﻿Electricity on Streams of Water Drops. 163 



For the particular case 



£_:!? -.31 



... E = --31D= +'62 (J Yr). 



Again, i£ drop B came off singly, it would have a potential 

 approximately equal to — 2/3 V and consequently a charge 

 equal to — 2/3 Yr induced by the inductor tube. That is, 

 the net charge, E', on the drop B is 



_gv,) + .62(|v;)=-38(?V,} 



The ratio of the charges D and E'on A and B will be 



D:E' = 2: -38 = 5-3:1. 



Reversing the order in which A and B leave the stream so 

 as to have the smaller one escape first, the effect of B on the 

 potential of A will be proportionally much smaller. From 

 equations (10), putting u = we have 



T> = -Jvr, E=+Ki;r, 



D =-4- E =+l-(14 



For this particular case 



o * Q , J -79 _ 



a = 2r, b = r, c=3'5r, g = JT34 ='° 9 - 



The charge given to A by the inductor ring is —2/3 . Y . 2r, 

 therefore net charge, D', on A is 



-»gvr) + -59(!v,)--wigv,), 



which leaves the ratio of D' : E = l*41 : 1. 



It is apparent from these facts that in consideration of 

 accidental causes which influence the breaking of the streams 

 (Rayleigh *) we may expect to have in the stream drops 

 large and small bearing charges of same sign (or it is con- 

 ceivable of opposite sign) having almost any ratio : as a 

 consequence in general for small distances we shall have 

 attractions of greater or less force and at groat distances 

 repulsions. 



* Sc. Papers, ii. p. 10-°>. 



M2 



