A Treatise on Electricity and Magnetism/Part I/Chapter XII

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A Treatise on Electricity and Magnetism by James Clerk Maxwell Part I, Chapter XII: Conjugate Functions in Two Dimensions

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CHAPTER XII.

THEORY OF CONJUGATE FUNCTIONS IN TWO DIMENSIONS.

182.] The number of independent cases in which the problem of electrical equilibrium has been solved is very small. The method of spherical harmonics has been employed for spherical conductors, and the methods of electrical images and of inversion are still more powerful in the cases to which they can be applied. The case of surfaces of the second degree is the only one, as far as I know, in which both the equipotential surfaces and the lines of force are known when the lines of force are not plane curves.

But there is an important class of problems in the theory of electrical equilibrium, and in that of the conduction of currents, in which we have to consider space of two dimensions only.

For instance, if throughout the part of the electric field under consideration, and for a considerable distance beyond it, the surfaces of all the conductors are generated by the motion of straight lines parallel to the axis of $z$, and if the part of the field where this ceases to be the case is so far from the part considered that the electrical action of the distant part on the field may be neglected, then the electricity will be uniformly distributed along each gene rating line, and if we consider a part of the field bounded by two planes perpendicular to the axis of $z$ and at distance unity, the potential and the distribution of electricity will be functions of $x$ and $y$ only.

If $\rho\ dx\ dy$ denotes the quantity of electricity in an element whose base is $dx\ dy$ and height unity, and $\rho\ ds$ the quantity on an element of area whose base is the linear element $ds$ and height unity, then the equation of Poisson may be written

$\frac{d^{2}V}{dx^{2}}+\frac{d^{2}V}{dy^{2}}+4\pi\rho=0$.

When there is no free electricity, this is reduced to the equation of Laplace,

$\frac{d^{2}V}{dx^{2}}+\frac{d^{2}V}{dy^{2}}=0$

The general problem of electric equilibrium may be stated as follows:–

A continuous space of two dimensions, bounded by closed curves $C_{1},\ C_{2}$, &c. being given, to find the form of a function, $V$, such that at these boundaries its value may be $V_{1},\ V_{2}$, &c. respectively, being constant for each boundary, and that within this space $V$ may be everywhere finite, continuous, and single valued, and may satisfy Laplace's equation.

I am not aware that any perfectly general solution of even this question has been given, but the method of transformation given in Art. 190 is applicable to this case, and is much more powerful than any known method applicable to three dimensions.

The method depends on the properties of conjugate functions of two variables.

Definition of Conjugate Functions.

183.] Two quantities $\alpha$ and $\beta$ are said to be conjugate functions of $x$ and $y$, if $\alpha+\sqrt{-1}\beta$ is a function of $x+\sqrt{-1}y$.

It follows from this definition that

$\frac{d\alpha}{dx}=\frac{d\beta}{dy},\ \mathrm{and}\ \frac{d\alpha}{dy}+\frac{d\beta}{dx}=0$;

(1)

$\frac{d^{2}\alpha}{dx^{2}}+\frac{d^{2}\alpha}{dy^{2}}=0,\ \frac{d^{2}\beta}{dx^{2}}+\frac{d^{2}\beta}{dy^{2}}=0$

(2)

Hence both functions satisfy Laplace's equation. Also

$\frac{d\alpha}{dx}\frac{d\beta}{dy}-\frac{d\alpha}{dy}\frac{d\beta}{dx}=\left(\frac{d\alpha}{dx}\right)^{2}+\left(\frac{d\alpha}{dy}\right)^{2}=\left(\frac{d\beta}{dx}\right)^{2}+\left(\frac{d\beta}{dy}\right)^{2}=R^{2}$.

(3)

If $x$ and $y$ are rectangular coordinates, and if $ds_{1}$ is the intercept of the curve ($\beta$ = constant) between the curves $\alpha$ and $\alpha+d\alpha$, and $ds_{2}$ the intercept of a between the curves $\beta$ and $\beta+d\beta$, then

$\frac{ds_{1}}{d\alpha}=\frac{ds_{2}}{d\beta}=\frac{1}{R}$

(4)

and the curves intersect at right angles.

If we suppose the potential $V=V_{0}+k\alpha$, where $k$ is some constant, then $V$ will satisfy Laplace's equation, and the curves ($\alpha$) will be equipotential curves. The curves ($\beta$) will be lines of force, and the surface-integral of a surface whose projection on the plane of $x<math>$y</math> is the curve $AB$ will be $k\left(\beta_{B}-\beta_{A}\right)$, where $\beta_{A}$ and $\beta_{B}$ are the values of $\beta$ at the extremities of the curve.

If a series of curves corresponding to values of $\alpha$ in arithmetical progression is drawn on the plane, and another series corresponding to a series of values of $\beta$ having the same common difference, then the two series of curves will everywhere intersect at right angles, and, if the common difference is small enough, the elements into which the plane is divided will be ultimately little squares, whose sides, in different parts of the field, are in different directions and of different magnitude, being inversely proportional to $R$.

If two or more of the equipotential lines ($\alpha$) are closed curves enclosing a continuous space between them, we may take these for the surfaces of conductors at potentials $\left(V_{0}+k\alpha_{1}\right)$, $\left(V_{0}+k\alpha_{2}\right)$, &c. respectively. The quantity of electricity upon any one of these between the lines of force $\beta_{1}$ and $\beta_{2}$ will be $\tfrac{k}{4\pi}\left(\beta_{2}-\beta_{1}\right)$.

The number of equipotential lines between two conductors will therefore indicate their difference of potential, and the number of lines of force which emerge from a conductor will indicate the quantity of electricity upon it.

We must next state some of the most important theorems relating to conjugate functions, and in proving them we may use either the equations (1), containing the differential coefficients, or the original definition, which makes use of imaginary symbols.

184.] THEOREM I. If $x'$ and $y'$ are conjugate functions with respect to $x$ and $y$, and if $x''$ and $y''$ are also conjugate functions with respect to $x$ and $y$, then the functions $x'+x''$ and $y'+y''$ will be conjugate functions with respect to $x$ and $y$.

For

$\frac{dx'}{dx}=\frac{dy'}{dy}$ and $\frac{dx''}{dx}=\frac{dy''}{dy}$;

therefore

$\frac{d(x'+x'')}{dx}=\frac{d(y'+y'')}{dy}$.

Also

$\frac{dx'}{dy}=-\frac{dy'}{dx}$ and $\frac{dx''}{dy}=-\frac{dy''}{dx}$;

therefore

$\frac{d(x'+x'')}{dy}=-\frac{d(y'+y'')}{dx}$;

or $x+x''$ and $y'+y''$ are conjugate with respect to $x$ and $y$.

Graphic Representation of a Function which is the Sum of Two Given Functions.

Let a function ($\alpha$) of $x$ and $y$ be graphically represented by a series of curves in the plane of $xy,$ each of these curves corresponding to a value of a which belongs to a series of such values increasing by a common difference, $\delta$.

Let any other function, $\beta$, of $x$ and $y$ be represented in the same way by a series of curves corresponding to a series of values of $\beta$ having the same common difference as those of $\alpha$.

Then to represent the function $\alpha+\beta$ in the same way, we must draw a series of curves through the intersections of the two former series from the intersection of the curves ($\alpha$) and ($\beta$) to that of the curves $(\alpha+\delta)$ and $(\beta-\delta)$, then through the intersection of $(\alpha+2\delta)$ and $(\beta-2\delta)$, and so on. At each of these points the function will have the same value, namely $\alpha+\beta$. The next curve must be drawn through the points of intersection of $\alpha$ and $\alpha+\delta$, of $\alpha+\delta$ and $\beta$, of $\alpha+2\delta$ and $\beta-\delta$, and so on. The function belonging to this curve will be $\alpha+\beta+\delta$.

In this way, when the series of curves ($\alpha$) and the series ($\beta$) are drawn, the series $(\alpha+\beta)$ may be constructed. These three series of curves may be drawn on separate pieces of transparent paper, and when the first and second have been properly superposed, the third may be drawn.

The combination of conjugate functions by addition in this way enables us to draw figures of many interesting cases with very little trouble when we know how to draw the simpler cases of which they are compounded. We have, however, a far more powerful method of transformation of solutions, depending on the following theorem.

185.] THEOREM II. If $x''$ and $y''$ are conjugate functions with respect to the variables $x'$ and $y'$, and if $x'$ and $y'$ are conjugate functions with respect to $x$ and $y$, then $x''$ and $y''$ will be con jugate functions with respect to $x$ and $y$.

For

$\begin{array}{ll} \frac{dx''}{dx} & =\frac{dx''}{dx'}\frac{dx'}{dx}+\frac{dx''}{dy'}\frac{dy'}{dx}\\ \\ & =\frac{dy''}{dy'}\frac{dy'}{dy}+\frac{dy''}{dx'}\frac{dx'}{dy}\\ \\ & =\frac{dy''}{dy}; \end{array}$

and

$\begin{array}{ll} \frac{dx''}{dy} & =\frac{dx''}{dx'}\frac{dx'}{dy}+\frac{dx''}{dy'}\frac{dy'}{dy}\\ \\ & =-\frac{dy''}{dy'}\frac{dy'}{dx}-\frac{dy''}{dx'}\frac{dx'}{dx}\\ \\ & =-\frac{dy''}{dx}; \end{array}$

and these are the conditions that $x''$ and $y''$ should be conjugate functions of $x$ and $y$.

This may also be shewn from the original definition of conjugate functions. For $x''+\sqrt{-1}y''$ is a function of $x'+\sqrt{-1}y'$, and $x'+\sqrt{-1}y'$ is a function of $x+\sqrt{-1}y$. Hence, $x''+\sqrt{-1}y''$ is a function of $x+\sqrt{-1}y$.

In the same way we may shew that if $x'$ and $y'$ are conjugate functions of $x$ and $y$, then $x$ and $y$ are conjugate functions of $x'$ and $y'$.

This theorem may be interpreted graphically as follows:–

Let $x', y'$ be taken as rectangular coordinates, and let the curves corresponding to values of $x''$ and of $y''$ taken in regular arithmetical series be drawn on paper. A double system of curves will thus be drawn cutting the paper into little squares. Let the paper be also ruled with horizontal and vertical lines at equal intervals, and let these lines be marked with the corresponding values of $x'$ and $y'$.

Next, let another piece of paper be taken in which $x$ and $y$ are made rectangular coordinates and a double system of curves $x', y'$ is drawn, each curve being marked with the corresponding value of $x'$ or $y'$. This system of curvilinear coordinates will correspond, point for point, to the rectilinear system of coordinates $x', y'$ on the first piece of paper.

Hence, if we take any number of points on the curve $x''$ on the first paper, and note the values of $x', y'$ at these points, and mark the corresponding points on the second paper, we shall find a number of points on the transformed curve $x''$. If we do the same for all the curves $x'', y''$ on the first paper, we shall obtain on the second paper a double series of curves $x'', y''$ of a different form, but having the same property of cutting the paper into little squares.

186.] THEOREM III. If $V$ is any function of $x'$ and $y'$, and if $x'$ and $y'$ are conjugate functions of $x$ and $y$, then

$\iint\left(\frac{d^{2}V}{dx^{2}}+\frac{d^{2}V}{dy^{2}}\right)dx\ dy=\iint\left(\frac{d^{2}V}{dx'^{2}}+\frac{d^{2}V}{dy'^{2}}\right)dx'\ dy'$,

integration being between the same limits.

For

$\begin{array}{c} \frac{dV}{dx}=\frac{dV}{dx'}\frac{dx'}{dx}+\frac{dV}{dy'}\frac{dy'}{dx},\\ \\ \frac{d^{2}V}{dx^{2}}=\frac{d^{2}V}{dx'^{2}}\left(\frac{dx'}{dx}\right)^{2}+2\frac{d^{2}V}{dx'dy'}\frac{dx'}{dx}\frac{dy'}{dx}+\frac{d^{2}V}{dy'^{2}}\left(\frac{dy'}{dx}\right)^{2}+\frac{dV}{dx'}\frac{d^{2}x'}{dx^{2}}+\frac{dV}{dy'}\frac{d^{2}y'}{dx^{2}}; \end{array}$;

and

$\frac{d^{2}V}{dy^{2}}=\frac{d^{2}V}{dx'^{2}}\left(\frac{dx'}{dy}\right)^{2}+2\frac{d^{2}V}{dx'dy'}\frac{dx'}{dy}\frac{dy'}{dy}+\frac{d^{2}V}{dy'^{2}}\left(\frac{dy'}{dy}\right)^{2}+\frac{dV}{dx'}\frac{d^{2}x'}{dy^{2}}+\frac{dV}{dy'}\frac{d^{2}y'}{dy^{2}}$.

Adding the last two equations, and remembering the conditions of conjugate functions (1), we find

$\begin{array}{ll} \frac{d^{2}V}{dx^{2}}+\frac{d^{2}V}{dy^{2}} & =\frac{d^{2}V}{dx'^{2}}\left(\left(\frac{dx'}{dx}\right)^{2}+\left(\frac{dx'}{dy}\right)^{2}\right)+\frac{dV}{dy'^{2}}\left(\left(\frac{dy'}{dx}\right)^{2}+\left(\frac{dy'}{dy}\right)^{2}\right)\\ \\ & =\left(\frac{d^{2}V}{dx'^{2}}+\frac{d^{2}V}{dy'^{2}}\right)\left(\frac{dx'}{dx}\frac{dy'}{dy}-\frac{dx'}{dy}\frac{dy'}{dx}\right) \end{array}$

Hence

$\begin{array}{ll} \iint\left(\frac{d^{2}V}{dx^{2}}+\frac{d^{2}V}{dy^{2}}\right)dx\ dy & =\iint\left(\frac{d^{2}V}{dx'^{2}}+\frac{d^{2}V}{dy'^{2}}\right)\left(\frac{dx'}{dx}\frac{dy'}{dy}-\frac{dx'}{dy}\frac{dy'}{dx}\right)dx'\ dy'\\ \\ & =\iint\left(\frac{d^{2}V}{dx'^{2}}+\frac{d^{2}V}{dy'^{2}}\right)dx'\ dy' \end{array}$

If $V$ is a potential, then, by Poisson's equation

$\frac{d^{2}V}{dx^{2}}+\frac{d^{2}V}{dy^{2}}+4\pi\rho=0$

and we may write the result

$\iint\rho\ dx\ dy=\iint\rho'\ dx'\ dy'$,

or the quantity of electricity in corresponding portions of two systems is the same if the coordinates of one system are conjugate functions of those of the other.

Additional Theorems on Conjugate Functions.

187.] THEOREM IV. If $x_{1}$ and $y_{1}$, and also $x_{2}$ and $y_{2}$ are conjugate functions of $x$ and $y$, then, if

$X=x_{1}x_{2}-y_{1}y_{2}$ and $Y=x_{1}y_{2}+x_{2}y_{1}$

$X$ and $T$ will be conjugate functions of $x$ and $y$.

For

$X+\sqrt{-1}Y=\left(x_{1}+\sqrt{-1}y_{1}\right)\left(x_{2}+\sqrt{-1}y_{2}\right)$

THEOREM V. If $\phi$ be a solution of the equation

$\frac{d^{2}\phi}{dx^{2}}+\frac{d^{2}\phi}{dy^{2}}=0$,

and if

$2R=\log\left(\left(\frac{d\phi}{dx}\right)^{2}+\left(\frac{d\phi}{dy}\right)^{2}\right)$, and $\Theta=\tan^{-1}\frac{\frac{d\phi}{dx}}{\frac{d\phi}{dy}}$,

$R$ and $\Theta$ will be conjugate functions of $x$ and $y$.

For $R$ and $\Theta$ are conjugate functions of $\tfrac{d\phi}{dx}$ and $\tfrac{d\phi}{dy}$, and these are conjugate functions of $x$ and $y$.

EXAMPLE I. – Inversion.

188.] As an example of the general method of transformation let us take the case of inversion in two dimensions.

If $O$ is a fixed point in a plane, and $OA$ a fixed direction, and if $r=OP=ae^{\rho}$, and $\theta=AOP$, and if $x, y$ are the rectangular coordinates of $P$ with respect to $O$,

$\left.\begin{array}{lll} \rho=\log\frac{1}{a}\sqrt{x^{2}+y^{2}}, & & \theta=\tan^{-1}\frac{y}{x},\\ x=ae^{\rho}\cos\theta, & & y=ae^{\rho}\sin\theta, \end{array}\right\}$

(5)

$\rho$ and $\theta$ are conjugate functions of $x$ and $y$.

If $\rho'=n\rho$ and $\theta'=n\theta$, $\rho'$ and $\theta'$ will be conjugate functions of $\rho$ and $\theta$. In the case in which $n=-1$ we have

$r'=\frac{a^{2}}{r},\ \mathrm{and}\ \theta'=-\theta$,

which is the case of ordinary inversion combined with turning the figure 180° round $OA$.

Inversion in Two Dimensions.

In this case if $r$ and $r'$ represent the distances of corresponding points from $O$, $e$ and $e'$ the total electrification of a body, $S$ and $S'$ superficial elements, $V$ and $V'$ solid elements, $\sigma$ and $\sigma'$ surface- densities, $\rho$ and $\rho'$ volume densities, $\phi$ and $\phi'$ corresponding potentials,

$\left.\begin{array}{c} \frac{r'}{r}=\frac{S'}{S}=\frac{a^{2}}{r^{2}}=\frac{r'^{2}}{a^{2}},\ \frac{V'}{V}=\frac{a^{4}}{r^{4}}=\frac{r'^{4}}{a^{4}},\\ \\ \frac{e'}{e}=1,\ \frac{\sigma'}{\sigma}=\frac{r^{2}}{a^{2}}=\frac{a^{2}}{r'^{2}},\ \frac{\rho'}{\rho}=\frac{r^{4}}{a^{4}}=\frac{a^{4}}{r'^{4}},\\ \\ \frac{\phi'}{\phi}=1. \end{array}\right\}$

(7)

EXAMPLE II. Electric Images in Two Dimensions.

189.] Let $A$ be the centre of a circle of radius $AQ=b$, and let $E$ be a charge at $A$, then the potential at any point $P$ is

$\phi=2E\log\frac{b}{AP}$

(8)

Fig. 17.

and if the circle is a section of a hollow conducting cylinder, the surface-density at any point $Q$ is $-\tfrac{E}{2\pi b}$.

Invert the system with respect to a point $O$, making

$A=mb$ and $a^{2}=\left(m^{2}-1\right)b^{2}$;

then we have a charge at $A'$ equal to that at $A$, where $AA'=\tfrac{b}{m}$.

The density at $Q'$ is

$-\frac{E}{2\pi b}\frac{b^{2}-(AA')^{2}}{A'Q'^{2}}$

and the potential at any point $P'$ within the circle is

$\begin{array}{ll} \phi'=\phi & =2E(\log b-\log AP),\\ & =2E(\log OP'-\log A'P'-\log m). \end{array}$

(9)

This is equivalent to a combination of a charge $E$ at $A'$, and a charge $-E$ at $O$, which is the image of $A'$, with respect to the circle. The imaginary charge at is equal and opposite to that at $A'$.

If the point $P'$ is defined by its polar coordinates referred to the centre of the circle, and if we put

$\rho=\log r-\log b$ and $\rho_{0}=\log AA'-\log b$

then

$AP=be^{\rho},\ AA'=be^{\rho_{0}},\ AO=be^{-\rho_{0}}$;

(10)

and the potential at the point $(\rho,\theta)$ is

$\begin{array}{l} \phi=E\log\left(e^{-2\rho_{0}}-2e^{-\rho_{0}}e^{\rho}\cos\theta+e^{2\rho}\right)\\ \qquad-E\log\left(e^{2\rho_{0}}-2e^{-\rho_{0}}e^{\rho}\cos\theta+e^{2\rho}\right)+2E\rho_{0} \end{array}$

(11)

This is the potential at the point $(\rho,\theta)$ due to a charge $E$, placed at the point $(\rho_{0},0)$, with the condition that when $\rho=0,\ \phi=0$0.

In this case $\rho$ and $\theta$ are the conjugate functions in equations (5): $\rho$ is the logarithm of the ratio of the radius vector of a point to the radius of the circle, and $\theta$ is an angle.

The centre is the only singular point in this system of coordinates, and the line-integral of $\int\tfrac{d\theta}{ds}ds$ round a closed curve is zero or $2\pi$, according as the closed curve excludes or includes the centre.

EXAMPLE III. Neumann's Transformation of this Case.

190.] Now let $\alpha$ and $\beta$ be any conjugate functions of $x$ and $y$, such that the curves ($\alpha$) are equipotential curves, and the curves ($\beta$) are lines of force due to a system consisting of a charge of half a unit at the origin, and an electrified system disposed in any manner at a certain distance from the origin.

Let us suppose that the curve for which the potential is a is a closed curve, such that no part of the electrified system except the half-unit at the origin lies within this curve.

Then all the curves ($\alpha$) between this curve and the origin will be closed curves surrounding the origin, and all the curves ($\beta$) will meet in the origin, and will cut the curves ($\alpha$) orthogonally.

The coordinates of any point within the curve ($\alpha_{0}$) will be determined by the values of $\alpha$ and $\beta$ at that point, and if the point travels round one of the curves $\alpha$ in the positive direction, the value of $\beta$ will increase by $2\pi$ for each complete circuit.

If we now suppose the curve ($\alpha_{0}$) to be the section of the inner surface of a hollow cylinder of any form maintained at potential zero under the influence of a charge of linear density $E$ on a line of which the origin is the projection, then we may leave the external electrified system out of consideration, and we have for the potential at any point ($\alpha$) within the curve

$\phi=2E\left(\alpha-\alpha_{0}\right)$

(12)

and for the quantity of electricity on any part of the curve $\alpha_{0}$ between the points corresponding to $\beta_{1}$ and $\beta_{2}$,

$Q=2E\left(\beta_{1}-\beta_{2}\right)$

(13)

If in this way, or in any other, we have determined the distribution of potential for the case of a given curve of section when the charge is placed at a given point taken as origin, we may pass to the case in which the charge is placed at any other point by an application of the general method of transformation.

Let the values of $\alpha$ and $\beta$ for the point at which the charge is placed be $\alpha_{1}$ and $\beta_{1}$, then substituting in equation (11) $\alpha-\alpha_{0}$ for $\rho$, and $\beta-\beta_{1}$ for $\theta$, we find for the potential at any point whose coordinates are $\alpha$ and $\beta$,

$\begin{array}{ll} \phi= & E\log\left(1-2e^{\alpha-\alpha_{1}}\cos\left(\beta-\beta_{1}\right)+e^{2\left(\alpha-\alpha_{1}\right)}\right)\\ & -E\log\left(1-2e^{\alpha-\alpha_{1}-2\alpha_{0}}\cos\left(\beta-\beta_{1}\right)+e^{2\left(\alpha+\alpha_{1}-2\alpha_{0}\right)}\right)+2E\left(\alpha_{1}-\alpha_{0}\right) \end{array}$

(14)

This expression for the potential becomes zero when $\alpha=\alpha_{0}$, and is finite and continuous within the curve $\alpha_{0}$ except at the point $\alpha_{1}\beta_{1}$, at which point the first term becomes infinite, and in its immediate neighbourhood is ultimately equal to $2E\log r'$, where $r'$ is the distance from that point.

We have therefore obtained the means of deducing the solution of Green's problem for a charge at any point within a closed curve when the solution for a charge at any other point is known.

The charge induced upon an element of the curve $\alpha_{0}$ between the points $\beta$ and $\beta+d\beta$ by a charge $E$ placed at the point $\alpha_{1}\beta_{1}$ is

$\frac{E}{2\pi}\frac{1-e^{2\left(\alpha_{1}-\alpha_{0}\right)}}{1-2e^{\left(\alpha_{1}-\alpha_{0}\right)}\cos\left(\beta-\beta_{1}\right)+e^{2\left(\alpha_{1}-\alpha_{0}\right)}}d\beta$

(15)

From this expression we may find the potential at any point $\alpha_{1}\beta_{1}$ within the closed curve, when the value of the potential at every point of the closed curve is given as a function of $\beta$, and there is no electrification within the closed curve.

For, by Theorem II of Chap. Ill, the part of the potential at $\alpha_{1}\beta_{1}$, due to the maintenance of the portion $d\beta$ of the closed curve at the potential $V$, is $nV$, where $n$ is the charge induced on $d\beta$ by unit of electrification at $\alpha_{1}\beta_{1}$. Hence, if $V$ is the potential at a point on the closed curve defined as a function of $\beta$, and $\phi$ the potential at the point $\alpha,\beta$ within the closed curve, there being no electrification within the curve,

$\phi=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{\left(1-e^{2\left(\alpha_{1}-\alpha_{0}\right)}\right)V\ d\beta}{1-2e^{\left(\alpha_{1}-\alpha_{0}\right)}\cos\left(\beta-\beta_{1}\right)+e^{2\left(\alpha_{1}-\alpha_{0}\right)}}$

(16)

EXAMPLE IV. – Distribution of Electricity near an Edge of a Conductor formed by Two Plane Faces.

191.] In the case of an infinite plane face of a conductor charged with electricity to the surface-density $\sigma_{0}$, we find for the potential at a distance $y$ from the plane

$V=C-4\pi\sigma_{0}y$

where $C$ is the value of the potential of the conductor itself.

Assume a straight line in the plane as a polar axis, and transform into polar coordinates, and we find for the potential

$V=C-4\pi\sigma_{0}ae^{\rho}\sin\theta$

and for the quantity of electricity on a parallelogram of breadth unity, and length $ae^{\rho}$ measured from the axis

$E=\sigma_{0}ae^{\rho}$

Now let us make $\rho=n\rho'$ and $\theta=n\theta'$, then, since $\rho'$ and $\theta'$ are conjugate to $\rho$ and $\theta$, the equations

$V=C-4\pi\sigma_{0}ae^{n\rho'}\sin n\theta'$

and

$E=\sigma_{0}ae^{n\rho'}$

express a possible distribution of electricity and of potential.

If we write $ae^{\rho'}$, $r$ will be the distance from the axis, and $\theta$ the angle, and we shall have

$\begin{array}{l} V=C-4\pi\sigma_{0}\frac{r^{n}}{a^{n-1}}\sin n\theta\\ \\ E=\sigma_{0}\frac{r^{n}}{a^{n-1}} \end{array}$

$V$ will be equal to $C$ whenever $n\theta=\pi$ or a multiple of $\pi$.

Let the edge be a salient angle of the conductor, the inclination of the faces being $\alpha$, then the angle of the dielectric is $2\pi-\alpha$, so that when $\theta=2\pi-\alpha$ the point is in the other face of the conductor. We must therefore make

$n(2\pi-\alpha)=\pi$ or $n=\frac{\pi}{2\pi-\alpha}$.

Then

$\begin{array}{l} V=C-4\pi\sigma_{0}a\left(\frac{r}{a}\right)^{\frac{\pi}{2\pi-\alpha}}\sin\frac{\pi\theta}{2\pi-a}\\ \\ E=\sigma_{0}a\left(\frac{r}{a}\right)^{\frac{\pi}{2\pi-\alpha}} \end{array}$

The surface-density $\sigma$ at any distance $r$ from the edge is

$\sigma=\frac{dE}{dr}=\frac{\pi}{2\pi-\alpha}\sigma_{0}\left(\frac{r}{a}\right)^{\frac{\alpha-\pi}{2\pi-\alpha}}$

When the angle is a salient one $\alpha$ is less than $\pi$, and the surface-density varies according to some inverse power of the distance from the edge, so that at the edge itself the density becomes infinite, although the whole charge reckoned from the edge to any finite distance from it is always finite.

Thus, when $\alpha=0$ the edge is infinitely sharp, like the edge of a mathematical plane. In this case the density varies inversely as the square root of the distance from the edge.

When $\alpha=\tfrac{\pi}{3}$ the edge is like that of an equilateral prism, and the density varies inversely as the $\tfrac{2}{5}$ power of the distance.

When $\alpha=\tfrac{\pi}{2}$ the edge is a right angle, and the density is inversely as the cube root of the distance.

When $\alpha=\tfrac{2\pi}{3}$ the edge is like that of a regular hexagonal prism, and the density is inversely as the fourth root of the distance.

When $\alpha=\pi$ the edge is obliterated, and the density is constant.

When $\alpha=\tfrac{4}{3}\pi$ the edge is like that in the inside of the hexagonal prism, and the density is directly as the square root of the distance from the edge.

When $\alpha=\tfrac{3}{2}\pi$ the edge is a re-entrant right angle, and the density is directly as the distance from the edge.

When $\alpha=\tfrac{5}{3}\pi$ the edge is a re-entrant angle of 60°, and the density is directly as the square of the distance from the edge.

In reality, in all cases in which the density becomes infinite at any point, there is a discharge of electricity into the dielectric at that point, as is explained in Art. 55.

EXAMPLE V. Ellipses and Hyperbolas. Fig. X.

192.] We have seen that if

$x_{1}=e^{\phi}\cos\psi,\ y_{1}=e^{\phi}\sin\psi$.

(1)

$x$ and $y$ will be conjugate functions of $\phi$ and $\psi$.

Also, if

$x_{2}=e^{-\phi}\cos\psi,\ y_{2}=-e^{-\phi}\sin\psi$

(2)

$x_{2}$ and $y_{2}$ will be conjugate functions. Hence, if

$2x=x_{1}+x_{2}=\left(e^{\phi}+e^{-\phi}\right)\cos\psi,\ 2y=y_{1}+y_{2}=\left(e^{\phi}-e^{-\phi}\right)\sin\psi,$

(3)

$x$ and $y$ will also be conjugate functions of $\phi$ and $\psi$.

In this case the points for which $\phi$ is constant lie in the ellipse whose axes are $e^{\phi}+e^{-\phi}$ and $e^{\phi}-e^{-\phi}$.

The points for which $\psi$ is constant lie in the hyperbola whose axes are $2\cos\psi$ and $2\sin\psi$.

On the axis of $x$, between $x=-1$ and $x=+1$,

$\phi=0,\ \psi=\cos^{-1}x$

(4)

On the axis of $x$, beyond these limits on either side, we have

$\begin{array}{lllll} x>1, & & \psi=0, & & \phi=\log\left(x+\sqrt{x^{2}-1}\right),\\ x<-1, & & \psi=\pi, & & \phi=\log\left(\sqrt{x^{2}-1}-x\right), \end{array}$

(5)

Hence, if $\phi$ is the potential function, and $\psi$ the function of flow, we have the case of electricity flowing from the negative to the positive side of the axis of $x$ through the space between the points -1 and +1 , the parts of the axis beyond these limits being impervious to electricity.

Since, in this case, the axis of $y$ is a line of flow, we may suppose it also impervious to electricity.

We may also consider the ellipses to be sections of the equipotential surfaces due to an indefinitely long flat conductor of breadth 2, charged with half a unit of electricity per unit of length.

If we make $\psi$ the potential function, and $\phi$ the function of flow, the case becomes that of an infinite plane from which a strip of breadth 2 has been cut away and the plane on one side charged to potential $\pi$ while the other remains at zero.

These cases may be considered as particular cases of the quadric surfaces treated of in Chapter X. The forms of the curves are given in Fig. X.

Fig. X.

EXAMPLE VI. Fig. XI.

193.] Let us next consider $x'$ and $y'$ as functions of $x$ and $y$, where

$x'=b\log\sqrt{x^{2}+y^{2}},\ y'=b\tan^{-1}\frac{y}{x}$

(6)

$x'$ and $y'$ will be also conjugate functions of $\phi$ and $\psi$.

The curves resulting from the transformation of Fig. X with respect to these new coordinates are given in Fig. XI.

Fig. XI.

If $x'$ and $y'$ are rectangular coordinates, then the properties of the axis of $x$ in the first figure will belong to a series of lines parallel to $x'$ in the second figure for which $y'=bn'\pi$, where $n'$ is any integer.

The positive values of $x'$ on these lines will correspond to values of $x$ greater than unity, for which, as we have already seen,

$\psi=n\pi,\ \phi=\log\left(x+\sqrt{x^{2}-1}\right)=\log\left(e^{\frac{x'}{b}}+\sqrt{e^{\frac{2x'}{b}}-1}\right)$

(7)

The negative values of $x'$ on the same lines will correspond to values of $x$ less than unity, for which, as we have seen,

$\phi=0,\ \psi=\cos^{-1}x=\cos^{-1}e^{\frac{x'}{b}}$

(8)

The properties of the axis of $y$ in the first figure will belong to a series of lines in the second figure parallel to $x'$, for which

$y'=b\pi\left(n'+\frac{1}{2}\right)$

(9)

The value of $\psi$ along these lines is $\psi=\pi\left(n'+\frac{1}{2}\right)$ for all points both positive and negative, and

$\phi=\log\left(y+\sqrt{y^{2}+1}\right)=\log\left(e^{\frac{x'}{b}}+\sqrt{e^{\frac{2x'}{b}}+1}\right)$

(10)

194.] If we consider $\phi$ as the potential function, and $\psi$ as the function of flow, we may consider the case to be that of an in definitely long strip of metal of breadth $\pi b$ with a non-conducting division extending from the origin indefinitely in the positive direction, and thus dividing the positive part of the strip into two separate channels. We may suppose this division to be a narrow slit in the sheet of metal.

If a current of electricity is made to flow along one of these divisions and back again along the other, the entrance and exit of the current being at an indefinite distance on the positive side of the origin, the distribution of potential and of current will be given by the functions $\phi$ and $\psi$ respectively.

If, on the other hand, we make $\psi$ the potential, and $\phi$ the function of flow, then the case will be that of a current in the general direction of $y$, flowing through a sheet in which a number of non-conducting divisions are placed parallel to $x$, extending from the axis of $y$ to an indefinite distance in the negative direction.

195.] We may also apply the results to two important cases in statical electricity.

(1) Let a conductor in the form of a plane sheet, bounded by a straight edge but otherwise unlimited, be placed in the plane of $xz$ on the positive side of the origin, and let two infinite conducting planes be placed parallel to it and at distances $\tfrac{1}{2}\pi b$ on either side. Then, if $\psi$ is the potential function, its value is for the middle conductor and $\tfrac{1}{2}\pi$ for the two planes.

Let us consider the quantity of electricity on a part of the middle conductor, extending to a distance 1 in the direction of $z$, and from the origin to $x = a$.

The electricity on the part of this strip extending from $x_{1}$ to $x_{2}$ is $\tfrac{1}{4\pi}\left(\phi_{2}-\phi_{1}\right)$.

Hence from the origin to $x'=a$ the amount is

$E=\frac{1}{4\pi}\log\left(e^{\frac{a}{b}}+\sqrt{e^{\frac{2a}{b}}-1}\right)$

(11)

If $a$ is large compared with $b$, this becomes

$\begin{array}{ll} E & =\frac{1}{4\pi}\log2e^{\frac{a}{b}}\\ \\ & =\frac{a+b\log_{e}2}{4\pi b} \end{array}$

(12)

Hence the quantity of electricity on the plane hounded by the straight edge is greater than it would have been if the electricity had been uniformly distributed over it with the same density that it has at a distance from the boundary, and it is equal to the quantity of electricity having the same uniform surface-density, but extending to a breadth equal to $b\log_{e}2$ beyond the actual boundary of the plate.

This imaginary uniform distribution is indicated by the dotted straight lines in Fig. XI. The vertical lines represent lines of force, and the horizontal lines equipotential surfaces, on the hypo thesis that the density is uniform over both planes, produced to infinity in all directions.

196.] Electrical condensers are sometimes formed of a plate placed midway between two parallel plates extending considerably beyond the intermediate one on all sides. If the radius of curvature of the boundary of the intermediate plate is great compared with the distance between the plates, we may treat the boundary as approximately a straight line, and calculate the capacity of the condenser by supposing the intermediate plate to have its area extended by a strip of uniform breadth round its boundary, and assuming the surface-density on the extended plate the same as it is in the parts not near the boundary.

Thus, if $S$ be the actual area of the plate, $L$ its circumference, and $B$ the distance between the large plates, we have

$b=\frac{1}{\pi}B$

(13)

and the breadth of the additional strip is

$\alpha=\frac{\log_{e}2}{\pi}\cdot B$

(14)

so that the extended area is

$S'=S+BL\frac{1}{\pi}\log_{e}2$

(15)

The capacity of the middle plate is

$\frac{1}{2\pi}\frac{S'}{B}=\frac{1}{2\pi}\left\{ \frac{S}{B}+L\frac{1}{\pi}\log_{e}2\right\}$

(16)

Correction for the Thickness of the Plate.

Since the middle plate is generally of a thickness which cannot be neglected in comparison with the distance between the plates, we may obtain a better representation of the facts of the case by supposing the section of the intermediate plate to correspond with the curve $\psi=\psi'$.

The plate will be of nearly uniform thickness, $\beta=2b\psi'$, at a distance from the boundary, but will be rounded near the edge.

The position of the actual edge of the plate is found by putting $y'=0$, whence

$x'=b\log\cos\psi'$

(17)

The value of $\phi$ at this edge is 0, and at a point for which $x'=a$ it is

$\frac{a+b\log_{e}2}{b}$

Hence the quantity of electricity on the plate is the same as if a strip of breadth

$\alpha'=\frac{B}{\pi}\log_{e}\left(2\cos\frac{\pi\beta}{2B}\right)$

(18)

had been added to the plate, the density being assumed to be every where the same as it is at a distance from the boundary.

Density near the Edge.

The surface-density at any point of the plate is

$\begin{array}{ll} \frac{1}{4\pi}\frac{d\phi}{dx'} & =\frac{1}{4\pi b}\frac{e^{\frac{x'}{b}}}{\sqrt{e^{\frac{2x'}{b}}-1}}\\ \\ & =\frac{1}{4\pi b}\left(1+\frac{1}{2}e^{-\frac{2x'}{b}}+\frac{3}{8}e^{-\frac{4x'}{b}}-\mathrm{etc}.\right) \end{array}$

(19)

The quantity within brackets rapidly approaches unity as $x'$ increases, so that at a distance from the boundary equal to $n$ times the breadth of the strip $\alpha$, the actual density is greater than the normal density by about $\tfrac{1}{2^{2n+1}}$ of the normal density.

In like manner we may calculate the density on the infinite planes

$=\frac{1}{4\pi b}\frac{e^{\frac{x}{b}}}{\sqrt{e^{\frac{2x'}{b}}+1}}$

(20)

When $x'=0$, the density is $2^{-\tfrac{1}{2}}$ of the normal density.

At $n$ times the breadth of the strip on the positive side, the density is less than the normal density by about $\tfrac{1}{2^{2n+1}}$.

At $n$ times the breadth of the strip on the negative side, the density is about $\tfrac{1}{2^{n}}$ of the normal density.

These results indicate the degree of accuracy to be expected in applying this method to plates of limited extent, or in which irregularities may exist not very far from the boundary. The same distribution would exist in the case of an infinite series of similar plates at equal distances, the potentials of these plates being alternately $+V$ and $-V$. In this case we must take the distance between the plates equal to $B$.

197.] (2) The second case we shall consider is that of an infinite series of planes parallel to $xz$ at distances $B=\pi b$, and all cut off by the plane of $yz$, so that they extend only on the negative side of this plane. If we make $\phi$ the potential function, we may regard these planes as conductors at potential zero.

Let us consider the curves for which $\phi$ is constant.

When $y'=n\pi b$, that is, in the prolongation of each of the planes, we have

$x'=b\log\frac{1}{2}\left(e^{\phi}+e^{-\phi}\right)$

(21)

when $y'=\left(n+\frac{1}{2}\right)b\pi$, that is, in the intermediate positions

$x'=b\log\frac{1}{2}\left(e^{\phi}-e^{-\phi}\right)$

(22)

Hence, when $\phi$ is large, the curve for which is constant is an undulating line whose mean distance from the axis of $y'$ is approximately

$a=b\left(\phi-\log_{e}2\right)$

(23)

and the amplitude of the undulations on either side of this line is

$\frac{1}{2}b\log\frac{e^{\phi}+e^{-\phi}}{e^{\phi}-e^{-\phi}}$

(24)

When $\phi$ is large this becomes $be^{-2\phi}$, so that the curve approaches to the form of a straight line parallel to the axis of $y'$ at a distance $a$ from $ab$ on the positive side.

If we suppose a plane for which $x'=a$, kept at a constant potential while the system of parallel planes is kept at a different potential, then, since $b\phi=a+b\log_{e}2$, the surface-density of the electricity induced on the plane is equal to that which would have been induced on it by a plane parallel to itself at a potential equal to that of the series of planes, but at a distance greater than that of the edges of the planes by $b\log_{e}2$.

If $B$ is the distance between two of the planes of the series, $B=\pi b$, so that the additional distance is

$\alpha=B\frac{\log_{e}2}{\pi}$

(25)

198.] Let us next consider the space included between two of the equipotential surfaces, one of which consists of a series of parallel waves, while the other corresponds to a large value of $\phi$, and may be considered as approximately plane.

If $D$ is the depth of these undulations from the crest to the trough of each wave, then we find for the corresponding value of $\phi$,

$\phi =\frac{1}{2}\log\frac{e^{\frac{D}{b}}+1}{e^{\frac{D}{b}}-1}$

(26)

The value of $x'$ at the crest of the wave is

$b\log\frac{1}{2}\left(e^{\phi}+e^{-\phi}\right)$

(27)

Hence, if $A$ is the distance from the crests of the waves to the opposite plane, the capacity of the system composed of the plane surface and the undulated surface is the same as that of two planes at a distance $A+\alpha'$ where

$\alpha'=\frac{B}{\pi}\log_{e}\frac{2}{1+e^{-\pi\frac{D}{B}}}$

199.] If a single groove of this form be made in a conductor having the rest of its surface plane, and if the other conductor is a plane surface at a distance $A$, the capacity of the one conductor with respect to the other will be diminished. The amount of this diminution will be less than the $\tfrac{1}{n}$th part of the diminution due to $n$ such grooves side by side, for in the latter case the average electrical force between the conductors will be less than in the former case, so that the induction on the surface of each groove will be diminished on account of the neighbouring grooves.

If $L$ is the length, $B$ the breadth, and $D$ the depth of the groove, the capacity of a portion of the opposite plane whose area is $S$ will be

$\frac{S}{4\pi A}-\frac{LB}{4\pi A}\frac{\alpha'}{A+\alpha'}$

(29)

If $A$ is large compared with $B$ or $\alpha'$, the correction becomes

$\frac{L}{4\pi^{2}}\frac{B^{2}}{A^{2}}\log_{e}\frac{2}{1+e^{-\pi\frac{D}{B}}}$

(30)

and for a slit of infinite depth, putting $D=\infty$, the correction is

$\frac{L}{4\pi^{2}}\frac{B^{2}}{A^{2}}\log_{e}2$

(31)

To find the surface-density on the series of parallel plates we must find $\sigma=\frac{1}{4\pi}\frac{d\psi}{dx'}$ when $\phi=0$. We find

$\sigma=\frac{1}{4\pi b}\frac{1}{\sqrt{e^{-2\frac{x'}{b}}-1}}$

(32)

The average density on the plane plate at distance $A$ from the edges of the series of plates is $\bar{\sigma}=\frac{1}{4\pi b}$. Hence, at a distance from the edge of one of the plates equal to $n\alpha$ the surface-density is $\tfrac{1}{\sqrt{2^{2n}-1}}$ of this average density.

200.] Let us next attempt to deduce from these results the distribution of electricity in the figure formed by rotating the plane of the figure about the axis $y'=-R$. In this case, Poisson s equation will assume the form

$\frac{d^{2}V}{dx'^{2}}+\frac{d^{2}V}{dy'^{2}}+\frac{1}{R+y'}\frac{dV}{dy'}+4\pi\rho=0$

(33)

Let us assume $V=\phi$, the function given in Art. 193, and determine the value of $\rho$ from this equation. We know that the first two terms disappear, and therefore

$\rho=-\frac{1}{4\pi}\frac{1}{R+y'} \frac{d\phi}{dy'}$

(34)

If we suppose that, in addition to the surface-density already investigated, there is a distribution of electricity in space according to the law just stated, the distribution of potential will be represented by the curves in Fig. XI.

Now from this figure it is manifest that $\frac{d\phi}{dy'}$ is generally very small except near the boundaries of the plates, so that the new distribution may be approximately represented by what actually exists, namely a certain superficial distribution near the edges of the plates.

If therefore we integrate $\iint\rho\ dx'dy'$ between the limits $y'=0$ and $y'=\frac{\pi}{2}b$, and from $x'=-\infty$ to $x=+\infty$, we shall find the whole additional charge on one side of the plates due to the curvature.

Since

$\begin{array}{ll} \frac{d\phi}{dy'}=\frac{d\psi}{dx'},\\ \\ \int_{-\infty}^{+\infty}\rho\ dx' & =\int_{-\infty}^{+\infty}\frac{1}{4\pi}\frac{1}{(R+y')}\frac{d\psi}{dx}dx,\\ \\ & =\frac{1}{8}\frac{1}{R+y'}\left(\frac{y'}{B}-1\right). \end{array}$

(35)

Integrating with respect to $y'$, we find

$\int_{0}^{B}\int_{-\infty}^{+\infty}\rho\ dx'dy'=\frac{1}{8}-\frac{1}{8}\frac{R+B}{B}\log\frac{R+B}{R},$

(36)

$=\frac{1}{16}\frac{B}{R}+\frac{1}{48}\frac{B^{2}}{R^{2}}-\mathrm{etc.}$

(37)

This is the total quantity of electricity which we must suppose distributed in space near the positive side of one of the cylindric plates per unit of circumference. Since it is only close to the edge of the plate that the density is sensible, we may suppose it all condensed on the surface of the plate without altering sensibly its action on the opposed plane surface, and in calculating the attraction between that surface and the cylindric surface we may suppose this electricity to belong to the cylindric surface.

The superficial charge on the positive surface of the plate per unit of length would have been $-\frac{1}{8}$, if there had been no curvature. Hence this charge must be multiplied by the factor $\left(1+\frac{1}{2}\frac{B}{R}\right)$ to get the total charge on the positive side.

In the case of a disk of radius $R$ placed midway between two infinite parallel plates at a distance $B$, we find for the capacity of the disk

$\frac{R^{2}}{B}+2\frac{\log_{e}2}{\pi}R+\frac{1}{2}B$

(38)

Theory of Thomson's Guard-ring.

201.] In some of Sir W. Thomson's electrometers, a large plane surface is kept at one potential, and at a distance $a$ from this surface is placed a plane disk of radius $R$ surrounded by a large plane plate called a Guard-ring with a circular aperture of radius $R'$ concentric with the disk. This disk and plate are kept at potential zero.

The interval between the disk and the guard-plate may be regarded as a circular groove of infinite depth, and of breadth $R'-R$, which we denote by $B$.

The charge on the disk due to unit potential of the large disk, supposing the density uniform, would be $\frac{R^{2}}{4A}$.

The charge on one side of a straight groove of breadth $B$ and length $L=2\pi R$, and of infinite depth, would be

$\frac{1}{4}\frac{RB}{A+\alpha'}$

But since the groove is not straight, but has a radius of curvature $R$, this must be multiplied by the factor $\left(1+\frac{1}{2}\frac{B}{R}\right)$.

The whole charge on the disk is therefore

$\frac{R^{2}}{4A}+\frac{1}{4}\frac{RB}{A+\alpha'}\left(1+\frac{B}{2R}\right)$

(39)

$=\frac{R^{2}+R'^{2}}{8A}-\frac{R'^{2}-R^{2}}{8A}\cdot\frac{\alpha'}{A+\alpha'}$

(40)

The value of a cannot be greater than

$\alpha'=\frac{B\log2}{\pi},\ =0.22B$ nearly.

If $B$ is small compared with either $A$ or $R$ this expression will give a sufficiently good approximation to the charge on the disk due to unity of difference of potential. The ratio of $A$ to $R$ may have any value, but the radii of the large disk and of the guard-ring must exceed $R$ by several multiples of $A$.

EXAMPLE VII. – Fig. XII.

202.] Helmholtz, in his memoir on discontinuous fluid motion, has pointed out the application of several formulae in which the coordinates are expressed as functions of the potential and its conjugate function.

One of these may be applied to the case of an electrified plate of finite size placed parallel to an infinite plane surface connected with the earth.

Since

$x_{1}=A\phi$ and $y_{1}=A\psi$,

and also

$x_{2}=Ae^{\phi}\cos\psi$ and $y_{2}=Ae^{\phi}\sin\psi$,

are conjugate functions of $\phi$ and $\psi$, the functions formed by adding $x_{1}$ to $x_{2}$ and $y_{1}$ to $y_{2}$ will be also conjugate. Hence, if

$\begin{array}{l} x=A\phi+Ae^{\phi}\cos\psi\\ y=A\psi+Ae^{\phi}\sin\psi \end{array}$

then $x$ and $y$ will be conjugate with respect to $\phi$ and $\psi$, and $\phi$ and $\psi$ will be conjugate with respect to $x$ and $y$.

Now let $x$ and $y$ be rectangular coordinates, and let $k\psi$ be the potential, then $k\phi$ will be conjugate to $k\psi$, $k$ being any constant.

Let us put $\psi=\pi$, then $y=A\pi$, $x=A\left(\phi-e^{\phi}\right)$.

If $\phi$ varies from $-\infty$ to 0, and then from 0 to $+\infty$, $x$ varies from $-\infty$ to $-A$ and from $-A$ to $-\infty$. Hence the equipotential surface for which $b=\pi A$ is a plane parallel to $x$ at a distance $b=\pi A$ from the origin, and extending from $-\infty$ to $x=-A$.

Let us consider a portion of this plane, extending from

$x=-(A+a)$ to $x=-A$ and from $z=0$ to $z=c$,

let us suppose its distance from the plane of $xz$ to be $y=B=A\pi$, and its potential to be $V=k\psi=k\pi$.

The charge of electricity on any portion of this part of the plane is found by ascertaining the values of $\phi$ at its extremities.

If these are $\phi_{1}$ and $\phi_{2}$, the quantity of electricity is

$\frac{1}{4\pi}ck\left(\phi_{2}-\phi_{1}\right)$

We have therefore to determine $\phi$ from the equation

$x=-(A+a)=A\left(\phi-e^{\phi}\right)$

$\phi$ will have a negative value $\phi_{1}$ and a positive value $\phi_{2}$ at the edge of the plane, where $x=-A,\ \phi=0$.

Hence the charge on the negative side is $-ck\phi_{1}$, and that on the positive side is $ck\phi_{2}$.

If we suppose that $a$ is large compared with $A$,

$\begin{array}{l} \phi_{1}=-\frac{a}{A}-1+e^{-\frac{a}{A}-1+e^{-\frac{a}{A}-+\mathrm{etc.}}}\\ \\ \phi_{e}=\log\left\{ \frac{a}{A}+1+\log\left(\frac{a}{A}+1+\mathrm{etc.}\right)\right\} \end{array}$

If we neglect the exponential terms in $\phi_{1}$ we shall find that the charge on the negative surface exceeds that which it would have if the superficial density had been uniform and equal to that at a distance from the boundary, by a quantity equal to the charge on a strip of breadth $A=\frac{b}{\pi}$ with the uniform superficial density.

The total capacity of the part of the plane considered is

$C=\frac{c}{4\pi^{2}}\left(\phi_{2}-\phi_{1}\right)$

The total charge is $CV$, and the attraction towards the infinite plane is

$\begin{array}{l} -\frac{1}{2}V^{2}\frac{dC}{db}=V^{2}\frac{ac}{4\pi b^{2}}\left(1+\frac{\frac{A}{a}}{1+\frac{A}{a}\log\frac{a}{A}}+e^{-\frac{a}{A}}+\mathrm{etc.}\right)\\ \\ \qquad=\frac{V^{2}c}{4\pi b^{2}}\left\{ a+\frac{b}{\pi}-\frac{b^{2}}{\pi^{2}a}\log\frac{a}{A}+\mathrm{etc.}\right\} \end{array}$

The equipotential lines and lines of force are given in Fig. XII.

Fig. XII.

EXAMPLE VIII. – Theory of a Grating of Parallel Wires. Fig. XIII.

203.] In many electrical instruments a wire grating is used to prevent certain parts of the apparatus from being electrified by induction. We know that if a conductor be entirely surrounded by a metallic vessel at the same potential with itself, no electricity can be induced on the surface of the conductor by any electrified body outside the vessel. The conductor, however, when completely surrounded by metal, cannot be seen, and therefore, in certain cases, an aperture is left which is covered with a grating of fine wire. Let us investigate the effect of this grating in diminishing the effect of electrical induction. We shall suppose the grating to consist of a series of parallel wires in one plane and at equal intervals, the diameter of the wires being small compared with the distance between them, while the nearest portions of the electrified bodies on the one side and of the protected conductor on the other are at distances from the plane of the screen, which are considerable compared with the distance between consecutive wires.

204.] The potential at a distance $r'$ from the axis of a straight wire of infinite length charged with a quantity of electricity $\lambda$ per unit of length is

$V=-2\lambda\log r'+C$

(1)

We may express this in terms of polar coordinates referred to an axis whose distance from the wire is unity, in which case we must make

$r'^{2}=1+2r\cos\theta+r^{2}$

(2)

and if we suppose that the axis of reference is also charged with the linear density $\lambda'$, we find

$V=-\lambda\log\left(1-2r\cos\theta+r^{2}\right)-2\lambda'\log r+C$

(3)

If we now make

$r=e^{2\pi\frac{y}{a}},\ \theta=\frac{2\pi x}{a}$

(4)

then, by the theory of conjugate functions,

$V=-\lambda\log\left(1-2e^{\frac{2\pi y}{a}}\cos\frac{2\pi x}{a}+e^{\frac{4\pi y}{a}}\right)-2\lambda'\log e^{\frac{2\pi y}{a}}+C$

(5)

where $x$ and $y$ are rectangular coordinates, will be the value of the potential due to an infinite series of fine wires parallel to $z$ in the plane of $yz$, and passing through points in the axis of $x$ for which $x$ is a multiple of $a$.

Each of these wires is charged with a linear density $\lambda$.

The term involving $\lambda'$ indicates an electrification, producing a constant force $-\frac{4\pi\lambda'}{a}$ in the direction of $y$.

The forms of the equipotential surfaces and lines of force when $\lambda'=0$ are given in Fig. XIII. The equipotential surfaces near the wires are nearly cylinders, so that we may consider the solution approximately true, even when the wires are cylinders of a diameter which is finite but small compared with the distance between them.

Fig. XIII.

The equipotential surfaces at a distance from the wires become more and more nearly planes parallel to that of the grating.

If in the equation we make $y=b_{1}$, a quantity large compared with $a$, we find approximately,

$V_{1}=-\frac{4\pi b_{1}}{a}(\lambda+\lambda')+C$

(6)

If we next make $y=-b_{2}$ where $b_{2}$ is a negative quantity large compared with $a$, we find approximately,

$V_{2}=-\frac{4\pi b_{2}}{a}(\lambda-\lambda')+C\ \mathrm{nearly}.$

If $c$ is the radius of the wires of the grating, $c$ being small compared with $a$, we may find the potential of the grating itself by supposing that the surface of the wire coincides with the equipotential surface which cuts the plane of $yz$ at a distance $c$ from the axis of $z$. To find the potential of the grating we therefore put $x = c$, and $y=0$, whence

$V=-2\lambda\log2\sin\frac{\pi c}{a}+C$

(8)

205.] We have now obtained expressions representing the electrical state of a system consisting of a grating of wires whose diameter is small compared with the distance between them, and two plane conducting surfaces, one on each side of the grating, and at distances which are great compared with the distance between the wires.

The surface-density $\sigma_{1}$ on the first plane is got from the equation (6)

$4\pi\sigma_{1}=\frac{dV_{1}}{db_{1}}=-\frac{4\pi}{a}(\lambda+\lambda')$

(9)

That on the second plane $\sigma_{2}$ from the equation (7)

$4\pi\sigma_{2}=-\frac{dV_{2}}{db_{2}}=-\frac{4\pi}{a}(\lambda-\lambda')$

(10)

If we now write

$\alpha=-\frac{a}{2\pi}\log_{e}\left(2\sin\frac{\pi c}{a}\right)$

(11)

and eliminate $\lambda$ and $\lambda'$ from the equations (6), (7), (8), (9), (10), we find

$4\pi\sigma_{1}\left(b_{1}+b_{2}+\frac{2b_{1}b_{2}}{\alpha}\right)=V_{1}\left(1+2\frac{b_{2}}{\alpha}\right)-V_{2}-V\frac{2b_{2}}{\alpha}$

(12)

$4\pi\sigma_{2}\left(b_{1}+b_{2}+\frac{2b_{1}b_{2}}{\alpha}\right)=-V_{1}+V_{2}\left(1+2\frac{b_{1}}{\alpha}\right)-V\frac{2b_{1}}{\alpha}$

(13)

When the wires are infinitely thin, $\alpha$ becomes infinite, and the terms in which it is the denominator disappear, so that the case is reduced to that of two parallel planes without a grating interposed.

If the grating is in metallic communication with one of the planes, say the first, $V=V_{1}$, and the right-hand side of the equation for $\sigma_{1}$ becomes $V_{1}-V_{2}$. Hence the density $\sigma_{1}$ induced on the first plane when the grating is interposed is to that which would have been induced on it if the grating were removed, the second plane being maintained at the same potential, as 1 to $1+\tfrac{2b_{1}b_{2}}{\alpha\left(b_{1}+b_{2}\right)}$.

We should have found the same value for the effect of the grating in diminishing the electrical influence of the first surface on the second, if we had supposed the grating connected with the second surface. This is evident since $b_{1}$ and $b_{2}$ enter into the expression in the same way. It is also a direct result of the theorem of Art. 88.

The induction of the one electrified plane on the other through the grating is the same as if the grating were removed, and the distance between the planes increased from $b_{1}+b_{2}$ to

$b_{1}+b_{2}+2\frac{b_{1}b_{2}}{\alpha}$

If the two planes are kept at potential zero, and the grating electrified to a given potential, the quantity of electricity on the grating will be to that which would be induced on a plane of equal area placed in the same position as

$2b_{1}b_{2}$ is to $2b_{1}b_{2}+\alpha\left(b_{1}+b_{2}\right)$.

This investigation is approximate only when $b_{1}$ and $b_{2}$ are large compared with $a$, and when $a$ is large compared with $c$. The quantity $a$ is a line which may be of any magnitude. It becomes infinite when $c$ is indefinitely diminished.

If we suppose $c=\tfrac{1}{2}a$ there will be no apertures between the wires of the grating, and therefore there will be no induction through it. We ought therefore to have for this case $\alpha=0$. The formula (11), however, gives in this case

$\alpha=-\frac{a}{2\pi}\log_{e}2,\ =-0.11a$,

which is evidently erroneous, as the induction can never be altered in sign by means of the grating. It is easy, however, to proceed to a higher degree of approximation in the case of a grating of cylindrical wires. I shall merely indicate the steps of this process.

Method of Approximation.

206.] Since the wires are cylindrical, and since the distribution of electricity on each is symmetrical with respect to the diameter parallel to $y$, the proper expansion of the potential is of the form

$V=C_{0}\log r+\Sigma C_{i}r^{i}\cos i\theta$

(14)

where $r$ is the distance from the axis of one of the wires, and $\theta$ the angle between $r$ and $y$, and, since the wire is a conductor, when $r$ is made equal to the radius $V$ must be constant, and therefore the coefficient of each of the multiple cosines of $\theta$ must vanish.

For the sake of conciseness let us assume new coordinates $\xi,\eta$, &c. such that

$a\xi=2\pi x,\ a\eta=2\pi y,\ a\rho=2\pi r,\ a\beta=2\pi b,\ \mathrm{etc.}$

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and let

$F_{\beta}=\log\left(\epsilon^{\eta+\beta}+\epsilon^{-(\eta+\beta)}-2\cos\xi\right)$

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Then if we make

$V=A_{0}F+A_{1}\frac{dV}{d\eta}+A_{e}\frac{d^{2}F}{d\eta^{2}}+\mathrm{etc.}$

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by giving proper values to the coefficients $A$ we may express any potential which is a function of $\eta$ and $\cos\xi$, and does not become infinite except when $\eta+\beta=0$ and $\cos\xi=1$.

When $\beta=0$ the expansion of $F$ in terms of $\rho$ and $\theta$ is

$F_{0}=2\log\rho+\frac{1}{12}\rho^{2}\cos2\theta-\frac{1}{1440}\rho^{4}\cos4\theta+\mathrm{etc.}$

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For finite values of $\beta$ the expansion of $F$ is

$F_{\beta}=\beta+2\log\left(1-e^{-\beta}\right)+\frac{1+e^{-\beta}}{1-e^{-\beta}}\rho\cos\theta-\frac{e^{-\beta}}{\left(1-e^{-\beta}\right)^{2}}\rho^{2}\cos2\theta+\mathrm{etc}.$

(19)

In the case of the grating with two conducting planes whose equations are $\eta=-\beta_{1}$ and $\eta=-\beta_{2}$, that of the plane of the grating being $\eta=0$, there will be two infinite series of images of the grating. The first series will consist of the grating itself together with an infinite series of images on both sides, equal and similarly electrified. The axes of these imaginary cylinders lie in planes whose equations are of the form

$\eta=\pm2n\left(\beta_{1}+\beta_{2}\right)$

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$n$ being an integer.

The second series will consist of an infinite series of images for which the coefficients $A_{0},\ A_{1},\ A_{2}$, &c. are equal and opposite to the same quantities in the grating itself, while $A_{1},\ A_{3}$ &c. are equal and of the same sign. The axes of these images are in planes whose equations are of the form

$\eta=2\beta_{2}\pm2m\left(\beta_{1}+\beta_{2}\right)$

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$m$ being an integer.

The potential due to any finite series of such images will depend on whether the number of images is odd or even. Hence the potential due to an infinite series is indeterminate, but if we add to it the function $B\eta+C$, the conditions of the problem will be sufficient to determine the electrical distribution.

We may first determine $V_{1}$ and $V_{2}$, the potentials of the two conducting planes, in terms of the coefficients $A_{0},\ A_{1}$, &c., and of $B$ and $C$. We must then determine $\sigma_{1}$ and $\sigma_{2}$, the surface-density at any point of these planes. The mean values of $\sigma_{1}$ and $\sigma_{2}$ are given by the equations

$4\pi\sigma_{1}=A_{0}-B,\ 4\pi\sigma_{2}=A_{0}+B,$

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We must then expand the potentials due to the grating itself and to all the images in terms of $\rho$ and cosines of multiples of $\theta$ adding to the result

$B\rho\cos\theta+C$

The terms independent of $\theta$ then give $V$ the potential of the grating, and the coefficient of the cosine of each multiple of $\theta$ equated to zero gives an equation between the indeterminate coefficients.

In this way as many equations may be found as are sufficient to eliminate all these coefficients and to leave two equations to determine $\sigma_{1}$ and $\sigma_{2}$ in terms of $V_{1},\ V_{2}$, and $V$.

These equations will be of the form

$\begin{array}{l} V_{1}-V=4\pi\sigma_{1}\left(b_{1}+\alpha-\gamma\right)+4\pi\sigma_{2}(\alpha+\gamma),\\ V_{2}-V=4\pi\sigma_{1}\left(\alpha+\gamma\right)+4\pi\sigma_{2}(b_{2}+\alpha-\gamma), \end{array}$

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The quantity of electricity induced on one of the planes protected by the grating, the other plane being at a given difference of potential, will be the same as if the plates had been at a distance

$\frac{(\alpha-\gamma)\left(b_{1}+b_{2}\right)+\beta_{1}\beta_{2}-4\alpha\gamma}{\alpha+\gamma}$ instead of $b_{1}+b_{2}$.

The values of $\alpha$ and $\gamma$ are approximately as follows,

$\alpha=\frac{a}{2\pi}\left\{ \log\frac{a}{2\pi c}-\frac{5}{3}\cdot\frac{\pi^{4}c^{4}}{15a^{4}+\pi^{4}c^{4}}+2e^{-4\pi\frac{b_{1}+b_{2}}{a}}\left(1+e^{-4\pi\frac{b_{1}}{a}}+e^{-4\pi\frac{b_{2}}{a}}+\mathrm{etc}.\right)+\mathrm{etc}.\right\}$

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$\gamma=\frac{3\pi ac^{2}}{3a^{2}+\pi^{2}c^{2}}\left(\frac{e^{-4\pi\frac{b_{1}}{a}}}{1-e^{-4\pi\frac{b_{1}}{a}}}-\frac{e^{-4\pi\frac{b_{2}}{a}}}{1-e^{-4\pi\frac{b_{2}}{a}}}\right)+\mathrm{etc}.$

(25)

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