﻿162 Prof. Barton and Mr. Wiegand : Effect of 



then given by D=-2/3Vr. 



But from equation (6), if B is distant, centre to centre, 

 cr from A, the charge E on B due to charge D on A will be 

 given by 



D = (Ii*-Jr>, 



E=(-Jtt + l!>)r. 



But jnst as it breaks away B is at zero potential : putting- 

 then v — we have 





T) = Iur, E = -Jw. 



rom which 







E=-f.D= + ^.V, 



For the distance from centre to centre equal to 2*5 r, just as 

 B breaks away, 



J_ -52537 



I ~ 1-25324" 



••• E= + -42(|V,). 



But if the drop B came off singly, it would have approxi- 

 mately a potential — 2/3 V and consequently a charge 

 — 2/3 W, induced by the inductor tube. The total 

 charge, E', on B will then be 



E'=-|Vr + -42Qv^=--58Qvr). 



Ratio of charges D and E' on drops A and B is 1*7 : 1. 



As our photographs show there is a variation in the sizes 

 of the drops leaving the stream. The effect of a large drop 

 on a succeding small one is much more noticeable than the 

 above case of equal drops. Let radii of A and B be 

 respectively 2r and r and the potential of A as it leaves be 

 the same as before. Then if the nearest point of the drop A 

 is *5r distant from boundary of B, just as the latter breaks 

 away, we have the case of two drops (equation 10), in which 



a = 2r, b = r, and c = 3*5r. 



If potential of A is — 2/3V, the value of charge D on A is 



_2/3V.2r=-2(^vA 



From equations (10), putting potential of B equal zero 

 (?< = 0). we have 



D=i«r, E= J 



E=— our. I 



