504  The  Electrical  Conductivity  of  Flames. 
Equation  (5)  may  be  written 
i=  —pepYpsm  (pt  —  cc) 
{ (m  Dp2  -Air  de  p)2  +  A2  D2 }*" 
It  is  easy  to  show  that  the  term  A2D2  is  negligible  compared 
with  (mDp2  —  ±7rdep)2.     We  have  D  =  0-6  cm.  and  Xe  =  Av, 
where  X= electric  intensity  and  v  =  velocity  of  negative  ions 
duetoX.     If  X=l  E.S.unit,  v  =  3xl05— ;  so  that  D2.A2 
sec. 
is  equal  to  0*6  x  9  x  10"20  _  0       n_99 
9xlOw    -~6X1U  -• 
In  (mDp2 — Att  de  p)2  the  term  ±irdep  is  about  6  x  10-9  for 
6?  =  3  x  10-10,  and  Air  dp  is  about  20.     Also  when  p  is  say  1067 
mDp2  must  be  small  compared  with  A^dep,  because,  as  v*e 
have  seen,  changing  p  does   not  much  affect  the  apparent 
capacity.     Hence  (mDp2  — Circle  p)2  is  of  the  order  10-1S. 
The  expression  for  i  becomes  therefore  on  putting  a  =  —90°, 
._    pepY0  cos  pt 
4zirdep  —  mDp2' 
Hence^the  apparent  capacity  per  unit  area  is 
n-  eP 
4zirdep—mDp2 
Therefore  if  L\  and  C2  are  values  of  C  corresponding  to 
values  2h>  Vi  ot V->  we  have 
±*dp--Dp22=:^; 
Also  approximately  p  =  87rV0C2  when  p  is  small  enough 
for  mDjt?2  to  be  small  compared  with  ±irdep,  so  that 
«  = D^lra2) ' 
If  C^nearly  =C2  we  can  put  C2  =  C102;  and  get 
g  _    D(^22—  j?i2) 
m      87rV0(C2  —  C,) 
Thus  if  the  variation  of  C  \vith_p  were  known  with  sufficient 
accuracy,  —  for  the  negative  ions  could  be  calculated.     It  is 
hoped  that  further  experiments  will  enable  this  to  be  done  for 
the  negative  ions  of  different  salts. 
