Begin2.DVI

In a similar fashion one can verify that

∂f

∂y=

−3(y−y′)

|~r−~r′|^5

and

∂f

∂z=

−3(z−z′)

|~r−~r′|^5

Using the above results verify that

∇

(

1

|~r−~r′|^3

)

= −^3

|~r−~r′|^5

(~r−~r′) (9.157)

and show the right-hand side of equation (9.156) is zero because(~r−~r′)×(~r−~r′) =~ 0.

It then follows that

∇·B~= 0

which is the third equation in the Maxwell’s equations (9.134).

Example 9-13. If the charge densityρ(coul/m^3 )moves with velocity~v(m/s),

then the current density is given byJ~=ρ~v(amp/m^2 ). Surround the current density

field with a simple closed surfaceSwhich encloses a volumeV. The flux across the

surfaceSis then given by

∫ ∫

S

J~·dS~=

∫ ∫

S

J~·eˆndS=

∫ ∫

V

∇·J dV~

where the divergence theorem of Gauss has been employed to express the flux surface

integral with a volume integral. The charge must be conserved so that the flux out

of the volume must be accounted for by the time rate of change of the charge density

within the volume so that one can write

∫ ∫

S

J~·dS~=

∫ ∫ ∫

V

∇·J dV~ =−

d

dt

∫ ∫ ∫

V

ρ dV=−

∫ ∫ ∫

V

∂ρ

∂tdV

or ∫ ∫ ∫

V

[

∇·J~+∂ρ

∂t

]

dV= 0 (9.158)

The equation (9.158) must hold for all volumesV and consequently one must require

∇·J~+∂ρ

∂t

= 0 (9.159)

The equation (9.159) is know asthe continuity equationfor the charge density.