Page 131 - Engineering Electromagnetics, 8th Edition

CHAPTER 5 Conductors and Dielectrics 113

Selecting an instant of time t = 1s,we may calculate the total outward current at
r = 5m:

1 −1 2
I = J r S = 5 e (4π5 ) = 23.1A
At the same instant, but for a slightly larger radius, r = 6m,wehave
1 −1 2

I = J r S = 6 e 4π6 = 27.7A
Thus, the total current is larger at r = 6 than it is at r = 5.
To see why this happens, we need to look at the volume charge density and the
velocity. We use the continuity equation first:

∂ρ ν 1 −t 1 ∂ 2 1 −t 1 −t
− =∇ · J =∇ · e a r = r e = e
2
∂t r r ∂r r r 2
We next seek the volume charge density by integrating with respect to t. Because ρ ν
is given by a partial derivative with respect to time, the “constant” of integration may
be a function of r:
1 −t 1 −t

ρ ν =− e dt + K(r) = e + K(r)
r 2 r 2
If we assume that ρ ν → 0as t →∞, then K(r) = 0, and
1 −t 3
ρ ν = e C/m
r 2
We may now use J = ρ ν v to find the velocity,

1
e −t
J r r
ν r = = 1 = r m/s
ρ ν e −t
r 2
The velocity is greater at r = 6 than it is at r = 5, and we see that some (unspecified)
force is accelerating the charge density in an outward direction.
In summary, we have a current density that is inversely proportional to r,a charge
2
density that is inversely proportional to r , and a velocity and total current that are
−t
proportional to r. All quantities vary as e .
6 1.5
D5.2. Current density is given in cylindrical coordinates as J =−10 z a z
2
A/m in the region 0 ≤ ρ ≤ 20 µm; for ρ ≥ 20 µm, J = 0. (a) Find the total
current crossing the surface z = 0.1min the a z direction. (b)If the charge
velocity is 2 × 10 m/s at z = 0.1m,find ρ ν there. (c)If the volume charge
6
3
density at z = 0.15mis −2000 C/m , find the charge velocity there.
3
Ans. −39.7 µA; −15.8 mC/m ; 29.0 m/s