( 50 ) 
1 _- € cos (D €) 
NE 
Pon D 
if D is in the direction of this radius vector, as is indicated by the 
index D of r,. If an electric induction in the direction I does not 
require a component of electric force in the direction II, then I and 
IL are conjugate diameters of the electric ellipsoid, or briefly electrically 
conjugate diameters; we may also say that an electric induction in 
the direction II does not require a component of electric force in 
the direction I. 
2. If we solve f, g and ’ from the equations (1), (2) and (3), 
we find: 
f= We PAU, Otho ia oe GON 
I= hijs P+ ky Q + hye Bi eel ige NN 
hl PC Bey Oot BiyRieg Ge 
where kiy= kje Maz == ken Kye =z. I shall call the surface: 
oe ad “fF hijg y° |. kes 2 -+ 2 Ry: ye |. 2 eee zv + 2 K zy Uy = 1 
the reciprocal electric ellipsoid. Its axes have the same direction 
as those of the electrie ellipsoid, but the axes of the one are the 
reciprocal values of the axes of the other. If € is a radius vector 
: ase 1 
of the reciprocal electric ellipsoid, then U, = En. and D is normal 
TL 
to the diametral plane that is conjugate to the direction of €. 
The radius vector #', of the reciprocal electric ellipsoid is: 
1 2 cos (D €) 
Ten OE 
KS 
ee 
if € fails in the direction of that radius vector. If the directions 
I’ and IL are conjugate diameters of the reciprocal electric ellipsoid, 
then an electric force in one direction does not produce an electric 
induction in the other direction. 
3. We take the medium as not permanently magnetizable. If 
a, /?, y are the components of the magnetic force 9, and a, 5, c those 
