Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/180

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,