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

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

Now let a small displacement be given to $$A$$, so that

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

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,