and thereby simplify the integral (9.55) to the form
∂V 2
∂x
−
∂V 1
∂y
=
∫z
z 0
∂F 3
∂z
dz +
∂f 2
∂x
−
∂f 1
∂y
=F 3 (x, y, z)−F 3 (x, y, z 0 ) + ∂f^2
∂x
−∂f^1
∂y
.
(9 .56)
This equation tells us that if f 1 , f 2 are selected to satisfy
∂f 2
∂x
−∂f^1
∂y
=F 3 (x, y, z 0 ),
then the last equation of (9.53) is satisfied. One choice of f 1 and f 2 which satisfies
the required condition is f 1 (x, y ) = 0 and
f 2 (x, y ) =
∫x
x 0
F 3 (x, y, z 0 )dx.
For the special conditions assumed, the constructed vector potential V has the com-
ponents
V 1 =
∫z
z 0
F 2 (x, y, z)dz
V 2 =−
∫z
z 0
F 1 (x, y, z )dz +
∫x
x 0
F 3 (x, y, z 0 )dx
V 3 = 0.
(9 .57)
Other vector potential functions may be constructed by utilizing different assump-
tions on the components of V and performing similar integrations to those illus-
trated. Alternatively one could add the gradient of any arbitrary scalar function to
the vector potential V and obtain other potential functions V∗=V +∇ψ.
Example 9-2.
By Newton’s inverse square law, the force of attraction between two masses m 1
and m 2 is given by
F =Gm^1 m^2
ρ^2
ˆeρ (9 .58)
where G= 6 .6730 (10)−^11 m^3 kg −^1 s−^2 is called the gravitational constant, ρis the dis-
tance between the center of mass of each body and ˆeρis a unit vector pointing along
the line connecting the center of mass of the two bodies. The force is an attractive
force and so the direction of ˆeρdepends upon which center of mass is selected to
sketch this force.
