Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/178

 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

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

or

the solution of which equation in ascending powers of $$r$$ is