Page:MortonCharge.djvu/5

184 (x y ζ) on the surface, are

$$-\frac{d\phi}{dx},\ -\frac{d\phi}{dy},\ -\frac{d\phi}{d\zeta}$$.

This force acts in the normal to the surface, and is proportional to the surface-density at (x y ζ), which we shall call σ'. Therefore

$$\left(\frac{d\phi}{dx},\ \frac{d\phi}{dy},\ \frac{d\phi}{d\zeta}\right)=-A\sigma'\left(\frac{dF}{dx},\ \frac{dF}{dy},\ \frac{dF}{d\zeta}\right)/\sqrt{\left(\frac{dF}{dx}\right)^{2}+\left(\frac{dF}{dy}\right)^{2}+\left(\frac{dF}{d\zeta}\right)^{2}}$$

But

$$\frac{d}{d\zeta}=k\frac{d}{dz}$$;

therefore, denoting differentiation with respect to x y z by subscripts 1 2 3,

$$(\phi_{1},\ \phi_{2},\ \phi_{3})=-A\sigma'(F_{1},\ F_{2},\ F_{3})/\sqrt{F_{1}^{2}+F_{2}^{2}+F_{3}^{2}}$$.

Now let σ be the surface-density at (x y z) on the moving conductor F(x y z)=C, then equating σ to the normal component of (f g h),

$$\sigma=\frac{fF_{1}+gF_{2}+hF_{3}}{\sqrt{F_{1}^{2}+F_{2}^{2}+F_{3}^{2}}}$$,

or putting in the values we have found for (f g h) in terms of φ

Now the perpendicular from the origin on the tangent plane to F(x y z)=C at the point (x y z) is

$$p=\frac{xF_{1}+yF_{2}+zF_{3}}{\sqrt{F_{1}^{2}+F_{2}^{2}+F_{3}^{2}}}$$,

and the perpendicular from the origin on the tangent plane to F(x, y, kζ)=C at (x y ζ) is

$$p'=\frac{x\frac{dF}{dx}+y\frac{dF}{dy}+\zeta\frac{dF}{d\zeta}}{\sqrt{\left(\frac{dF}{dx}\right)^{2}+\left(\frac{dF}{dy}\right)^{2}+\left(\frac{dF}{d\zeta}\right)^{2}}}$$