Problem 6.15 A coaxial capacitor of length l = 6 cm uses an insulating dielectric
material with εr = 9. The radii of the cylindrical conductors are 0.5 cm and 1 cm. If
the voltage applied across the capacitor is
V (t) = 50 sin(120π t) (V)
what is the displacement current?
Solution:
l
Id
r
+
V(t)
2a
2b
-
Figure P6.15:
To find the displacement current, we need to know E in the dielectric space between
the cylindrical conductors. From Eqs. (4.114) and (4.115),
Q
,
2πε rl
µ ¶
b
Q
V=
ln
.
2πε l
a
E = −r̂
Hence,
E = −r̂
D = εE
72.1
50 sin(120π t)
V
¡ b ¢ = −r̂
= −r̂
sin(120π t) (V/m),
r ln 2
r
r ln a
= εr ε0 E
= −r̂ 9 × 8.85 × 10−12 ×
72.1
sin(120π t)
r
5.75 × 10−9
sin(120π t) (C/m2 ).
r
The displacement current flows between the conductors through an imaginary
cylindrical surface of length l and radius r. The current flowing from the outer
conductor to the inner conductor along −r̂ crosses surface S where
= −r̂
S = −r̂ 2π rl.
Hence,
∂D
∂
Id =
· S = −r̂
∂t
∂t
µ
¶
5.75 × 10−9
sin(120π t) · (−r̂ 2π rl)
r
= 5.75 × 10−9 × 120π × 2π l cos(120π t)
= 0.82 cos(120π t) (µ A).
Alternatively, since the coaxial capacitor is lossless, its displacement current has to
be equal to the conduction current flowing through the wires connected to the voltage
sources. The capacitance of a coaxial capacitor is given by (4.116) as
C=
The current is
I =C
2πε l
¡ ¢.
ln ba
2πε l
dV
= ¡ b ¢ [120π × 50 cos(120π t)] = 0.82 cos(120π t) (µ A),
dt
ln a
which is the same answer we obtained before.