Law of the Induction of an Electric Current upon itself. 35 
And first, if we divide the charge between the two wires, we 
: must now integrate (a) considering P as a function of the time, 
and two successive integrations give us 
aQ’. AzR 
dt ~ Cx? w fia we 
and ; 
Aydt . : 
V=Ga y Pat. ‘ 
Now as the discharge is divided between the wires, the electro- 
motive force P at any given instant, is the same obviously in both 
Wires and consequently /fPd@? is the same in both, and the 
quantity of statical electricity Q’ discharged by each is conse- 
3 d2 Q’ 
_ the value of oe Hence equation (a) shows that 
R 
— P 
S 
_ must have the same value in both wires, so that P the resistance 
___ to discharge, is inversely as the diameter of the wire and directly 
as its length. 
have proceeded on the tacit supposition, that its surface isnot = 
extensive surface capable of itself receiving or absorbing the 
urface of Fyaes 
