498     Prof.  Wilson  and  Mr.  Gold :  Electrical  Conductivity  of 
remains  n  except  within  a  distance  d  o£  each  electrode, 
d  being  twice  the  amplitude  of  vibration  of  the  negative 
ions.  It  is  easy  to  see  that  on  these  assumptions  the  negative 
ions  will  practically  all  be  contained  in  a  slab  of  thickness 
~D—d,  which  will  vibrate  between  the  plates  so  as  just  not  to 
touch  either  of  them.  For  if  a  new  negative  ion  is  formed 
outside  the  slab,  it  will  almost  immediately  strike  the  electrode 
near  it  ;  whereas  a  new  negative  ion  formed  in  the  slab  cannot 
reach  either  electrode  except  by  diffusion,  which  we  shall 
neglect.  Thus  we  may  regard  the  whole  space  between  the 
plates  as  filled  with  positive  electricity  of  density  -\-ne,  and 
the  vibrating  slab  of  thickness  D  —  d  as  containing  also 
negative  electricity  of  density  —  ne.  Thus  inside  the  slab 
the  total  density  is  zero  and  outside  is  +ne. 
Let  X  denote  the  electric  intensity   between   the   plates 
at  a  distance  x  from   one  of  them.     Then,  inside  the  slab 
-^— =0,  and  outside  -f-  ^iirne. 
dx  dx 
Let  A  and  B  be  the  two  plates,  and  let  the  slab  be  repre- 
sented by  the  space  between  the 
two  dotted  lines  E,  F.  Let  AB  =  D, 
AE  =  £l3  FB=^25  anJ  suppose  the 
potential  of  A  kept  zero  while  that 
of  B=V.  Let  the  rise  of  potential 
in  AE  be  Yl5  in  EF  be  V3,  and  in 
FB  be  V2.     In  EF  ^7X  =  0  so  that 
dx 
X   is    constant.       Let   its  value    be 
X0.     Then 
VB=-X0(D-<0, 
where  d  =  t1  +  t2.     In  AE  we  have 
d2Y 
tl 
va 
D 
E 
B 
where 
dV  = 
dx 
dx°- 
From  this  we  get 
Airpx  +  C     and     Y  =  —  2'rrpx'1  +  CU'  +  D, 
where  C  and  D  are  constants  to  be  determined.    When  x  =  0 
Hence 
Y  =  D  =  0, 
and  when  x  =  t1        jy 
da 
=  -4:7rpt1  +  G-X0. 
Y1  =  27r/of12--X(/, 
