Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/176

 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