We may now write the hyperbolic sines in terms of
p
,
q
,
r
{\displaystyle p,q,r}
q
a
a
=
∑
s
=
0
s
=
∞
2
k
r
s
p
−
p
r
s
,
q
a
b
=
−
∑
s
=
1
s
=
∞
2
k
r
s
−
1
r
s
,
q
b
b
=
∑
s
=
0
s
=
∞
2
k
q
r
s
−
1
q
r
s
.
{\displaystyle {\begin{array}{ll}q_{aa}=&\sum _{s=0}^{s=\infty }{\frac {2k}{{\frac {r^{s}}{p}}-{\frac {p}{r^{s}}}}},\\q_{ab}=-&\sum _{s=1}^{s=\infty }{\frac {2k}{r^{s}-{\frac {1}{r^{s}}}}},\\\\q_{bb}=&\sum _{s=0}^{s=\infty }{\frac {2k}{qr^{s}-{\frac {1}{qr^{s}}}}}.\end{array}}}
Proceeding to the actual calculation we find, either by this process or by the direct calculation of the successive images as shewn in Sir W. Thomson's paper, which is more convenient for the earlier part of the series,
q
a
a
=
a
+
a
2
b
c
2
−
b
2
+
a
3
b
2
(
c
2
−
b
2
+
a
c
)
(
c
2
−
b
2
−
a
c
)
+
e
t
c
.
,
q
a
b
=
−
a
b
c
−
a
2
b
2
c
(
c
2
−
a
2
−
b
2
)
−
a
3
b
3
c
(
c
2
−
a
2
−
b
2
+
a
b
)
(
c
2
−
a
2
−
b
2
−
a
b
)
−
e
t
c
.
q
b
b
=
b
+
a
b
2
c
2
−
a
2
+
a
2
b
3
(
c
2
−
a
2
+
b
c
)
(
c
2
−
a
2
−
b
c
)
+
e
t
c
.
{\displaystyle {\begin{array}{l}q_{aa}=a+{\frac {a^{2}b}{c^{2}-b^{2}}}+{\frac {a^{3}b^{2}}{\left(c^{2}-b^{2}+ac\right)\left(c^{2}-b^{2}-ac\right)}}+etc.,\\\\q_{ab}=-{\frac {ab}{c}}-{\frac {a^{2}b^{2}}{c\left(c^{2}-a^{2}-b^{2}\right)}}-{\frac {a^{3}b^{3}}{c\left(c^{2}-a^{2}-b^{2}+ab\right)\left(c^{2}-a^{2}-b^{2}-ab\right)}}-etc.\\\\q_{bb}=b+{\frac {ab^{2}}{c^{2}-a^{2}}}+{\frac {a^{2}b^{3}}{\left(c^{2}-a^{2}+bc\right)\left(c^{2}-a^{2}-bc\right)}}+etc.\end{array}}}
174.] We have then the following equations to determine the charges
E
a
{\displaystyle E_{a}}
E
b
{\displaystyle E_{b}}
V
a
{\displaystyle V_{a}}
V
b
{\displaystyle V_{b}}
E
a
=
V
a
q
a
a
+
V
b
q
a
b
,
{\displaystyle E_{a}=V_{a}q_{aa}+V_{b}q_{ab},}
E
b
=
V
a
q
a
b
+
V
b
q
b
b
.
{\displaystyle E_{b}=V_{a}q_{ab}+V_{b}q_{bb}.}
If we put
q
a
a
q
b
b
−
q
a
b
2
=
D
=
1
D
′
,
{\displaystyle q_{aa}q_{bb}-q_{ab}^{2}=D={\frac {1}{D'}},}
and
p
a
a
=
q
b
b
D
′
,
p
a
b
=
−
q
a
b
D
′
,
p
b
b
=
q
a
a
D
′
,
{\displaystyle p_{aa}=q_{bb}D',\ p_{ab}=-q_{ab}D',\ p_{bb}=q_{aa}D',}
whence
p
a
a
p
b
b
−
p
a
b
2
=
D
′
;
{\displaystyle p_{aa}p_{bb}-p_{ab}^{2}=D';}
then the equations to determine the potentials in terms of the charges are
V
a
=
p
a
a
E
a
+
p
a
b
E
b
,
{\displaystyle V_{a}=p_{aa}E_{a}+p_{ab}E_{b},}
V
b
=
p
a
b
E
a
+
p
b
b
E
b
,
{\displaystyle V_{b}=p_{ab}E_{a}+p_{bb}E_{b},}
and
p
a
a
,
p
a
b
{\displaystyle p_{aa},p_{ab}}
p
b
b
{\displaystyle p_{bb}}
The total energy of the system is, by Art. 85,
Q
=
1
2
(
E
a
V
a
+
E
b
V
b
)
,
=
1
2
(
V
a
2
q
a
a
+
2
V
a
V
b
q
a
b
+
V
b
2
q
b
b
)
,
=
1
2
(
E
a
2
p
a
a
+
2
E
a
E
b
p
a
b
+
E
b
2
p
b
b
)
.
{\displaystyle {\begin{array}{ll}Q&={\frac {1}{2}}\left(E_{a}V_{a}+E_{b}V_{b}\right),\\\\&={\frac {1}{2}}\left(V_{a}^{2}q_{aa}+2V_{a}V_{b}q_{ab}+V_{b}^{2}q_{bb}\right),\\\\&={\frac {1}{2}}\left(E_{a}^{2}p_{aa}+2E_{a}E_{b}p_{ab}+E_{b}^{2}p_{bb}\right).\end{array}}}
