# A Treatise on Electricity and Magnetism/Volume 1/Part 1/Chapter 6

CHAPTER VI.
ON POINTS AND LINES OF EQUILIBRIUM.
112.] If at any point of the electric field the resultant force is zero, the point is called a Point of equilibrium.
If every point on a certain line is a point of equilibrium, the line is called a Line of equilibrium.
The conditions that a point shall be a point of equilibrium are that at that point
${\frac {dV}{dx}}=0,\ {\frac {dV}{dy}}=0,\ {\frac {dV}{dz}}=0$
At such a point, therefore, the value of $V$ is a maximum, or a minimum, or is stationary, with respect to variations of the coordinates. The potential, however, can have a maximum or a minimum value only at a point charged with positive or with negative electricity, or throughout a finite space bounded by a surface electrified positively or negatively. If, therefore, a point of equilibrium occurs in an unelectrified part of the field it must be a stationary point, and not a maximum or a minimum.
In fact, the first condition of a maximum or minimum is that
${\frac {d^{2}V}{dx^{2}}},\ {\frac {d^{2}V}{dy^{2}}},$ and ${\frac {d^{2}V}{dz^{2}}}$
must be all negative or all positive, if they have finite values.
Now, by Laplace’s equation, at a point where there is no electrification, the sum of these three quantities is zero, and therefore this condition cannot be fulfilled.
Instead of investigating the analytical conditions for the cases in which the components of the force simultaneously vanish, we shall give a general proof by means of the equipotential surfaces.
If at any point, $P$, there is a true maximum value of $V$, then, at all other points in the immediate neighbourhood of $P$, the value of $V$ is less than at $P$. Hence $P$ will be surrounded by a series of ​closed equipotential surfaces, each outside the one before it, and at all points of any one of these surfaces the electrical force will be directed outwards. But we have proved, in Art. 76, that the surface-integral of the electrical force taken over any closed surface gives the total electrification within that surface multiplied by 4$\pi$. Now, in this case the force is everywhere outwards, so that the surface-integral is necessarily positive, and therefore there is positive electrification within the surface, and, since we may take the surface as near to $P$ as we please, there is positive electrification at the point $P$.
In the same way we may prove that if $V$ is a minimum at $P$, then $P$ is negatively electrified.
Next, let $P$ be a point of equilibrium in a region devoid of electrification, and let us describe a very small closed surface round $P$, then, as we have seen, the potential at this surface cannot be everywhere greater or everywhere less than at $P$. It must there fore be greater at some parts of the surface and less at others. These portions of the surface are bounded by lines in which the potential is equal to that at $P$. Along lines drawn from $P$ to points at which the potential is less than that at $P$ the electrical force is from $P$, and along lines drawn to points of greater potential the force is towards $P$. Hence the point $P$ is a point of stable equilibrium for some displacements, and of unstable equilibrium for other displacements.
113.] To determine the number of the points and lines of equilibrium, let us consider the surface or surfaces for which the potential is equal to $C$, a given quantity. Let us call the regions in which the potential is less than $C$ the negative regions, and those in which it is greater than $C$ the positive regions. Let $V_{0}$ be the lowest, and $V_{1}$ the highest potential existing in the electric field. If we make $C=V_{0}$, the negative region will include only the electrified point or conductor of lowest potential, and this is necessarily electrified negatively. The positive region consists of the rest of space, and since it surrounds the negative region it is periphractic. See Art. 18.
If we now increase the value of $C$ the negative region will expand, and new negative regions will be formed round negatively electrified bodies. For every negative region thus formed the surrounding positive region acquires one degree of periphraxy.
As the different negative regions expand, two or more of them may meet in a point or a line. If $n+1$ negative regions meet, the positive region loses $n$ degrees of periphraxy, and the point ​or the line in which they meet is a point or line of equilibrium of the $n$th degree.
When $C$ becomes equal to $V_{1}$ the positive region is reduced to the electrified point or conductor of highest potential, and has therefore lost all its periphraxy. Hence, if each point or line of equilibrium counts for one, two, or $n$ according to its degree, the number so made up by the points or lines now considered will be one less than the number of negatively electrified bodies.
There are other points or lines of equilibrium which occur where the positive regions become separated from each other, and the negative region acquires periphraxy. The number of these, reckoned according to their degrees, is one less than the number of positively electrified bodies.
If we call a point or line of equilibrium positive when it is the meeting-place of two or more positive regions, and negative when the regions which unite there are negative, then, if there are $p$ bodies positively and $n$ bodies negatively electrified, the sum of the degrees of the positive points and lines of equilibrium will be $p-1$, and that of the negative ones $n-1$.
But, besides this definite number of points and lines of equilibrium arising from the junction of different regions, there may be others, of which we can only affirm that their number must be even. For if, as the negative region expands, it meets itself, it becomes a cyclic region, and it may acquire, by repeatedly meeting itself, any number of degrees of cyclosis, each of which corresponds to the point or line of equilibrium at which the cyclosis was established. As the negative region continues to expand till it fills all space, it loses every degree of cyclosis it has acquired, and becomes at last acyclic. Hence there is a set of points or lines of equilibrium at which cyclosis is lost, and these are equal in number of degrees to those at which it is acquired. .
If the form of the electrified bodies or conductors is arbitrary, we can only assert that the number of these additional points or lines is even, but if they are electrified points or spherical conductors, the number arising in this way cannot exceed ($n-l$)($n-2$), where $n$ is the number of bodies.
114.] The potential close to any point $P$ may be expanded in the series
$V=V_{0}+H_{1}+H_{2}+\mathrm {etc} .;$
where $H_{1},\ H_{2}$, &c. are homogeneous functions of $x,y,z$, whose dimensions are 1, 2, &c. respectively.
​Since the first derivatives of $V$ vanish at a point of equilibrium, $H_{1}=0$, if $P$ be a point of equilibrium.
Let $H_{i}$ be the first function which does not vanish, then close to the point $P$ we may neglect all functions of higher degrees as compared with $H_{i}$.
Now
$H_{i}=0$
is the equation of a cone of the degree $i$, and this cone is the cone of closest contact with the equipotential surface at $P$.
It appears, therefore, that the equipotential surface passing through $P$ has, at that point, a conical point touched by a cone of the second or of a higher degree.
If the point $P$ is not on a line of equilibrium this cone does not intersect itself, but consists of $i$ sheets or some smaller number.
If the nodal line intersects itself, then the point $P$ is on a line of equilibrium, and the equipotential surface through $P$ cuts itself in that line.
If there are intersections of the nodal line not on opposite points of the sphere, then $P$ is at the intersection of three or more lines of equilibrium. For the equipotential surface through $P$ must cut itself in each line of equilibrium.
115.] If two sheets of the same equipotential surface intersect, they must intersect at right angles.
For let the tangent to the line of intersection be taken as the axis of $z$, then ${\tfrac {d^{2}V}{dz^{2}}}=0$. Also let the axis of $x$ be a tangent to one of the sheets, then ${\tfrac {d^{2}V}{dx^{2}}}=0$. It follows from this, by Laplace’s equation, that ${\frac {d^{2}V}{dy^{2}}}=0$, or the axis of $y$ is a tangent to the other sheet.
This investigation assumes that $H_{2}$ is finite. If $H_{2}$ vanishes, let the tangent to the line of intersection be taken as the axis of $z$, and let $x=r\cos \theta$, and $y=r\sin \theta$, then, since
${\frac {d^{2}V}{dz^{2}}}=0,\ {\frac {d^{2}V}{dx^{2}}}+{\frac {d^{2}V}{dy^{2}}}=0;$
or
${\frac {d^{2}V}{dr^{2}}}+{\frac {1}{r}}{\frac {dV}{dr}}+{\frac {1}{r^{2}}}{\frac {d^{2}V}{d\theta ^{2}}}=0;$
the solution of which equation in ascending powers of $r$ is
$V=V_{0}+A_{1}r\cos(\theta +\alpha )+A_{2}r^{2}\cos(2\theta +\alpha _{2})+\mathrm {etc} .+A_{i}r^{i}\cos(i\theta +\alpha _{i}).$
$A_{1}$
$r^{i}$
$V-V_{0}=A_{i}r^{i}\cos(i\theta +\alpha _{i})+$ terms in higher powers of r.
This gives $i$ sheets of the equipotential surface $V=V_{0}$, intersecting at angles each equal to ${\tfrac {\pi }{i}}$. This theorem was given by Rankine.
It is only under certain conditions that a line of equilibrium can exist in free space, but there must be a line of equilibrium on the surface of a conductor whenever the electrification of the conductor is positive in one portion and negative in another.
In order that a conductor may be oppositely electrified in different portions of its surface, there must be in the field some places where the potential is higher than that of the body and others where it is lower. We must remember that at an infinite distance the potential is zero.
Let us begin with two conductors electrified positively to the same potential. There will be a point of equilibrium between the two bodies. Let the potential of the first body be gradually raised. The point of equilibrium will approach the other body, and as the process goes on it will coincide with a point on its surface. If the potential of the first body be now increased, the equipotential surface round the first body which has the same potential as the second body will cut the surface of the second body at right angles in a closed curve, which is a line of equilibrium.
Earnshaw’s Theorem.
116.] An electrified body placed in a field of electric force cannot be in stable equilibrium.
First, let us suppose the electricity of the moveable body ($A$), and also that of the system of surrounding bodies ($B$), to be fixed in those bodies.
Let $V$ be the potential at any point of the moveable body due to the action of the surrounding bodies ($B$), and let $e$ be the electricity on a small portion of the moveable body $A$ surrounding this point. Then the potential energy of $A$ with respect to $B$ will be
$M=\sum (Ve)$
where the summation is to be extended to every electrified portion of $A$.
​Let a, b, c be the coordinates of any electrified part of $A$ with respect to axes fixed in $A$, and parallel to those of x, y, z. Let the coordinates of the point fixed in the body through which these axes pass be $\xi ,\eta ,\zeta$.
Let us suppose for the present that the body $A$ is constrained to move parallel to itself, then the absolute coordinates of the point $a,b,c$ will be
$x=\xi +a,\ y=\eta +b,\ z=\zeta +c.$
The potential of the body $A$ with respect to $B$ may now be expressed as the sum of a number of terms, in each of which $V$ is expressed in terms of a, b, c and $\xi ,\eta ,\zeta$, and the sum of these terms is a function of the quantities a, b, c, which are constant for each point of the body, and of $\xi ,\eta ,\zeta$, which vary when the body is moved.
Since Laplace’s equation is satisfied by each of these terms it is satisfied by their sum, or
${\frac {d^{2}M}{d\xi ^{2}}}+{\frac {d^{2}M}{d\eta ^{2}}}+{\frac {d^{2}M}{d\zeta ^{2}}}=0$
Now let a small displacement be given to $A$, so that
$d\xi =l\ dr,\ d\eta =m\ dr,\ d\zeta =n\ dr;$
then ${\tfrac {dM}{dr}}dr$ will be the increment of the potential of $A$ with respect to the surrounding system $B$.
If this be positive, work will have to be done to increase $r$, and there will be a force ${\tfrac {dM}{dr}}$ tending to diminish $r$ and to restore $A$ to its former position, and for this displacement therefore the equilibrium will be stable. If, on the other hand, this quantity is negative, the force will tend to increase $r$, and the equilibrium will be unstable.
Now consider a sphere whose centre is the origin and whose radius is $r$, and so small that when the point fixed in the body lies within this sphere no part of the moveable body $A$ can coincide with any part of the external system $B$. Then, since within the sphere $\nabla ^{2}M=0$, the surface-integral
$\iint {\frac {dM}{dr}}dS=0,$
taken over the surface of the sphere.
Hence, if at any part of the surface of the sphere ${\tfrac {dM}{dr}}$ is positive, there must be some other part of the surface where it is negative, ​and if the body $A$ be displaced in a direction in which ${\tfrac {dM}{dr}}$ is negative, it will tend to move from its original position, and its equilibrium is therefore necessarily unstable.
The body therefore is unstable even when constrained to move parallel to itself, à fortiori it is unstable when altogether free.
Now let us suppose that the body $A$ is a conductor. We might treat this as a case of equilibrium of a system of bodies, the moveable electricity being considered as part of that system, and we might argue that as the system is unstable when deprived of so many degrees of freedom by the fixture of its electricity, it must à fortiori be unstable when this freedom is restored to it.
But we may consider this case in a more particular way, thus—
First, let the electricity be fixed in $A$, and let it move through the small distance $dr$. The increment of the potential of $A$ due to this cause is ${\tfrac {dM}{dr}}dr$.
Next, let the electricity be allowed to move within $A$ into its position of equilibrium, which is always stable. During this motion the potential will necessarily be diminished by a quantity which we may call C dr.
Hence the total increment of the potential when the electricity is free to move will be
$\left({\frac {dM}{dr}}-C\right)dr$
and the force tending to bring $A$ back towards its original position will be
${\frac {dM}{dr}}-C$
where $C$ is always positive.
Now we have shewn that ${\tfrac {dM}{dr}}$ is negative for certain directions of $r$, hence when the electricity is free to move the instability in these directions will be increased.