Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/188

of each. These are indicated by the dotted circles on the right hand of Fig. 6.

If there be another centre of force, we may in the same way draw the equipotential surfaces belonging to it, and if we now wish to find the form of the equipotential surfaces due to both centres together, we must remember that if $V_{1}$ be the potential due to one centre, and $V_{2}$ that due to the other, the potential due to both will be $V_{1}+V_{2}=V$. Hence, since at every intersection of the equipotential surfaces belonging to the two series we know both $V_{1}$ and $V_{2}$, we also know the value of $V$. If therefore we draw a surface which passes through all those intersections for which the value of $V$ is the same, this surface will coincide with a true equipotential surface at all these intersections, and if the original systems of surfaces be drawn sufficiently close, the new surface may be drawn with any required degree of accuracy. The equipotential surfaces due to two points whose charges are equal and opposite are represented by the continuous lines on the right hand side of Fig. 6.

This method may be applied to the drawing of any system of equipotential surfaces when the potential is the sum of two potentials, for which we have already drawn the equipotential surfaces.

The lines of force due to a single centre of force are straight lines radiating from that centre. If we wish to indicate by these lines the intensity as well as the direction of the force at any point, we must draw them so that they mark out on the equipotential surfaces portions over which the surface-integral of induction has definite values. The best way of doing this is to suppose our plane figure to be the section of a figure in space formed by the revolution of the plane figure about an axis passing through the centre of force. Any straight line radiating from the centre and making an angle $\theta$ with the axis will then trace out a cone, and the surface-integral of the induction through that part of any surface which is cut off by this cone on the side next the positive direction of the axis, is $2\pi E(1-\cos \theta )$.

If we further suppose this surface to be bounded by its inter section with two planes passing through the axis, and inclined at the angle whose arc is equal to half the radius, then the induction through the surface so bounded is

$E(1-\cos \theta )=2\Psi$ say;

and $\theta =\cos ^{-1}\left(1-2{\frac {\Psi }{E}}\right).$

If we now give to $\Psi$ a series of values 1, 2, 3 ... $E$, we shall find