6. Ch. 2, part 3

Lecture 6 - Fields in two dimensions, complex variable methods

We have already seen in Lecture 5 an example of two-dimensional solution of Laplace’s equation that is conveniently written in terms of the complex variable z = x+iy. Actually, any analytic function of z is a solution of Laplace’s equation in two dimensions. Let us prove this and show how the well-developed theory of functions of a complex variable is of great help in two-dimensional electrostatics.

6.1. Analytic functions and the Laplacian

If we make the coordinate transformation

z = x + iy
z^{∗} = x − iy

which has the inverse

x = \frac{1}{2}(z + z^{∗})
y = \frac{1}{2i}(z − z^{∗})

it is easy to show that for any twice-differentiable function F

\frac{∂^{2}F}{∂x^{2}} + \frac{∂^{2}F}{∂y^{2}} = 4\frac{∂^{2}F}{∂z∂z^{∗}}

Thus any function F (z) that depends only on z (and not on z^{∗}) is a solution of Laplace’s equation. So is its complex conjugate, as is shown by taking the complex conjugate of ∇^{2}F = 0. The real and imaginary parts of a complex F (z) are in general two distinct real solutions of Laplace’s equation. The physical potential of interest to us could be either the real part Φ or the imaginary part Ψ. Let us say it is Φ.

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Simplest examples

• If we take F = −Ez, we get Φ = −Ex, representing a uniform field E in the x direction. Note that Ψ gives a field E in the y direction. The lines of constant Φ are horizontal, the lines of constant Ψ are vertical: they are mutually orthogonal.

• If we take F = −2λ ln z, we get Φ = −2λ ln ρ, representing the field of a line charge λ. We also get Ψ = −2λθ (and what does it represent?). Ψ is constant on radial lines out of the origin: these lines are orthogonal to the lines of constant Φ, which are circles.

——————————————— 6.1.1. Field lines and the complex field

We see from the examples that the lines of constant Ψ are orthogonal to the lines of constant Φ (equipotentials). Thus the lines of constant Ψ are the field lines. This is a general property, due to the relations

\frac{∂F}{∂x} = \frac{dF}{dz}
\frac{∂F}{∂y} = i\frac{dF}{dz}

which imply

\frac{∂Φ}{∂x} = Re\frac{dF}{dz}  and  \frac{∂Ψ}{∂x} = Im\frac{dF}{dz}
\frac{∂Φ}{∂y} = −Im\frac{dF}{dz}  and  \frac{∂Ψ}{∂y} = Re\frac{dF}{dz}

from which it follows that

\frac{∂Φ}{∂x} \frac{∂Ψ}{∂x} + \frac{∂Φ}{∂y} \frac{∂Ψ}{∂y} = 0

i.e. ∇Φ · ∇Ψ = 0.

If we consider the line element dl, the flux through it (from left to right) is

\frac{∂Φ}{∂y} dx − \frac{∂Φ}{∂x} dy = −∇Ψ · dl

Hence the flux through a line extending from point A to point B is simply Ψ(A) − Ψ(B). As a check, for a line charge Ψ(θ) = −2λθ and Ψ(0) − Ψ(2π) = 4πλ, in agreement with Gauss. In fluid dynamics, Ψ is called the stream function.

We also see that

\frac{dF}{dz} = \frac{∂Φ}{∂x} − i\frac{∂Φ}{∂y}

so that we can find the cartesian components of E by computing −dF/dz. Finally

|E|^{2} = |\frac{dF}{dz} |^{2}

6.2. Edges and corners

If we examine the complex potential F = −iz^{ν} = −iρ^{ν} e^{iνθ} we see that

Re F = ρ^{ν} sin νθ

vanishes for θ = 0 and θ = π/ν. This is then the potential within a sector bounded by conductors making an angle β = π/ν. A complete set of solutions in the same corner region is of the form −iz^{nν} with n integer. Only the solutions with positive n are acceptable, unless the tip of the corner (the origin of coordinates) is excluded. An expansion of the (real) potential in this set is of the form

Φ(ρ, θ) = V + ∑_{n=1}^{∞} a_{n}ρ^{nπ/β} sin(nπθ/β)

as discussed by Jackson on page 77.

The basic solution −iz^{π/β} gives the complex field

\frac{iπ}{β} z^{(π/β)−1}

which has magnitude

|E| = \frac{π}{β} ρ^{(π/β)−1}

We see that the field does not penetrate well into a corner (β < π) , but becomes infinite at the tip of a wedge (β > π). The charge density is

σ = \frac{|E|}{4π} = \frac{1}{4β} ρ^{(π/β)−1}

and also diverges (but is integrable) at the tip of a wedge. See Fig. 2.13. In particular for the edge of a conducting half-plane (β = 2π), we have

σ ≈ |E| ≈ ρ^{−1/2}

To see how far from the edge this approximation extends, let us solve the physically interesting problem of a strip.

6.3. The conducting strip

For a charged metal strip, there is a simple solution. Consider a strip of width 2a, linear charge density λ, located between x = −a and x = a. The complex potential is F = −2λf, where

f = i arcsin(z/a)  (6.1)

The real potential is Φ = −2λϕ, where ϕ = Re f. Clearly ϕ vanishes on the strip, where z = x is real and x/a is between -1 and 1.

The complex field is 2λdf/dz, where

\frac{df}{dz} = \frac{i}{\sqrt{a^{2} − z^{2}}} = \frac{1}{\sqrt{z^{2} − a^{2}}}

Thus it has magnitude

|E| = \frac{2λ}{\sqrt[4]{(a^{2} − z^{2}) (a^{2} − z^{∗2})}}

where the radicand can be written as

a^{4} + |z|^{4} − a^{2}(z^{2} + z^{∗2}) = a^{4} + ρ^{4} − 2a^{2} (x^{2} − y^{2})

For large r/a we see that |E| approaches 2λ/ρ, as if the strip were just a line. On the strip, we get

E_{y} = 2λ Im \frac{df}{dz} |_{z=x±i0} = ± \frac{2λ}{\sqrt{a^{2} − x^{2}}}

where the + sign applies to the upper face. We have been careful in our approach to the real axis in the −a < x < a interval where the complex square root has a “cut”.

The charge density per unit area is

σ = \frac{λ}{π\sqrt{a^{2} − x^{2}}}

The integral of σ over the strip is λ as it should be. Note however that far from the edges σ is not close to its average value λ/a, as one may naively expect. Instead, it rises smoothly from a low value of λ/πa at x = 0 to an infinitely large value at x = ±a

.

Near the z = a edge we have

\frac{df}{dz} = \frac{1}{\sqrt{2a(z − a)}}

which can be integrated back to give

f = \frac{2}{\sqrt{2a}}\sqrt{z − a}

(Easier to do it this way than to tangle with the complex arcsin). We have recovered the known form of the potential near an edge, and we also have the strength:

Φ = −\frac{4λ}{\sqrt{a}}Re (ρ′e^{iθ′})^{1/2} = −4λ\sqrt{\frac{ρ′}{a}} cos (θ′/2)

where the primed coordinates are centered at the tip. Note that Φ is not proportional to the average charge per unit area λ/a.

6.3.1. The equipotentials

Plots of the equipotential lines, which are ellipses, are conveniently obtained from the equation

\frac{x^{2}}{cosh^{2} ϕ} + \frac{y^{2}}{sinh^{2} ϕ} = a^{2}  (6.2)

To show that this follows from f = i arcsin(z/a), rewrite the latter as

sin(−iϕ + ψ) = −i sinh ϕ cos ψ + cosh ϕ sin ψ = \frac{x}{a} + i \frac{y}{a}

from which

cos ψ = −\frac{y}{a sinh ϕ}
sin ψ = \frac{x}{a cosh ϕ}

Square and add to obtain (6.2) It is possible, but not instructive, to solve (6.2) explicitly for ϕ. Similarly, one can show that the field lines (constant ψ) are confocal hyperbolae.

As a bonus, we have obtained the solution for a charged ellipse or hyperbola. The problem could also be solved by using confocal elliptical coordinates, which are separable.

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Exercise

Obtain explicit expressions for the electric field components of a charged conducting strip of width 2a. (Put the strip, in cross section, on the (−a, a) segment of the x axis and give it a charge λ per unit length.) Find the limiting values of the field at large distances, on the two faces of the strip, and on a very small circle centered at x = a.

Exercise

Solve the problem of a slit of width 2a in a conducting plane having charge density σ far from the slit. (Put the slit, in cross section, on the (−a, a) segment of the x axis). Find the field near the edges.

Far below a conducting plane with a slit there is a uniform field E_{0} perpendicular to the plane. How much of the field leaks through the slit? Give explicit expressions for the field on the axial plane of the slit. (As above, put the slit on the (−a, a) segment of the x axis).

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6.4. Fringing fields

6.4.1. The plane capacitor edge

y
↑

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a ↕ − − − − − − − − − − − + −− → x

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If a is the spacing between the plates and V is the potential difference, the complex potential is

F = −\frac{iV}{2π} ln w

where w is given implicitly by

\frac{2πz}{a} = 1 + w + ln w  (6.3)

This solution seems difficult to handle, but the complex field can be expressed simply if we note that

\frac{2π}{a} \frac{dz}{dw} = 1 + \frac{1}{w}

We find

E_{x} − iE_{y} = −\frac{dF/dw}{dz/dw} = \frac{iV}{2π} \frac{1}{w + 1}

The charge on the outside of the plates due to the fringing field decreases inversely with the distance ℓ from the edge, for large ℓ/a. The total extra charge due to fringing is not finite: it grows logarithmically with ℓ. However, the percentage increase of capacitance due to fringing is still small for large ℓ/a. It is given approximately by

\frac{δC}{C} = \frac{1 + ln(2πℓ/a)}{2πℓ/a}  (6.4)

—————————— Proof of the capacitance formula

The charge on a segment of conducting contour from z_{1} to z_{2} is given by

4πQ = Im ∫ \frac{dF}{dz} dz = Ψ(z_{1}) − Ψ(z_{2})

It is enough to consider the top plate, at y = a/2, x ≤ 0. On the upper surface of this plate w varies from −∞ to −1, on the lower surface from −1 to 0. It is convenient to put z = ia/2−ℓ and w = −u′, so that we deal with positive quantities and Ψ = −(V/2π) ln u′. Equation (6.3) becomes

x′ + 1 = u′ − ln u′

where x′ = 2πℓ/a. The edge is at x′ = 0, u′ = 1 and Ψ = 0. Large distances from the edge (x′ >> 1) correspond to u′ >> 1 on the upper surface, u′ << 1 on the lower surface.

• On the upper surface for x′ >> 1 we have u′ ≈ x′ = 2πℓ/a, Ψ ≈ −(V/2π) ln(2πℓ/a). Recalling that Ψ = 0 at the edge, we see that the charge is Q_{upper} = (V/8π^{2}) ln(2πℓ/a). Surprisingly, this is not finite as ℓ → ∞, or in other words the charge density that “leaks” to the upper surface falls off slowly, like 1/ℓ.

• On the lower surface we have − ln u′ ≈ x′ + 1 = 2πℓ/a + 1 and Ψ ≈ (ℓ/a + 1/2π)V. Thus the charge is Q_{lower} = ℓV/4πa + V/8π^{2}. We see that an extra charge V/8π^{2} accumulates on the lower surface near the edge.

• The total charge on the plate is Q = Q_{upper} + Q_{lower} and the capacitance is

\frac{Q}{V} = \frac{ℓ}{4πa} + \frac{1}{8π^{2}} (1 + ln \frac{2πℓ}{a})

Since C = ℓ/4πa is the standard result when fringe effects are neglected, we see that the excess capacitance due to fringing fields is as given in eq. (6.4).

6.4.2. The interleaved capacitor

We look at the edge of one (movable) plate at potential V that is equidistant from two large (fixed) plates at zero potential.

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b ↕ —————————————————

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The extra capacitance due to fringing is as if the middle plate were moved inward by (b/π) ln 2, where b is the spacing between the fixed plates.

6.4.3. The cylindrical trough

The ratio of the charge inside to the charge outside is

\frac{π − α}{π + α}

where 2α is the subtended angle. For the half-cylinder, 1/4 of the charge is on the inside, 3/4 on the outside. For comparison, the formula for the spherical cap is (Weber, page 252):

\frac{sin α + α − (π/2) (1 − cos α)}{sin α + α + (π/2) (1 − cos α)}

For the half-sphere, we get 1/ (1 + π) = .241453