Discharge of a cylindrical capacitor
1
z
R
−Q0
b
a
+Q0
A cylindrical capacitor has internal radius a, external radius b > a, and
length ℓ ≫ b. For t < 0, the two cylindrical plates have total charges ±Q0 ,
respectively, and are electrically isolated and disconnected. At t = 0 the
plates are connected via a resistance R as in the figure. We assume that
during the discharge 1) the slowly varying current approximation holds, 2)
the surface charge density on the plates is uniform, 3) the effects of the
boundaries and of the external circuit and the resistance of the plates are
assumed to be negligible.
a) Calculate the current I = I(z, t) flowing along the plates and the magnetic
field B = Bφ (r, z, t)φ inside the capacitor.
b) Calculate Poynting’s vector S and verify that its flux through a cylindrical surface coaxial with the capacitor equals the temporal variation of the
electrostatic energy inside the surface.
c) Discuss the validity of the slowly varying current approximation and of
the assumption of uniform charge density.
O
ℓ
r
Solution
a) For symmetry reasons (in the assumption of a very long capacitor) the
electric field inside the capacitor is radial and easily obtained via Gauss’s theorem as
Er = Er (r) =
1 Q0
.
2πε0 ℓ r
(1)
By integrating Er over a < r < b to find the potential drop V , we obtain the capacity:
C=
Q0
2πε0 ℓ
=
.
V
ln(b/a)
(2)
The initial electrostatic energy is Ues = Q20 /2C.
For t > 0, the system behaves as a RC circuit, and the charge decays as Q(t) = Q0 e−t/τ with
τ = RC. Within our assumptions of uniform charge density at any time, the charge contained
between the bottom of the capacitor (z = 0) and the position z < ℓ is ∆Q(z, t) = Q(t)z/ℓ. Thus,
the current flowing through the surface at z is
I(z, t) = −
d(∆Q)
Q(t) z
Q0 z −t/τ
=
=
e
.
dt
τ ℓ
τ ℓ
(3)
To obtain this result we may also integrate directly the continuity equation ∂z I = −∂t Q/ℓ.
In order to find the magnetic field B, we use the equation ∇ × B = µ0 (J + ε0 ∂t E) in cylindrical
coordinates, which gives for the two non-vanishing components
−∂z Bϕ =
1
∂t Er ,
c2
1
∂r (rBϕ ) = µ0 Jz .
r
(4)
Both these equations can be used to find Bϕ . The second equation is equivalent to calculate the line
integral of B along a circle of radius r (with a < r < b) and perpendicular to the axis; only the
current density J contributes to the flux through the circle, since the displacement current ∂t E/c2 is
parallel to the surface of the circle. Thus
I
B(r, z, t) · dl = µ0 I(z, t) ,
(5)
from which we obtain
Bϕ (r, z, t) =
µ0 Q0 z −t/τ
µ0 I(z, t)
=
e
≡ B0ϕ (r, z) e−t/τ .
2πr
2πr τ ℓ
(6)
The same result is also obtained by integrating along z the first equation.
d) Poynting’s vector is given by
S=
1
z Q2
E × B = 2 20 2 e−2t/τ ẑ .
µ0
4π ε0 ℓ τ r
2
(7)
The flux of S through a section of the capacitor at height z is
Z b
z Q20
1
z ln(b/a) Q20 −2t/τ
−2t/τ
Φs (z, t) = 2 2 e
2πr
dr
=
e
.
2
4π ε0 ℓ τ
2πε0 ℓ2 τ
a r
(8)
The electrostatic energy associated to the volume between the bottom of the capacitor (z = 0) and
the height z is given by
z
Q2 e−2t/τ z
ln(b/a) 2 z −2t/τ
Ues (z, t) = Ues (t) = 0
=
Q e
ℓ
2C
ℓ
4πε0 ℓ 0 ℓ
(9)
because the electric field does not depend in z. Thus
2
dUes (z, t)
= − Ues (z, t) = −Φs (z, t) .
dt
τ
(10)
e) The assumptions of slowly varying currents and of uniform charge density are closely related. In
fact, the capacitor can be viewed as a portion of a coaxial cable along which charge and current
signal can propagate at velocity c (in TEM mode). Thus the density may be assumed as uniform if
the propagation of the signals is “instantaneous” with respect to the duration of the discharge, i.e. if
the propagation time ℓ/c ≪ τ . This is equivalent to state that the wavelength values corresponding
to the frequency spectrum of the signal are much larger than ℓ, thus the field can be considered as
uniform along z.
We obtain the same conclusion by checking that the electric field generated by magnetic induction
must be small with respect to the electrostatic field: from the equation ∇ × E1 ≃ −∂t B, we obtain
the inductive field
µ0 Q0 z 2
1 z 2
E1r ≃ −
=−
E0r ,
(11)
4πr ℓ τ
2 cτ
where E0r = (Q0 /2πε0 ℓr)e−t/τ . Thus E1r ≪ E0r if ℓ ≪ cτ .
3