602 
ME. J. HOPKINSON ON THE 
If ^/((y) is determined for all values of a, the properties of the glass, as regards con- 
duction and residual charge, are completely expressed. 
Suppose that in equation (2) ^=0 till £=0, and that after that time # 4 =X a constant, 
y,=X(l+JVw*»), 
%=*m: 
now when t is very great, ^ is the steady flow of electricity through the glass divided 
by the capacity. Hence 
4<«)=B (3) 
B is the reciprocal of the specific resistance multiplied by 4<r and divided by the 
electrostatic capacity of the substance. 
We have no practicable method of determining y t ; but we may proceed thus: — 
During insulation y t is constant ; we have then 
r,=A-j\_ 
^{co)dco ; 
(4) 
x t and x t _ a alone can be measured ; (4) is, then, the equation by aid of which /(«) must 
be determined. 
(a) Let x t be maintained constant=X from time 0 to time t, then insulate ; differ- 
entiating (4), 
= -X^ f 
= — BX when t is very great. J 
( 5 ) 
To find B, charge for a long time to a constant potential, insulate and instantly 
observe the rate of decrease of the potential. 
(/3) Let the flask be charged for a very short time r and then be insulated ; at the 
instant of insulation we have ^=— X\J/(r). Hence an approximation may be made 
to an inferior limit of ^(O). 
(y) Let x t be constant =X for a long time from t=— T to £=0; discharge and, 
after a further time t, insulate : — 
r* +T =A— X f 
T+i+ 
| ®t-a+ T $(®)da , | 
~z=Xty(t) — B) when t vanishes. 
. . . ( 6 ) 
To find ip(t) in terms of t charge for a very long time, discharge and from time to time 
insulate and determine 
