An introduction to the mathematical theory of attraction ...

Induced Distribution on Sphere. 199
due to e at A and to e' at B must be along the normal at
every point of S, but at the
// v NP point D in which CA meets S,
-/ } ^ /k / the force due to e is normal; so
D {-' g - — ' A also, therefore, is that due to
- - ')A ée', and consequently B must lie
\.: ~/ ~ in the line CA. Again, if AC
produced meet S in D', we have
e e' e e'
AD) + = = AD' + BD7"
whence D'D is cut harmonically in B and A, and therefore
CA.CB = CD2. Accordingly, B is a point in CA determined
by this equation if A and B be images of each other in S.
Let P be any point of the sphere S, and let CA =f,
CB =f', CP = a; then, from the similar triangles ACP and
PCB, we have AP: BP =: a. Hence, if we assume
ea e el
e' = - e, we obtain e+ = 0,
and B is the image of A, and the charge e' at B is given by
the equation
(1)
111. Induced Distribution on Sphere.-The distribution of electricity on a sphere at potential zero under the
influence of an external electrified point A can now be readily
determined.
Adopting the same notation as that of the preceding
Article, let o denote the density of the surface distribution
at any point P of the sphere whose distances from A and B are
r an'd r'. Then, if R be the resultant force at P, this force
is in the direction of the normal; and therefore, if we resolve
the forces due to e at A, and to e' at B along CP and any
other direction, the resultant along this latter direction is zero,
and R is the sum of the components along CP. Hence,
resolving along CP and CA, if the direction of R be from C
to P, we have
e a e' a
R=J;+ - r2
r' r