DOUBLE-STREAM AMPLIFIERS 



663 



We shall treat only a special case, that in which 



Ji Ji 



/o 



3 — 3 — 3 ' 

 U\ «2 Wo 



(16.10) 



Here /o is a sort of mean current which, together with wo , specifies the ratios 

 J\.lu\ and J-Jui^ which appear in (4) and (5). 



In terms of these new quantities, the expression for the total a-c charge 

 density p is, from (16.2) and (16.3) and (16.6) 



d2 



P = Pi + P2 = 



/or 



(16.11) 



Equation (16.11) is a ballistic equation telling what charge density p is 

 produced when the flow is bunched by a voltage V. To solve our problem, 

 that is, to solve for the phase constant /3, we must associate (16.11) with a 

 circuit equation which tells us what voltage V the charge density produces. 

 We assume that the electron flow takes place in a tube too narrow to propa- 

 gate a wave of the frequency considered. Further, we assume that the wave 

 velocity is much smaller than the velocity of light. Under these circumstances 

 the circuit problem is essentially an electrostatic problem. The a-c voltage 

 will be of the same sign as, and in phase with the a-c charge density p. In 

 other words the "circuit effect" is purely capacitive. 



Let us assume at first that the electron stream is very narrow compared 

 with the tube through which it flows, so that V may be assumed to be con- 

 stant over its cross section. We can easily obtain the relation between V 

 and p in two extreme cases. If the wavelength in the stream is very short 

 (J3 large), so that transverse a-c fields are negligible, then, from Poisson's 

 equation, we have 



^z (16.12) 



p = e^W 



If, on the other hand, the wavelength is long compared with the tube radius 

 (S small) so that the fields are chiefly transverse, the lines of force running 

 from the beam outward lo the surrounding tube, we may write 



= CV 



(16.13) 



Here C is a constant expressing the capacitance per unit length between the 

 region occupied by the electron flow and the tube wall. 



