Page 78 - Engineering Electromagnetics, 8th Edition

60 ENGINEERING ELECTROMAGNETICS

An identical result would be obtained for ρ< a. Thus the coaxial cable or
capacitor has no external field (we have proved that the outer conductor is a “shield”),
and there is no field within the center conductor.
Our result is also useful for a finite length of coaxial cable, open at both ends, pro-
vided the length L is many times greater than the radius b so that the nonsymmetrical
conditions at the two ends do not appreciably affect the solution. Such a device is
also termed a coaxial capacitor. Both the coaxial cable and the coaxial capacitor will
appear frequently in the work that follows.

EXAMPLE 3.2
Let us select a 50-cm length of coaxial cable having an inner radius of 1 mm and an
outer radius of 4 mm. The space between conductors is assumed to be filled with air.
The total charge on the inner conductor is 30 nC. We wish to know the charge density
on each conductor, and the E and D fields.
Solution. We begin by finding the surface charge density on the inner cylinder,

Q inner cyl 30 × 10 −9 2
ρ S,inner cyl = = = 9.55 µC/m
−3
2πaL 2π(10 )(0.5)
The negative charge density on the inner surface of the outer cylinder is

Q outer cyl −30 × 10 −9 2
ρ S,outer cyl = = =−2.39 µC/m
2πbL 2π(4 × 10 )(0.5)
−3
The internal fields may therefore be calculated easily:

−3
−6
aρ S 10 (9.55 × 10 ) 9.55 2
D ρ = = = nC/m
ρ ρ ρ
and
9.55 × 10 −9 1079
D ρ
E ρ = = −12 = V/m
0 8.854 × 10 ρ ρ
Both of these expressions apply to the region where 1 <ρ < 4 mm. For ρ< 1mm
or ρ> 4 mm, E and D are zero.

D3.5. A point charge of 0.25 µCis located at r = 0, and uniform surface
2
charge densities are located as follows: 2 mC/m at r = 1 cm, and −0.6 mC/m 2
at r = 1.8 cm. Calculate D at: (a) r = 0.5 cm; (b) r = 1.5 cm; (c) r = 2.5 cm.
(d) What uniform surface charge density should be established at r = 3cmto
cause D = 0at r = 3.5 cm?

2
2
2
Ans. 796a r µC/m ; 977a r µC/m ;40.8a r µC/m ; −28.3 µC/m 2