# cilindro electromagnetico

```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
&micro; &para;
b
Q
V=
ln
.
2πε l
a
E = −r̂
Hence,
E = −r̂
D = εE
72.1
50 sin(120π t)
V
&iexcl; b &cent; = −r̂
= −r̂
sin(120π t) (V/m),
r ln 2
r
r ln a
= εr ε0 E
= −r̂ 9 &times; 8.85 &times; 10−12 &times;
72.1
sin(120π t)
r
5.75 &times; 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 =
&middot; S = −r̂
∂t
∂t
&micro;
&para;
5.75 &times; 10−9
sin(120π t) &middot; (−r̂ 2π rl)
r
= 5.75 &times; 10−9 &times; 120π &times; 2π l cos(120π t)
= 0.82 cos(120π t) (&micro; 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
&iexcl; &cent;.
ln ba
2πε l
dV
= &iexcl; b &cent; [120π &times; 50 cos(120π t)] = 0.82 cos(120π t) (&micro; A),
dt
ln a
which is the same answer we obtained before.
```