for
H
{\displaystyle \mathrm {H} }
d
F
d
x
+
d
G
d
y
+
d
H
d
z
=
0.
{\displaystyle {\dfrac {d\mathrm {F} }{dx}}+{\dfrac {d\mathrm {G} }{dy}}+{\dfrac {d\mathrm {H} }{dz}}=0.}
Inside the sphere the differential equations for
F
1
{\displaystyle \mathrm {F} _{1}}
G
1
{\displaystyle \mathrm {G} _{1}}
−
σ
4
π
μ
′
∇
2
F
1
=
−
i
p
F
;
{\displaystyle -{\dfrac {\sigma }{4\pi \mu '}}\nabla ^{2}\mathrm {F} _{1}=-ip\mathrm {F} ;}
if
λ
1
2
=
−
4
π
μ
′
i
p
/
σ
{\displaystyle \lambda _{1}^{2}=-4\pi \mu 'ip/\sigma }
F
1
=
D
ϵ
i
p
t
S
2
(
λ
1
r
)
r
3
d
2
d
x
d
z
1
r
,
{\displaystyle \mathrm {F} _{1}=\mathrm {D} \epsilon ^{ipt}\mathrm {S} _{2}\left(\lambda _{1}r\right)r^{3}{\dfrac {d^{2}}{dx\ dz}}{\dfrac {1}{r}},}
where
S
2
(
λ
1
r
)
=
3
sin
λ
1
r
(
λ
1
r
)
3
−
3
cos
λ
1
r
(
λ
1
r
)
2
−
sin
λ
1
r
λ
1
r
.
{\displaystyle \mathrm {S} _{2}\left(\lambda _{1}r\right)={\dfrac {3\sin \lambda _{1}r}{\left(\lambda _{1}r\right)^{3}}}-{\dfrac {3\cos \lambda _{1}r}{\left(\lambda _{1}r\right)^{2}}}-{\dfrac {\sin \lambda _{1}r}{\lambda _{1}r}}.}
Similarly
G
1
=
D
ϵ
i
p
t
S
2
(
λ
r
)
r
3
d
2
d
x
d
z
1
r
;
{\displaystyle \mathrm {G} _{1}=\mathrm {D} \epsilon ^{ipt}\mathrm {S} _{2}\left(\lambda r\right)r^{3}{\dfrac {d^{2}}{dx\ dz}}{\dfrac {1}{r}};}
the differential equation for
H
1
{\displaystyle \mathrm {H} _{1}}
−
σ
4
μ
′
∇
2
H
1
=
−
i
p
H
1
+
B
a
3
.
{\displaystyle -{\dfrac {\sigma }{4\mu '}}\nabla ^{2}\mathrm {H} _{1}=-ip\mathrm {H} _{1}+{\dfrac {\mathrm {B} }{a^{3}}}.}
So that
H
1
=
D
ϵ
i
p
t
S
2
(
λ
1
r
)
r
3
d
2
d
z
2
1
r
+
2
D
ϵ
i
p
t
S
0
(
λ
1
r
)
+
B
i
p
a
3
ϵ
i
p
t
,
{\displaystyle \mathrm {H} _{1}=\mathrm {D} \epsilon ^{ipt}\mathrm {S} _{2}\left(\lambda _{1}r\right)r^{3}{\dfrac {d^{2}}{dz^{2}}}{\dfrac {1}{r}}+2\mathrm {D} \epsilon ^{ipt}\mathrm {S} _{0}\left(\lambda _{1}r\right)+{\dfrac {\mathrm {B} }{ipa^{3}}}\epsilon ^{ipt},}
where
S
0
(
λ
1
r
)
=
sin
λ
1
r
λ
1
r
,
{\displaystyle \mathrm {S} _{0}\left(\lambda _{1}r\right)={\dfrac {\sin \lambda _{1}r}{\lambda _{1}r}},}
and is introduced to make
d
F
d
x
+
d
G
d
y
+
d
H
d
z
=
0.
{\displaystyle {\dfrac {d\mathrm {F} }{dx}}+{\dfrac {d\mathrm {G} }{dy}}+{\dfrac {d\mathrm {H} }{dz}}=0.}
Since
F
1
,
G
1
,
H
1
{\displaystyle \mathrm {F} _{1},\mathrm {G} _{1},\mathrm {H} _{1}}
r
=
a
{\displaystyle r=a}
a
{\displaystyle a}
C
E
2
(
i
λ
a
)
−
D
S
2
(
λ
1
a
)
+
i
B
p
a
3
=
ω
e
p
2
a
3
k
,
−
2
C
E
0
(
i
λ
a
)
−
2
D
S
2
(
λ
1
a
)
+
i
B
p
a
3
=
0.
}
{\displaystyle \left.{\begin{array}{c}\mathrm {C} \mathrm {E} _{2}(i\lambda a)-\mathrm {DS} _{2}\left(\lambda _{1}a\right)+{\dfrac {i\mathrm {B} }{pa^{3}}}={\dfrac {\omega e}{p^{2}a^{3}k}},\\\\-2\mathrm {C} \mathrm {E} _{0}(i\lambda a)-2\mathrm {DS} _{2}\left(\lambda _{1}a\right)+{\dfrac {i\mathrm {B} }{pa^{3}}}=0.\end{array}}\right\}}
(5)
Since, on the assumption discussed above, the electrification on the surface of the moving sphere is equivalent to a tangential current-sheet whose intensity is
ω
ϵ
i
p
t
sin
θ
e
4
π
a
2
{\displaystyle \omega \epsilon ^{ipt}\sin \theta {\dfrac {e}{4\pi a^{2}}}}
