in Sec. 1.3. For two conductors, with charge separation leading to positive and negative charges +Q, −Q on them, the capacitance C(Q, −Q) is defined in terms of Q and the potential difference V between them:

One can also ask, what is the capacitance of two conductors kept at the same potential V? (They may be connected by a thin wire, or to a common voltage source, for example.) The total charge on the two conductors is then Q′, say, and the capacitance is

For a single isolated sphere with total charge Q on it, we can regard the second conductor as being at infinity, with charge −Q. Field lines diverge radially from the charged sphere. Outside the conducting sphere (for r > a where a is its radius), the potential is given by (1.6), so the potential on the sphere is V = Q/a. Hence the capacitance of an isolated sphere is equal to its radius:

Let C = C(Q, −Q) be the capacitance of a conducting body of maximum diameter D, surface area A, volume V, and integral of the mean curvature over the surface M. Maxwell shows that C ≤ D/2 (Maxwell, 1891 vol. 1, p 117). The other conductor is understood to be at infinity, as in the previous paragraph. It is known that A³ ≥ 36πV² for all solid bodies, and that M² ≥ 4πA and 2πD ≥ M ≥ 4πC for convex bodies (Szegö, 1945). Define the radii

Then R_M ≥ R_A ≥ R_V, the first inequality holding for convex bodies, the second for all bodies. Combined with Maxwell's inequality and the Poincaré theorem that the capacitance is greater or equal to R_V, we have the bounds

All these lengths, and R_A, are equal in the case of a spherical conductor.

The electrostatics of conductors is simple in that each is at a fixed potential: if the electric potential V(r) varied on the conductor, its gradient would cause an electric field $E=−∇V$⁠, and the mobile charges (which characterize a conductor) would flow until their redistribution gave a constant potential.

The energy of a system of N conductors with charges Q_i and potentials V_i is (Maxwell, 1891, Sec. 84; Landau and Lifshitz, 1960, Sec. 2; Jackson, 1975, Sec. 1.11)

The capacitance coefficients C_ij of a system of conductors are defined by the equations (Maxwell, 1891, Sec. 87)

If we know the capacitance coefficients, we can calculate the electrostatic energy for specified potentials V_i on the conductors:

Alternatively, we may specify the potentials, and invert the matrix C formed by the coefficients C_ij in Eq. (1.19), which we may write in vector form as Q = CV. Its inverse is V = C⁻¹Q, so the energy may be written as W = $12$Q · V = $12$(CV) · V = $12$Q · C⁻¹Q. In either case the energy will be a function of the relative positions of the conductors (through the position dependence of the capacitance coefficients), and by differentiations with respect to the coordinates we can find the forces acting on the conductors.

The work required to change the charges by amounts δQ_i is the work required to bring these charge increments from infinity (where the potential is zero) to the conductors:

The change in energy may also be expressed in terms of the change in potentials of the conductors (Landau and Lifshitz, 1960, Sec. 2):

From (1.21) and (1.22),

Hence

The reverse order of differentiation gives C_ij for the same quantity. Hence the capacitance matrix is symmetric:

In the case of two conductors a, b we need three coefficients: C_aa, C_ab, C_bb. From (1.19) the charges on the conductors are

When the conductors carry equal and opposite charges Q, −Q the potentials on them are

The capacitance of the two-conductor system is then the magnitude of the charge on each divided by the potential difference

Alternatively, the conductors may be kept at the same potential V. They then acquire charges

The total charge on the pair is then Q_a + Q_b = (C_aa + 2C_ab + C_bb)V. The capacitance of the pair is then

The capacitances C(Q, −Q) and C(V, V) are distinct physically; confusion between the two has led to errors, as discussed in Sec. 5 of Lekner (2011). We shall see in Chap. 4 that, for spheres in close approach, C(V, V) tends to a limit at contact (2a ln 2 in the case of equal radii), while C(Q, −Q) diverges logarithmically with the separation between the spheres. For spheres the capacitance coefficients depend on their radii a, b and on the distance between them. In general, the capacitance coefficients depend on the conductor geometric shapes together with the associated sizes, their separation, and their mutual orientation.

The electrostatic energy of a two-conductor system, with charges Q_a, Q_b, is, from (1.20),

Suppose now that the two conductors come close enough for an electrical short to occur, for example by electron tunneling at some local surface roughness. Charge will flow until the conductors are at the same potential V = (Q_a + Q_b)(C_aa + 2C_ab + C_bb)⁻¹, from (1.30). The new charges are, from (1.19),

The energy after contact is, on substituting the new charges into (1.31) and simplifying,

We expect $W~$ to be lower than the energy W before contact is made, otherwise charge would not flow. The electrostatic energy difference is

The factors in the denominator are both positive: if they were not, the physical capacitances C(Q, −Q) and C(V, V) defined in (1.28) and (1.30) would be negative. Hence

The capacitance inequalities in (1.35) are clearly true at large separation between the conductors, since then C_ab → 0 and C_aa, C_bb are positive. We shall verify the capacitance inequalities for two spheres, at any separation, in Sec. 4.2.

The equality of $W,W~$ occurs when the charges on the conductors are in the special ratio

Since charge is conserved, this implies that the energies before and after contact are equal when (and only when) $Qa=Qa~$ and $Qb=Qb~$⁠. The distribution of charge before and after contact can be very different. Nevertheless, the energy W can approach $W~$ when the capacitance coefficients are singular, as we shall see in Sec. 4.6.

We saw in Sec. 1.2 that a solitary sphere of radius a has polarizability α = a³: its dipole moment in external field E^ext is p = a³E^ext. Here we wish to consider the electric potential of a localized assembly of conductors carrying zero net charge. Instead of p = αE^ext with a scalar polarizability and dipole parallel to the external field, the polarizability becomes a tensor:

The polarization tensor is symmetric, α_ij = α_ji (Landau and Lifshitz, 1960, Sec. 11). Any non-spherical body has tensor polarizability, and so a localized assembly will also have tensor polarizability. The electrostatic energy of a neutral localized assembly of conductors is W = −$12$p · E^ext, as an extension of the argument of Landau and Lifshitz (1960) Sec. 2, for a single conductor shows:

A uniform field may be represented as due to a charge Q_∞ at infinity, and the energy of the system is W = $12$Q_∞V_∞, where V_∞ is the potential at the remote charge due to the uncharged conductors. The external field E^ext causes the conductors to acquire a dipole moment p, the potential of which is p · r/r³. Hence W = $12$Q_∞p · r/r³. However, −Q_∞r/r³ is just the field at the conductor due to Q_∞. Hence W = −$12$p · E^ext.

The dipole moment of the assembly is specified by the polarizability tensor α_ij, in (1.37). Hence the electrostatic energy of the assembly in the external field E^ext is

Only the dipole moment components of an uncharged system are needed to find the system energy in a uniform external field; the quadrupole and higher-order moments do not contribute. The higher multipoles are superpositions of equal and opposite pairs of dipoles, and thus have cancelling positive and negative terms.

The polarization tensor is real and symmetric and can thus be diagonalized. For simple systems it can be put in diagonal form by choice of coordinates. Suppose for example that for a two-sphere system we chose the z axis to coincide with the line through the sphere centers. Then the polarization tensor is

The transverse part is α_xx = α_yy = α_T (response to a field perpendicular to the line of centers) and the longitudinal part (response to a field along the line of centers) is α_zz = α_L. Suppose the angle between the external field and the line of centers is θ,

Then the electrostatic energy (1.38) is independent of the azimuthal angle ϕ:

There is no net force on the polarized two-sphere system, but there is a torque because of the angle-dependence. The magnitude of the torque is

The polarizability tensor of the two-sphere system will be discussed in detail in Secs. 3.3–3.5. We shall see that α_L > α_T and that the torque always acts to align the spheres with the field.

In cylindrical polar coordinates $(ρ,ϕ,z),ρ=x2+y2$ is the distance from the z axis. We shall consider a circular cylinder of radius a, with its axis coinciding with the z axis. The length 2b of the cylinder is taken to be much larger than a. The results of this section apply only to regions far (in comparison with a) from the ends of the cylinder, that is for |z| ≪ b, and also when ρ ≪ b. The dependence on the variable z can then be neglected. With this understood, we can derive a useful result for the polarizability per unit length. The capacitance per unit length is not defined, as we shall see. The difficult problems associated with a finite cylinder are left to Chap. 8.

The Laplace equation in cylindrical polar coordinates reads

If we neglect the dependence on z, as discussed above, the solutions are

First, we shall discuss the polarizability per unit length of the cylinder, because this is easier. The external field is taken to be perpendicular to the cylinder, say E^ext = [E₀, 0, 0]. The potential associated with the external field is −E₀ ρ cos ϕ, and the total potential, taken to be zero on the cylinder, is

To obtain the dipole moment and thus the polarizability, we need the concept of a two-dimensional dipole. This can be constructed from two parallel line charges, with charges per unit length ±q, as in Lekner (2013). From Gauss's law the field at distance ρ from a line charge +q is E = 2q/ρ, so the potential is V = constant −2q ln ρ. If the lines with charges ±q are at [±δx, 0], and $ρ±2=(x∓δx)2+y2=ρ2+2xδx+(δx)2$⁠, the potential due to the line charge pair is 2q ln (ρ₋/ρ₊). The line pair separation is 2δx and the two-dimensional dipole moment per unit length is p = 2qδx. We take the limit as δx → 0 of the potential, keeping p fixed:

Comparison with the potential (1.45) shows that the cylinder of radius a in an external field of strength E₀ has dipole moment p = $12$E₀a² and thus polarizability α per unit length

Figure 1.2 shows how an external field is modified by a long conducting cylinder (the external field is perpendicular to the cylinder axis).

Now we try to find the capacitance per unit length: let a cylinder of radius a carry a charge per unit length q. Because of the logarithmic nature of the potential the second conductor cannot be at infinity. Let it be a coaxial cylinder at some finite radius R, carrying a charge per unit length −q. The potential in a ≤ ρ < R is V(ρ) = 2q ln R/ρ; at the inner cylinder the potential is V₀ = 2q ln R/a, dependent on the radius of the large cylinder. The outer cylinder may be taken to be earthed, with potential zero. The capacitance per unit length of the cylinder pair, $qV0=12lnR/a$⁠, varies with R: there is no physical limit as R is increased to infinity. There is, however, a well-defined capacitance per unit length for a pair of parallel cylinders, as we have just seen for two coaxial cylinders, and as we shall find in Chap. 7 for the general case.

For the capacitance associated with an isolated cylinder, we need to treat the finite cylinder, as we shall in Chap. 8. Maxwell (1877) showed that for a cylinder of diameter 2a and length 2b the capacitance has the form, when b ≫ a,

The capacitance per unit length C/2b depends on the length to diameter ratio, logarithmically. The seemingly reasonable assumption of a constant charge per unit length is wrong, although it provides a first approximation when the cylinder is long. The charge density increases toward the ends of the cylinder, and this non-uniformity cannot be ignored (Griffiths and Ye 1995, Jackson 2000, 2001).

The surface of a conductor experiences an outward force per unit area (a negative pressure), equal in magnitude to the energy density of the field just outside the conductor (Landau and Lifshitz, 1960, Sec. 5; Jackson, 1975, Sec. 1.11; Griffiths, 1981, Sec. 2.5.3). The energy density is u = (8π)⁻¹E² in the immediate neighborhood of the surface. A small outward displacement Δr of an element of area ΔA produces an electrostatic energy decrease ΔU = −uΔrΔA. The outward force per unit area thus has magnitude u:

We illustrate with a simple but perhaps surprising example, based on the conducting sphere in an external field E₀ of Sec. 1.2. Suppose, as in Problem 3 of Chap. 1 of Landau and Lifshitz (1960), that the sphere is cut in half at the equator, the external field pointing south to north as in Fig. 1.1. The separated hemispheres retain their positive and negative surface charge densities (it is assumed that the separation is small on the scale of the sphere radius a). By Gauss's law the surface charge density is related to the field component normal to the surface by σ = E_n/4π. Using the expression for the normal field component given in (1.12), namely E_n(θ) = 3E₀ cos θ, the outward force per unit area is

The net force acting on the northern hemisphere is thus, using dA = 2πa² sin θ dθ,

The force acts northward (upward). An equal and opposite force acts on the southern hemisphere.

But wait, you say, surely there is an attractive force between the northern positive charges and the southern negative charges? Yes, but these charges also experience the pull of the external field. What we have calculated are the forces on the conductors, the two separated hemispheres.

Suppose now that the hemispheres move apart, the northern one moving north, the southern one moving south. The charges redistribute progressively, but each retains its total charge, the one on the northern hemisphere being

When the hemispheres are far apart (on the scale of the radius a) the force on the northern one tends to

This force is larger than the force (1.51) experienced when the hemispheres had barely separated, by the factor 4/3.

From a mathematical point of view, all of this book is devoted to solving Laplace's equation, with the boundary conditions that the solution (the electric potential) be constant on some surfaces, usually cylinders or spheres. The flow of heat and the flow of charge (in uniform media with constant conductivity) also obey Laplace's equation: let J denote the electric current density, or the heat flow vector, and K the electrical or thermal conductivity. For steady (time-independent) flow $∇⋅J=0$⁠, and $J=−Ke∇V$ or $J=−Kt∇T$⁠. Thus, if K is constant, $∇2V=0$ or $∇2T=0$⁠, where V and T are the electric potential and the temperature.

Suppose that we have solved the electrostatic problem of two oppositely charged conductors a and b, at potentials V_a, V_b, and found the capacitance C(Q, −Q) of (1.28), C(Q, −Q) = Q/(V_a − V_b). Then we have also solved the problem of steady flow between these two conductors (of electricity or heat), provided their conductivity is much larger than that of the medium surrounding them: The electric flux through a surface surrounding the conductor a with charge Q is, from Gauss's law, $∮dS⋅E=4πQ=4πC(Q,−Q)(Va−Vb)$⁠. By direct analogy, the total electric current through a surface surrounding the conductor a with potential V_a, which ends up going into the conductor b with potential V_b, is $I=∮dS⋅J$⁠.

Since $E=−∇V,J=−Ke∇V$⁠, the current is given by the same capacitance, I = 4πK_eC(Q, −Q) (V_a − V_b). In the heat flow case, the total heat current is H = 4πK_tC(Q, −Q)(T_a − T_b). The electrical or thermal resistances to the flow are thus

The larger the capacitance, the smaller the resistance. In Gaussian units the capacitance has the dimension of length. Roughly speaking, we expect the resistance to be proportional to the distance between the conductors, and inversely proportional to the effective area of flow. Equation (1.54) gives the precise dependence.

As a simple example of (1.54), consider the flow from a sphere of radius a to the sphere at infinity. The capacitance is a from Sec. 1.2, so the resistance to flow from the (highly conducting) sphere to infinity through a medium of conductivity K is (4πKa)⁻¹.

In problems involving long cylinders, the appropriate capacitance per unit length is C(q, −q) where ±q are charges per unit length on the two cylindrical surfaces, and the flows and resistances are then per unit length values also. In such cases, C(q, −q) is dimensionless. One example we shall meet shortly in Appendix  1B: two long confocal elliptic cylinders. The equipotentials are confocal ellipses, the field (and flow) lines lie on confocal hyperbolas. The capacitance per unit length is given in (1B.9). Another cylinder example will be considered in Secs. 1.8 and 7.2: two parallel circular cylinders of radii a, b with distance c between their axes. The capacitance per unit length is, from (7.7) and (7.15),

The first expression applies when the cylinders are exterior to each other, the second when one encloses the other. The formulas (1.54), with the capacitances (1.55) substituted, now give the resistance per unit length of the cylinders. Feynman et al. (1964), in Chap. 12, “Electrical analogs,” discusses flow and other problems that are equivalent to electrostatic ones.

The young William Thomson (later Baron Kelvin of Largs, often referred to as Kelvin) introduced the idea of fictitious charges arranged to make a chosen surface an equipotential (Thomson, 1872). In the simplest case of a charge near a conducting plane surface, the plane is made to be at zero potential by an equal and opposite charge placed equidistant on the normal to the plane at the mirror image point of the original charge; hence the name “image charge.” As in all examples to follow, what actually happens is that charge moves on the conductor to make it an equipotential. However, knowing the location and magnitude of the fictitious charges enables calculation of all electrostatic properties.

A two-dimensional problem, similar to the equal and opposite point charges above, is that of two parallel line charges of opposite sign. From Gauss's law, or from Sec. 1.5, the field due to a line charge (charge q per unit length) is equal to 2q/r at distance r from the line. The potential due to line charges q and −q at a point distant by r_a and r_b from the two line charges is 2q ln r_b/r_a. The potential is constant when r_b/r_a is constant. As we shall see, this is on circles in planes perpendicular to the parallel line charges. The potential is thus constant on cylinders with axes parallel to the line charges. We take the line charges to intersect the [x, y] plane at [±ℓ, 0]; then

For constant r_b/r_a = e^u, we find

Suppose the cylinders have radii a, b, and their axes are separated by distance c. The cylinder surfaces are at u_a and −u_b, where sinh u_a = ℓ/a, sinh u_b = ℓ/b; their centers are at $xa=ℓcothua,xb=−ℓcothub$⁠. Hence the distance between the cylinder axes is

The outer relation gives the value of ℓ in terms of a, b, c:

Figure 1.3 illustrates the two-cylinder geometry.

Since r_b/r_a = e^u, V = 2qu. The cylinder surfaces are at u_a and −u_b; the potential difference is therefore V_a − V_b = 2q(u_a + u_b) = 2q(arcsinh ℓ/a + arcsinh ℓ/b). Thus the capacitance per unit length of the cylinder pair is

This problem is discussed in detail in Sec. 7.2; the solution given there is in bicylindrical coordinates u, v, one of which was used here to simplify the algebra. The first of the expressions given in (1.55) is equivalent to (1.60), as shown in Sec. 7.2.

A related problem is that of a charged cylinder alongside an earthed conducting plane. Suppose that cylinder of radius a is near a conducting surface at x = 0. Line charges ±q at [±ℓ, 0] will give zero potential on the x = 0 plane, and potential V_a = 2qu_a = 2q arcsinh ℓ/a on the cylinder. Let d be the distance of the cylinder axis from the conducting plane (d = c/2 in our previous notation). The image of our cylinder is another cylinder of radius a with axis at [−d, 0], so ℓ² = d² − a² from (1.58). The capacitance per unit length of the cylinder-plane combination is therefore

When the cylinder is far from the plane on the scale of its radius, the capacitance tends to

The opposite limit is that of a cylinder close to the conducting plane. We set d = a + s and expand in powers of s, the closest cylinder-plane separation, to find

Suppose a positive charge Q is placed at distance d from the center of a conducting sphere of radius a, carrying zero charge. The external charge will polarize the sphere: negative charges will go to the side facing Q, leaving positive charge on the opposite side, and the sphere surface an equipotential. A charge Q′ = −(a/d)Q placed at distance a²/d from the center of the sphere on the line toward the external charge will make the sphere surface an equipotential also: if r, r′ are the distances from Q, Q′ to a point on the surface, the potential there is a constant plus $Qr+Q′r′=Qr(1−rr′ad)$⁠. But it follows from the lengths given that r′/r = a/d, so the potential is constant on the sphere. Just as for cylinders, discussed above, r′/r = constant corresponds to the surface of a sphere, in this case the sphere with radius a and center at the origin. [See Eq. (1.56), to which we can add a z² term in three dimensions.]

The same construction works if the charge Q is internal to the sphere, d < a.

The case of two spheres cutting orthogonally is the subject of Sec. 168 of Maxwell's Treatise (Maxwell, 1891) (see also Smythe, 1950, Sec. 5.103). Figure 1.4 gives the geometry.

We first note that the radius of the circle of intersection is $ℓ=ab/c=ab/a2+b2$⁠: the area of the right-angled triangle in the figure is $12$ℓc = $12$ab. Hence the center of the a circle is at z_a = a²/c, and the center of the b circle is at −z_b, where z_b = b²/c. It follows that the origin is the image point of the center of sphere a with respect to sphere b, and also the image point of the center of sphere b with respect to sphere a. (These relations are special to two spheres intersecting at right angles.) If charges Q_a = qa, Q_b = qb (that is, charges proportional to the sphere radii) are placed at the sphere centers, and Q′ = −q ab/c is placed at the origin, both spheres will be equipotential surfaces, at potential q. Note q is a charge per unit length, the same dimension as electric potential. The capacitance of the outer conducting surface formed by two spheres intersecting at right angles is thus

The equipotentials may be plotted as those of the three charges Q_a, Q_b, Q′. Figure 1.5 shows the equipotentials drawn for the configuration of Fig. 1.4, with a : b : c = 3 : 4 : 5.

The method of images is simplest in the case given, when the spheres intersect at right angles. One image charge Q′ and two charges Q_a, Q_b at the centers of the two spheres together make the composite body an equipotential. The method of images also works for angles of intersection π/n, n = 2, 3, 4, ….(n = 1 corresponds to the smaller sphere entirely inside the bigger sphere, and n → ∞ to touching spheres, in tangential contact). For finite n a finite number of images is needed (Tong, 1996; Palaniappan, 1999 and Palaniappan and Felderhof, 1999).

Two-dimensional electrostatic problems may often be most easily solved by complex analysis. We give the main results here, and shall illustrate some of the methods by considering a long conducting cylinder of elliptical cross-section in Appendix  1B.

Any differentiable function w of the complex variable z = x + iy satisfies Laplace's equation: let primes denote differentiation with respect to z, then

Cauchy–Riemann equations. Differentiation of w = u + iv with respect to x and y gives

The Cauchy–Riemann equations follow on equating the real and imaginary parts of the two expressions for w′ in (1.66):

That the equations $(∂x2+∂y2)(u,v)=0$ also hold follows by differentiation of (1.67).

The curves along which u is constant are orthogonal to the curves along which v is constant. Constant u implies dx ∂_xu + dy ∂_yu = 0; the slope is thus dy/dx = −∂_xu/∂_yu. Likewise, the slope of a v = constants curve is −∂_xv/∂_yv, and the product of the slopes equals −1, from the Cauchy–Riemann equations, proving that the curves cross at right angles.

If the dimensionless function v(x, y) is proportional to the potential, the curves of constant u(x, y) will give the electric field lines. This is because the electric field lines are given by dx/E_x = dy/E_y and E is proportional to the gradient of v, so dx/∂_xv = dy/∂_yv, giving the slope dy/dx = ∂_yv/∂_xv = −∂_xu/∂_yu, namely that of curves of constant u. Equipotentials and field lines are orthogonal, and orthogonality is preserved in the mapping from x, y to u, v. This is true of the angle of intersection of any pair of corresponding curves in the two spaces; hence the name “conformal mapping.”

There is a nice way of getting the electric field from the complex potential w = u + iv (see for example Smythe, 1950 Sec. 4.11): with z = x + iy as before,

The numerator on the right is equal to dx + idy times ∂_xu − i ∂_yu, or to dx + idy times ∂_yv + i ∂_xv, from the Cauchy–Riemann equations (1.67). Hence

Suppose v is proportional to the potential. Then the vector $e=−∇v$ is proportional to the electric field, with the same proportionality constant, e_x = −∂_xv, e_y = −∂_yv. Hence the electric field and its magnitude are given by