106 DOC.
90
JUNE
1915
At
infinity
E
=
cl+c2/r2
or c2/r2 as
usual.
If
you
name
the
charge
E,
then
also results
c1
+
c2
=
E/4n
c2
= E/4n
On
the other
hand,
the total
charge
must
be
determined
as
follows:
1
dCr2
r2
dr
E
=
4nr2drp
=
4n(Cf r2 ¡OO
ri
4n
\/Cier
+
c2e
r
k k
E
=
4n
\l
C1
sin h
c2
cos

r
r
If
charge
is to
remain
finite,
then
is
required
c1
=
0
ri
=
0
K
K
r
=
ri
c1
sin
k/ri+
c2
cos
K/ri
=
0
hence
E
=
4n/c2
and hence also
E
=
0
within the
cavity’s
boundary
condition
p=1/r2d/c2ek/2r/r2dr
p=1/r2d//c1sink/r+c2cosk/r/r2dr
=k/2r4ek/2r=8k3/c2(k/2r)4ek/2r
/sink/r
2
2/k
r/k