2
ON THE DISTRIBUTION OF ELECTRICITY
[706
(x, b, a). Hence, first, for a point interior to the sphere A, if x denote the distance
from A, and therefore c — x the distance of the same point from B, the potential of
the point in question is
= acf)X +
b 2
C — X
$
— x) ’
and, secondly, for a point interior to the sphere B, if x denote the distance from B
and therefore c—x the distance of the same point from A, the potential of the
point is
cx
c — x
</>
(T
c — x
+ bA> (x).
The two equations thus express that the potentials of a point interior to A and of
a point interior to B are =h and g respectively.
It is to be added that the potential of an exterior point, distances from the points
A and B — x and c — x respectively, is
cX
= — 9
x T
Ò 2 ru
+ ~ $
c—x
b 2
c — x.
and that, by the known properties of Legendre’s coefficients, when the potential upon
an axial point is given, it is possible to pass at once to the expression for the potential
of a point not on the axis, and also to the expression for the electrical density at a
point on the two spherical surfaces respectively. The determination of the functions
cf>(x) and <L(«) gives thus the complete solution of the question.
I obtain Poisson’s solution by a different process as follows:—Consider the two
functions
8XC + b
and
and let the nth functions be
a? (c — x)
c 2 — b- — ex’
b 2 (c — x)
c 2 — a? — ex'
a n x H - b,
cx + d ’
OLX + ft
yx + 8 5
a n x + /3,
suppose,
suppose ;
c n x + d,
and
y n x + B n
respectively.
Observing that the values of the coefficients are
(a, b ) = ( — a 2 , cXc ), and (ot, /3 ) = ( — b 2 ,
b 2 c
c,
— c ,
b 2
— c, & — cX
so that we have
a + d = a + B, = c 2 — a 2 — b 2 , ad — be = ctB — /3y, = cXb 2 ,
and consequently that the two equations
(A, + !)■’ (a + d) 2 (A. + 1) J (a. + B) 2
ad — be ’
aS — (3y ’
A,
A.