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IOWA ACADEMY OF SCIENCE Vol. XXIV, 1917 
face charge a between the limits r 1? r 2 , which results in an ex- 
pression for Q, the total charge on one of the plates. Omission 
of the potential difference from this expression leaves the equa- 
tion for capacity in terms of distance. The capacity for any 
shape of the opposed limited areas of the equipotentials can he 
'i/ 
v 
Figure 34 
similarly obtained by suitable integration of a over one of the 
areas, neglecting edge corrections for actual plates when their 
distance of separation is small compared with the other dimen- 
sions. 
In the experiment referred to the plates were non-rectangular, 
figure 36. Choose x-and y-axes as indicated in the figure. Con- 
-/4S- 
sidering the element of area, dy dx, its charge is dq— dy dx, 
where ff =1/4 7r dV 1 -V 2 )/© T/y. Integrate first over the 
strip parallel to the x-axis of width dy, and let the charge on 
the strip be represented by dQ. Thus, 
dQ=l/ 4- ^(Vi-Vg)/ ©’dy/ y*x (2) 
where x = l/m • (y + mx 1 -y 2 ), if m is the slope of the 
slant edge. The equation for this edge is, y=mx+b, where 
m = (r 2 -r, ) / (wi -w 2 ), and b=r 2 -mwi . (In figure 36, 
y 1 = r l5 y 2 =r 2 , = x 2 , w 2 = x t ). The next integration is 
that of the strips between the limits r 1} r 2 , and we get, after 
canceling the potential difference, 
0=1/4 7 r, [(w 2 r 2 -Wir 1 )/ (r 2 -r])' lo S ( I- 2/ r d'(w 2 -w 1 )]T/0 (3) 
