On the Electrostatic Deflection in a Cathode Ray Tube. 
23 
Take O ^ as origin, and O^Q as the positive direction of x. The value of 
/^^ at P is given by — 
- 4-40 = C (/5 — log - 1 + IS), where 27rC = 0-69 
/5 - log - (1 + /5) = - 40-06, 
which gives /ip = — 36"50, and log — (1 + P^) = 3*57. 
The value of at the origin is given by — 
ISo, = log - (1 + fSo,), 
l3o, = - 1-2785, and log - (1 + fh,) = - 1-2785. 
The deflection y-^ at 0| is given by — 
e 
mv- 
e 
= CD 1 + /ilogl + /i - 1 + /3 - i log-' 1 + -4-41ogl -f /^p 
> 
'y^ = 97-3 CD 15-9 D. 
^ 
A 
. Oi 
to '25 cm 
The velocity parallel to y is given by — 
= - 4-85 D. 
To find the deflection due to the field from 0^ to Q we now take 0. as 
origin and OoP as the positive direction of ,r. 
The value of /5 at Q is given bv — 
C{/3q -log{(l +/5q)}= -10-25, 
/?Q-log- (1 +/:^q)= -93-31, 
which gives /5q = — 88-82, and log I -\- [Sq = 4-47. 
An approximate value of fS at 0^ is — 1, say ( — 1 — y), 
then 3-95 = C(- 1-7 -logy), 
y + log 7 = — 36'96. 
An approximate solution is y = e- 
therefore Uo, = D log - (1 + /d'oJ = - 36-96 D. 
We may now regard the ray as projected from with velocities given by 
dx 
dt 
= — V, 
