Physics Department, Yarmouk University,
Irbid Jordan
Phys. 201 Mathematical Physics 1
Doc. A1
Dr. Nidal M. Ershaidat
Potential of a Uniform Sphere of Charge
The use of Poisson's and Laplace's equations will be explored for a uniform
sphere of charge. In spherical polar coordinates, Poisson's equation takes
the form:
∇2 V =
1 ∂  2 ∂V  1 ∂ 
∂V 
1
∂ 2V
ρ
r
+
sin
θ
+
=−




2
2
2
2
r ∂r  ∂r  r ∂θ 
∂θ  r sin θ ∂φ
ε0
(1)
but since there is full spherical symmetry here, the derivatives with respect
to θ and φ must be zero, leaving the form:
1 d  2 dV 
ρ
r
=−
2
r dr  dr 
ε0
(2)
Voutside
Examining first the region outside the sphere, Laplace's equation applies.
1 d  2 dVo 
=0
r
r 2 dr  dr 
(3)
which gives (r≠0)
1 d  2 dVo 
dr
a
r
 = a ⇒ Vot (r ) = a 2 = − + b
2
r dr 
dr 
r
r
∫
(4)
Since the zero of potential is arbitrary, it is reasonable to choose the zero of
potential at infinity, the standard practice with localized charges. This gives
the value b=0.
Since the sphere of charge will look like a point charge at large distances, i.e.
Vo (r )
=
r very l arg e
1 Q
, we may conclude that:
4 π ε0 r
a=
Q
4 π ε0
(5)
so the solution to Laplace's law outside the sphere is simply
Vo =
Q
4 π ε0 r
(6)
Vinside
Now examining the potential inside the sphere, the potential must have a
term of order r2 to give a constant on the left side of the equation, so the
solution is of the form
Vi = c r 2 + d
(7)
Substituting into Poisson's equation gives:
1 d  2 dVi  1 d  2 d
ρ
 1 d
c r2 + d  = 2
2c r3 = −
 = 2 r
r
2
r dr  dr  r dr  dr
ε0
 r dr
(
6c = −
)
(
)
(8)
ρ
ρ
giving c = −
6 ε0
ε0
(9)
ρ 2
r +d
6 ε0
(10)
The solution is thus:
Vi = −
In order to find d, equations 10 and 6 should meet the boundary conditions
at the surface of the sphere, r=R. In other words by equating Vi(R) = Vo(R),
we have:
ρ 2
Q
Q
ρ R2
−
R +d =
giving d =
+
(11)
6ε 0
4 π ε0 R
4 π ε0 R 6 ε0
The full solution for the potential inside the sphere from Poisson's equation is
Vi (r ) =
ρ
Q
ρ
ρ R2
ρ
R2 − r 2 +
=
R2 − r 2 +
=
3 R2 − r 2
6ε 0
4 π ε 0 R 6ε 0
3ε 0 6ε 0
[
]
[
]
[
]
(12)
And this is a unique solution.
See also: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/laplace.html#c2