( 436 ) 
= .( 52 ) 
Substituting the values of v x , v\ , u x , and u 9 according to (42), 
(43), (44) and (45) in (51) and (52), we find 
U = kJe^RX . . .( 53 ) 
tga=z{k,~ : k x )H .. (54) 
Considering the values of k x and k 2 to be known, and being able 
to measure U and X (A" roughly indeed), we may verify the 
formula (53). But if k x and k 9 are considered as unknown quantities 
the formula may serve to find the product k x L. 
hrom other experiments the values of k x and k. 2 are sufficiently 
well known. Substituting these values in the formula (54) we find 
that the angle u differs very little from zero, so that the inclination 
will not be visible. 
^Now one can ask whether positive and negative charges will not 
be formed in the front and at the back of the discharge, as in Hall’s 
phenomenon. 
We suppose until further notice that the path of the discharge is 
parallel to the axis of the cylinders and that the said charges have 
been formed, in consequence of which the ions are acted upon by 
a horizontal electric intensity Y, X being again the vertical electric 
intensity. The velocities v x and v 9 will be as great as in the former 
case, but the horizontal velocities of the positive and negative ions 
become resp. 
Ml = k x *HX + k x Y .(55) 
and 
u 2 — — k 9 Y .( 56 ) 
Both of them must now be equal to U, the velocity of the dis¬ 
charge. So we find 
• u = k x *BX + r = k 9 HX — k 9 Y .(57) 
whence 
Y = {k 9 -k x )HX . ..... ( 58 ) 
and 
U = k x k 9 HX .(59) 
The same value as before is found in this way for the velocity U. 
It is quite possible that the path of the discharge has an arbitrary 
