the surface of the ellipsoid, shall vanish at infinity, and shall satisfy (7). We see at once that if
f
(
x
,
y
,
z
)
{\displaystyle f(x,y,z)}
∇
2
f
=
0
{\displaystyle \nabla ^{2}f=0}
f
(
x
/
α
,
y
,
z
)
{\displaystyle f(x/{\sqrt {\alpha }},y,z)}
Φ
=
∫
λ
∞
d
λ
(
a
′
2
+
λ
)
(
b
2
+
λ
)
(
c
2
+
λ
)
{\displaystyle \Phi {=}\int _{\lambda }^{\infty }{\frac {d\lambda }{\sqrt {\left(a^{'2}+\lambda \right)\left(b^{2}+\lambda \right)\left(c^{2}+\lambda \right)}}}}
where
λ
{\displaystyle \lambda }
x
,
y
,
z
{\displaystyle x,y,z}
x
2
a
′
2
+
λ
+
y
2
b
2
+
λ
+
z
2
c
2
+
λ
=
1
{\displaystyle {\frac {x^{2}}{a^{'2}+\lambda }}+{\frac {y^{2}}{b^{2}+\lambda }}+{\frac {z^{2}}{c^{2}+\lambda }}{=}1}
satisfies
∇
2
Φ
=
0
{\displaystyle \nabla ^{2}\Phi =0}
Hence
Ψ
=
∫
λ
∞
A
d
λ
(
a
′
2
+
λ
)
(
b
2
+
λ
)
(
c
2
+
λ
)
{\displaystyle \mathbf {\Psi } =\int _{\lambda }^{\infty }{\frac {Ad\lambda }{\sqrt {\left(a^{'2}+\lambda \right)\left(b^{2}+\lambda \right)\left(c^{2}+\lambda \right)}}}}
(8)
where
λ
{\displaystyle \lambda }
x
,
y
,
z
{\displaystyle x,y,z}
x
2
(
a
′
2
+
λ
)
+
y
2
b
2
+
λ
+
z
2
c
2
+
λ
=
1
{\displaystyle {\frac {x^{2}}{\left(a^{'2}+\lambda \right)}}+{\frac {y^{2}}{b^{2}+\lambda }}+{\frac {z^{2}}{c^{2}+\lambda }}=1}
(9)
satisfies (7).
Writing
a
2
{\displaystyle a^{2}}
α
a
′
2
{\displaystyle \alpha a'^{2}}
Ψ
=
∫
λ
∞
A
d
λ
(
a
2
+
α
λ
)
(
b
2
+
λ
)
(
c
2
+
λ
)
{\displaystyle \mathbf {\Psi } =\int _{\lambda }^{\infty }{\frac {Ad\lambda }{\sqrt {\left(a^{2}+\alpha \lambda \right)\left(b^{2}+\lambda \right)\left(c^{2}+\lambda \right)}}}}
(10)
x
2
a
2
+
α
λ
+
y
2
b
2
+
λ
+
z
2
c
2
+
λ
=
1
{\displaystyle {\frac {x^{2}}{a^{2}+\alpha \lambda }}+{\frac {y^{2}}{b^{2}+\lambda }}+{\frac {z^{2}}{c^{2}+\lambda }}=1}
(11)
This value of
Ψ
{\displaystyle \mathbf {\Psi } }
a
,
b
,
c
{\displaystyle a,b,c}
λ
=
0
{\displaystyle \lambda =0}
Ψ
{\displaystyle \mathbf {\Psi } }
σ
{\displaystyle \sigma }
q
/
4
π
b
c
{\displaystyle q/4\pi bc}
a
{\displaystyle a}
Now
σ
=
K
4
π
E
n
=
K
4
π
E
1
{\displaystyle \sigma {=}{\frac {\mathrm {K} }{4\pi }}\mathbf {E} _{n}{=}{\frac {\mathrm {K} }{4\pi }}\mathbf {E} _{1}}
at the end of the axis.
But by (5)
E
1
=
−
d
Ψ
d
x
{\displaystyle \mathbf {E} _{1}{=}-{\frac {d\mathbf {\Psi } }{dx}}}
Again, at
x
=
a
,
y
=
z
=
0
{\displaystyle x=a,y=z=0}
d
λ
/
d
x
=
2
a
/
α
{\displaystyle d\lambda /dx=2a/\alpha }
d
Ψ
d
x
=
d
Ψ
d
λ
d
λ
d
x
=
−
A
a
b
c
⋅
2
a
α
{\displaystyle {\frac {d\mathbf {\Psi } }{dx}}{=}{\frac {d\mathbf {\Psi } }{d\lambda }}{\frac {d\lambda }{dx}}{=}-{\frac {A}{abc}}\cdot {\frac {2a}{\alpha }}}
Hence
A
=
q
α
2
K
{\displaystyle A{=}{\frac {q\alpha }{2\mathrm {K} }}}
