Chapter 24
Capacitance and Dielectrics
1
Capacitors and Capacitance
A capacitor is a device that stores electric potential energy and electric charge.
The simplest construction of a capacitor is two parallel plates with a dielectric
(insulating) material between the two plates. When placed in a circuit, equal and
opposite charges are stored on the two plates providing a source of energy for the
circuit.
We will see that electric potential energy can be stored in a capacitor in the form
of an electric field. The idea that the electric field contains energy forms our
fundamental understanding of electromagnetic waves and the nature of light.
Any two conductors separated by an insulator (or a vacuum) forma a capacitor.
Figure 1: This figure shows the simplest form of a capacitor–two charged conductors separated by
air (or vacuum or an insulating dielectric).
The capacitance is defined as the ratio of charge to potential difference, and this is
a property of the capacitor due to its physical construction.
1
C =
Q
Vab
(1)
The SI units of capacitance is called one farad, in honor of the 19th-century English
physicist Michael Faraday.
1 F = 1 farad = 1 C/V = 1 coulomb/volt
1.1
Calculating Capacitance: Capacitors in Vacuum
Figure 2: This figure shows the arrangement of capacitor plates in a parallel-plate capacitor. The
electric field produced by the accumulation of charge on the plates is also shown.
We can write the E-field between two capacitor plates as:
E =
σ
Q
=
o
A o
2
(2)
From the previous chapter we saw that the voltage between two capacitor plates
is:
Vab = E d =
1 Qd
o A
(3)
Using our definition of capacitance (Eq. 1) we can write the capacitance as:
C =
Q
o A
=
V
d
(4)
Units of Capacitance
1 F = 1 C2 /N·m = 1 C2 /J
Also,
o = 8.85 × 10−12 F/m
1.2
A Cylindrical Capacitor
Figure 3: This figure shows a cylindrical capacitor with linear charge density +λ on the inner
conductor and −λ on the outer conductor with an “air gap” between the two conductors.
We saw from the previous chapter that voltage for an infinitely-long line charge
was:
r λ
o
V =
`n
2πo
r
3
In order to calculate the capacitance of the two conductors, we must determine the
voltage between the two conductors, Vab .
λ
rb
Vab =
`n
2πo
ra
Using the electrical definition of capacitance we find:
C =
Q
=
Vab
λL
2πo L
=
rb
λ
`n (rb /ra )
2πo `n ra
The Capacitance per unit Length
2πo
55.6 pF/m
C
=
=
L
`n (rb /ra )
`n (rb /ra )
(5)
The capacitance per unit length of a coaxial cable is determined entirely by its
dimensions. Most coaxial cables have an insulating material instead of vacuum
between the conductors. A typical cable used for connecting a television has a
capacitance per unit length of 69 pF/m.
2
Capacitors in Series and Parallel
Capacitors are produced with standard values of capacitance; however, they may
not be the values you want. You can obtain the desired values by combining capacitors where the simplest combinations are series connections and parallel connections.
4
2.1
Capacitors in Series
Figure 4: This figure shows two capacitors in series (i.e., they share the same charge Q on both
capacitors. Also shown is the process for finding the equivalent capacitor that holds the same charge
Q as the individual capacitors C1 and C2 .
The equivalent capacitor for capacitors in series is:
1
1
1
=
+
+ ···
Ceq
C1
C2
5
(6)
2.2
Capacitors in Parallel
Figure 5: This figure shows two capacitors in parallel (i.e, they share the same voltage difference).
Also shown is the process for finding the equivalent capacitance having charge Q = Q1 + Q2 .
The equivalent capacitor for capacitors in parallel is:
Ceq = C1 + C2 + · · ·
2.3
(7)
Capacitors in Series and Parallel
Figure 6: This figure shows the procedure for reducing a circuit containing capacitors in series and
parallel into an equivalent circuit with only one capacitor.
6
3
Energy Storage in Capacitors and Electric-Field Energy
Most of the important applications of capacitors depend on their ability to store
energy. In order to calculate the amount of energy stored in a capacitor we have
must calculate how much work is required to charge a capacitor for a given voltage. If a battery with voltage V is connected to a capacitor, the total work W is
calculated by summing up the differential works dW for every dq of charge placed
on the capacitor plate. This is illustrated by the following figure.
Figure 7: This figure shows how a continuous stream of dq’s are added to the capacitor until it
reaches its total charge Q = V C.
In the above picture, it takes an increasing amount of work to move the same
charge dq on to the capacitor plate because the voltage V across the capacitor is
continuously increasing proportional to the increasing amount charge q deposited
on it. The amount work required to move a differential charge dq on to a capacitor
plate already containing a charge q is:
dW = dq V
where V is the instantaneous voltage on the capacitor due to the instantaneous
charge already on the capacitor: V = q/C.
dW = v dq =
7
q dq
C
(8)
Z
W
W =
0
1
dW =
C
Z
Q
q dq =
0
Q2
2C
(work to charge a capacitor)
(9)
Because Q = V C, we can write the energy stored in a capacitor U using one of the
three following expressions:
Q2
1
1
U =
= C V 2 = QV
2C
2
2
3.1
(10)
Electric-Field Energy
As we charge a capacitor, electrons move from one plate to the other, thus creating
an electric field between the plates. The volume between the plates is Ad where d
is the separation distance between the plates and A is the area.
u = Energy density =
1
2C
V2
Ad
(11)
where V = Ed.
However, we know the capacitance of the parallel-plate capacitor is: C = o A/d,
and we can write the following for the energy density:
u =
4
1
o E 2
2
[ joules/m3 ]
(12)
Dielectrics
When constructing capacitors, it is important to keep the separation distance between the plates d very uniform. This is accomplished by introducing an insulating
material of uniform thickness, known as a dielectric. While introducing a fix to
the construction tends to detract from its performance, in the case of a capacitor,
the dielectric actually enhances its performance by increasing its capacitance. The
new capacitance C in terms of its vacuum capacitance Co can be written as:
8
C = K Co
where K is called the dielectric constant having a value K > 1. Some of the
benefits of having a dielectric can be readily seen from the improved performance
characteristics of the capacitor:
• The charge stored on the capacitor is: Q = V C = K (V Co ) = KQo
• The energy density between the plates is: u = K 12 o Co V 2 = Kuo
Figure 8: This tables shows the dielectric constants for various materials.
Figure 9: This tables shows the dielectric constant and dielectric strength of some insulating
materials.
9
A Capacitor with a Fixed Charge Q
Figure 10: This figure shows the voltage across a capacitor with a fixed charge Q. The voltage
across the capacitor is reduced when a dielectric is introduced between the plates.
V =
10
Vo
K
4.1
Induced Charge and Polarization
A Capacitor with a Fixed Charge Q
Once again we look at a capacitor with a fixed charge Q before-and-after a dielectric
~ gives rise to
is introduced. The polarization of a dielectric in an electric field E
thin layers of bound charges on the surfaces of the dielectric creating a total charge
density of σ − σi where σi is the induced surface charge density on the dielectric.
~ across the capacitor.
This has an effector of reducing the electric field E
E =
Eo
K
(when Q is constant)
(13)
~ o and E
~ before-and-after a dielectric
Figure 11: This figure shows the initial and final electric field E
is introduced into a capacitor with fixed total charge Q.
Eo =
σ
o
E =
11
σ − σi
o
Using these equations along with Eq. 13 above, we can calculate the induced surface
charge density σi .
1
σi = σ 1 −
K
(induced surface charge density)
(14)
The produce Ko is called the permittivity of the dielectric, and it’s denoted
by :
= K o
(definition of permittivity)
E =
4.2
σ
A capacitor with and without a dielectric
C = KCo = Ko
4.3
(15)
A
A
= d
d
(16)
Energy storage with and without a dielectric
The energy density stored in a capacitor containing a dielectric is:
u =
1
1
Ko E 2 = E 2
2
2
. . . , and the energy density is reduced by the factor K as well:
u =
12
uo
K
(17)