II
INDUCTION IN ROTATING SPHERES
43
Hence these conditions follow for :—
In the matter of the shell
√² 4 = 2w=3%
дъ
(a)
for
ρ
(b)
= r and p = R
аф
W
=
nzXn
др Р
Əz
A solution of these equations is
For
φ
=
W
n+1
D³(po³xe - naxa).
მა
nzxn
Əz
= Xn
- 2(2n+1) X^ - 2n³X (§ 1, 4)
==
Əz
Əz
Determina-
tion of the
electric
potential.
=
2(n + 1) Xn
Əz
so that the equation for the interior is satisfied. Further,
is the product of p+1 by a function of the angles and w,
whence it follows that satisfies the boundary conditions.
The value of the constant which must be added to the
above expression to give the general solution depends in each
case on the electrostatic influences to which the shell is
subject. We may in any case charge the sphere with so
much free electricity as will just make the constant zero, and
this we shall in the sequel suppose to have been done.
From
we get at once
W
K
2 =
K
-
-
w =
ய
K
-
1 a
ρ
n+1dx
მ
1
n+1ǝy
2
όχι
Əz
nzxn+x-
-
dz
1
a
Əz
n+1əz
=) +
・nzxn+y-
nz Xn +2
Əz
Əz
Əz
}
Determina-
tion of u, v,
20.