An introduction to the mathematical theory of attraction ...

Examples. 211
curve subtends at A1, &c.; and E the total charge required;
then
47rE = 4 a f dS = f NdS = o J O c cos c s + &c. d
r2 r 2 2 )
= ei2 + e2Q2 + &c.; whence
elig + e2-22 + &c.
E= (05)
47r
If S be a closed surface, ~~ is equal to 47r or zero, according
as A is inside or outside the surface, and the total mass
on S is equal to the sum of the charges at the internal
points.
EXAMRPLES.
1. An insulated conductor formed of the larger segments of two spheres
cutting orthogonally is in electric equilibrium: what is the density of the distribution at a point on the circle of intersection of the spheres? Ans. O.
2. A conductor formed of the larger segments of two spheres cutting
orthogonally, whose centres are A and B, is at potential zero under the influence
of an external electrified point 0: find the potential at any point, and the distribution of mass on the conductor.
I XLet I and J be the images of O in the spheres, then AJ and BI intersect at
a point K such that
AK. AJ= a2, BBK. BI = b2
where a and b are the radii of the spheres.
For, if G be the point in which AB meets the
j/ \/\\ \ plane of intersection of the spheres, the quadri/ I -sKI\^ \ Klaterals BGIO and AGJO are cyclic; whence
angle GBI= GOI= AJG, and GKJB is cyclic,
\ CGJ B jand AK. AJ = AG. AB = a2. In like
manner BK. BI = b2
Hence charges - at, - B- at J,
AO BO
eab eab
AJ. BO BI. -AO
produce a potential which is zero at the surface of the conductor, the charge at
O being e. oIt is easy to see that AJ. BO = BI. AO, since the quadrilateral
IOJK is cyclic anc therefore the angle AJO = BIA.
If we put AO =fi, BO =f2, and express AJ. BO in terms offi, f2 a, and
b, we find AJ2. BO2 = a2f22 + b2fi2 - a2b2.
P 2