Page 28 - Electromagnetics Handbook

This expression involves the time derivative of ρ with r fixed. We can also find an
expression in terms of the material derivative by using the transport equation (A.67).
Enforcing conservation of charge by setting that expression to zero, we have
Dρ(r, t)
+ ρ(r, t) ∇· v(r, t) = 0. (1.12)
Dt
Here Dρ/Dt is the time rate of change of the charge density experienced by an observer
moving with the current.
We can state the large-scale form of the continuity equation in terms of a stationary
volume. Integrating (1.11) over a stationary volume region V and using the divergence
theorem, we find that
∂ρ(r, t)

dV =− J(r, t) · dS.
∂t
V S
Since V is not changing with time we have
dQ(t) d
= ρ(r, t) dV =− J(r, t) · dS. (1.13)
dt dt V S
Hence any increase of total charge within V must be produced by current entering V
through S.

Use of the continuity equation. As an example, suppose that in a bounded region
of space we have
ρ(r, t) = ρ 0 re −βt .
We wish to find J and v, and to verify both versions of the continuity equation in point
form. The spherical symmetry of ρ requires that J = ˆ rJ r . Application of (1.13) over a
sphere of radius a gives
d a −βt 2 2
4π ρ 0 re r dr =−4π J r (a)a .
dt 0
Hence
r 2 −βt
J = ˆ rβρ 0 e
4
and therefore
1 ∂ 2 −βt
∇· J = 2 (r J r ) = βρ 0 re .
r ∂r
The velocity is
J r
v = = ˆ rβ ,
ρ 4
and we have ∇· v = 3β/4. To verify the continuity equations, we compute the time
derivatives
∂ρ −βt
=−βρ 0 re ,
∂t

Dρ ∂ρ
= + v ·∇ρ
Dt ∂t
r −βt

−βt
=−βρ 0 re + ˆ rβ · ˆ rρ 0 e
4
3 −βt
=− βρ 0 re .
4
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