Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/181

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.