through Exhausted Tubes without Electrodes. 461 



In the metal the electromotive intensity satisfies the 

 equation 



d' 2 E _ AirfidE 



dx 2 " a dt ' 



so that the electromotive intensity E 2 inside the plate may 

 be represented by the equation 



where 



7YI = { 27TfjLp/<7}i. 



On the far side of the plate the electromotive intensity E 3 is 

 given by 



E 3 = ce t( ^ +ax) . 



To determine the coefficients b, c, A, B, we have the con- 

 ditions that at the surface of the plate, i.e. when x = 0, and 

 when x= — h, h being the thickness of the plate : — 



(1) The electromotive intensity is continuous ; 



(2) The magnetic force parallel to the plate is continuous. 



This implies that - -r- is continuous. 

 //, ax 



From (1) we get 



l + b = A + B, 



C6 -ax = J± e m(l + L )h + J* € -m{l + i)h . 



while from (2), 



a(l-6) = (l + 0-(A-B), 



ace- ah =—(l + i) — (Ae m ( l+ ^ h —Be- m ( 1 +^ h ). 



We can simplify these equations if we remember that for 

 electrical vibrations as rapid as those we are now considering 

 m/afi is exceedingly large. Introducing this condition we find 



B = 



m(l + l) 6 m ( 1+ -0*-_ e -m{l+i)h> 



2a/i 4-a+& 



Thus in the metal plate the electromotive intensity is given 



by the equation 



/ 



2 fw(l + 4 ) e»d+')* — e-»d+0*i € ^ '' ' 



