FIELD SOLUTIONS 635 



We recognize d as the half-width of the opening filled by electrons. Then 



-A = 1 - (^ (14.15) 



We can say something about the quantity .4. From purely d-c considera- 

 tions, the electron flow will cause a fall in d-c potential toward the center 

 of the beam. Indeed, this is so severe for large currents that it sets a limit 

 to the current density which can be transmitted. If we take Fo and Uq as 

 values at ^^ = ±d (the wall), the maximum value of A as defined by (14.12) 

 is 2/3, and at this maximum value the potential at y = is Fo/4. This is 

 inconsistent with the analysis, in which Vq and Mo are assumed to be con- 

 stant across the electron flow. Thus, for the current densities for which the 

 analysis is valid, which are the current densities such as are usually used in 

 traveling-wave tubes 



A « 1 (14.16) 



In the a-c analysis we will deal here only with the symmetrical type of 

 wave in which £e(+y) = Ez{—y). The work can easily be extended to 

 cover cases for which Ez{-\-y) = —E;{—y). We assume 



H,= Hosmhyye~'^' (14.17) 



From Maxwell's equations 



jojeEy = "^^ = —j^Hois'mh yy)e 



dHx .^„ / . , N -S0Z 



Ey = - ^ Hoisinh \y)e~'^' (14.18) 



coe 



Similarly 



jueiEz = — ~—^ = — 7//o(cosh 7v)e "'^^ 

 ay 



£, = ^ Fn(cosh yy)e~'^' (14.19) 



coei 



We must also have 



—joinUx = -— — ^— 

 ay dz 



jcofxHoe ■'^"' sinh yy = "^-^ H„e '^' cosh 7V — -- HqC ^^' sinh 7y 



coei ' we 



y = (€i/e)0S2 _ ^l) (14.20) 



/3o = aj2^€ = «Vc2 (14.21) 



