Lecture Note
Capacitance and Dielectrics
-
University:
Massachusetts Institute of Technology -
Course:
8.02 | Physics II: Electricity and Magnetism Academic year:
2019
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Views:
249
Pages:
44
Author:
Sporkz
Capacitance and Dielectrics
5.1 Introduction
A capacitor is a device which stores electric charge. Capacitors vary in shape and size,
but the basic configuration is two conductors carrying equal but opposite charges (Figure
5.1.1). Capacitors have many important applications in electronics. Some examples
include storing electric potential energy, delaying voltage changes when coupled with
resistors, filtering out unwanted frequency signals, forming resonant circuits and making
frequency-dependent and independent voltage dividers when combined with resistors.
Some of these applications will be discussed in latter chapters.
Figure 5.1.1 Basic configuration of a capacitor.
In the uncharged state, the charge on either one of the conductors in the capacitor is zero.
During the charging process, a charge Q is moved from one conductor to the other one,
giving one conductor a charge +Q , and the other one a charge −Q . A potential
difference ∆V is created, with the positively charged conductor at a higher potential than
the negatively charged conductor. Note that whether charged or uncharged, the net charge
on the capacitor as a whole is zero.
The simplest example of a capacitor consists of two conducting plates of area A , which
are parallel to each other, and separated by a distance d, as shown in Figure 5.1.2.
Figure 5.1.2 A parallel-plate capacitor
Experiments show that the amount of charge Q stored in a capacitor is linearly
proportional to ∆V , the electric potential difference between the plates. Thus, we may
write
Q = C | ∆V |
(5.1.1) where C is a positive proportionality constant called capacitance. Physically,
capacitance is a measure of the capacity of storing electric charge for a given potential
difference ∆V . The SI unit of capacitance is the farad (F) :
1 F = 1 farad = 1 coulomb volt = 1 C V
A typical capacitance is in the picofarad ( 1 pF = 10− 12 F ) to millifarad range,
( 1 mF = 10− 3 F=1000 µ F; 1 µ F = 10− 6 F ).
Figure 5.1.3(a) shows the symbol which is used to represent capacitors in circuits. For a
polarized fixed capacitor which has a definite polarity, Figure 5.1.3(b) is sometimes used.
(a)
(b)
Figure 5.1.3 Capacitor symbols.
5.2 Calculation of Capacitance
Let’s see how capacitance can be computed in systems with simple geometry.
Example 5.1: Parallel-Plate Capacitor
Consider two metallic plates of equal area A separated by a distance d, as shown in
Figure 5.2.1 below. The top plate carries a charge +Q while the bottom plate carries a
charge –Q. The charging of the plates can be accomplished by means of a battery which
produces a potential difference. Find the capacitance of the system.
Figure 5.2.1 The electric field between the plates of a parallel-plate capacitor
Solution:
To find the capacitance C, we first need to know the electric field between the plates. A
real capacitor is finite in size. Thus, the electric field lines at the edge of the plates are not
straight lines, and the field is not contained entirely between the plates. This is known as edge effects, and the non-uniform fields near the edge are called the fringing fields. In
Figure 5.2.1 the field lines are drawn by taking into consideration edge effects. However,
in what follows, we shall ignore such effects and assume an idealized situation, where
field lines between the plates are straight lines.
In the limit where the plates are infinitely large, the system has planar symmetry and we
can calculate the electric field everywhere using Gauss’s law given in Eq. (4.2.5):
JG JG q
E
w
∫∫ ⋅ d A = enc
ε0
S
By choosing a Gaussian “pillbox” with cap area A′ to enclose the charge on the positive
plate (see Figure 5.2.2), the electric field in the region between the plates is
EA' =
qenc
ε0
=
σ A'
ε0
⇒ E=
σ
ε0
(5.2.1)
The same result has also been obtained in Section 4.8.1 using superposition principle.
Figure 5.2.2 Gaussian surface for calculating the electric field between the plates.
The potential difference between the plates is
− G
G
∆V = V− − V+ = − ∫ E ⋅ d s = − Ed
+
(5.2.2)
where we have taken the path of integration to be a straight line from the positive plate to
the negative plate following the field lines (Figure 5.2.2). Since the electric field lines are
always directed from higher potential to lower potential, V− < V+ . However, in
computing the capacitance C, the relevant quantity is the magnitude of the potential
difference:
| ∆V |= Ed
and its sign is immaterial. From the definition of capacitance, we have
(5.2.3) C=
ε A
Q
= 0
| ∆V |
d
( parallel plate)
(5.2.4)
Note that C depends only on the geometric factors A and d. The capacitance C increases
linearly with the area A since for a given potential difference ∆V , a bigger plate can hold
more charge. On the other hand, C is inversely proportional to d, the distance of
separation because the smaller the value of d, the smaller the potential difference | ∆V |
for a fixed Q.
Interactive Simulation 5.1: Parallel-Plate Capacitor
This simulation shown in Figure 5.2.3 illustrates the interaction of charged particles
inside the two plates of a capacitor.
Figure 5.2.3 Charged particles interacting inside the two plates of a capacitor.
Each plate contains twelve charges interacting via Coulomb force, where one plate
contains positive charges and the other contains negative charges. Because of their
mutual repulsion, the particles in each plate are compelled to maximize the distance
between one another, and thus spread themselves evenly around the outer edge of their
enclosure. However, the particles in one plate are attracted to the particles in the other, so
they attempt to minimize the distance between themselves and their oppositely charged
correspondents. Thus, they distribute themselves along the surface of their bounding box
closest to the other plate.
Example 5.2: Cylindrical Capacitor
Consider next a solid cylindrical conductor of radius a surrounded by a coaxial
cylindrical shell of inner radius b, as shown in Figure 5.2.4. The length of both cylinders
is L and we take this length to be much larger than b− a, the separation of the cylinders,
so that edge effects can be neglected. The capacitor is charged so that the inner cylinder
has charge +Q while the outer shell has a charge –Q. What is the capacitance? (a)
(b)
Figure 5.2.4 (a) A cylindrical capacitor. (b) End view of the capacitor. The electric field
is non-vanishing only in the region a < r < b.
Solution:
To calculate the capacitance, we first compute the electric field everywhere. Due to the
cylindrical symmetry of the system, we choose our Gaussian surface to be a coaxial
cylinder with length A < L and radius r where a < r < b . Using Gauss’s law, we have
JG JG
λA
E
w
∫∫ ⋅ d A = EA = E ( 2π rA ) =
⇒
ε0
S
E=
λ
2πε 0 r
(5.2.5)
where λ = Q / L is the charge per unit length. Notice that the electric field is nonvanishing only in the region a < r < b . For r < a , the enclosed charge is qenc = 0 since
any net charge in a conductor must reside on its surface. Similarly, for r > b , the enclosed
charge is qenc = λ A − λ A = 0 since the Gaussian surface encloses equal but opposite
charges from both conductors.
The potential difference is given by
b
∆ V = Vb − Va = − ∫ Er dr = −
a
λ
2πε 0
b
∫a
dr
λ
⎛b⎞
ln ⎜ ⎟
=−
2πε 0 ⎝ a ⎠
r
(5.2.6)
where we have chosen the integration path to be along the direction of the electric field
lines. As expected, the outer conductor with negative charge has a lower potential. This
gives
C=
2πε 0 L
Q
λL
=
=
| ∆V | λ ln(b / a ) / 2πε 0 ln(b / a )
(5.2.7)
Once again, we see that the capacitance C depends only on the geometrical factors, L, a
and b. Example 5.3: Spherical Capacitor
As a third example, let’s consider a spherical capacitor which consists of two concentric
spherical shells of radii a and b, as shown in Figure 5.2.5. The inner shell has a charge
+Q uniformly distributed over its surface, and the outer shell an equal but opposite
charge –Q. What is the capacitance of this configuration?
Figure 5.2.5 (a) spherical capacitor with two concentric spherical shells of radii a and b.
(b) Gaussian surface for calculating the electric field.
Solution:
The electric field is non-vanishing only in the region a < r < b . Using Gauss’s law, we
obtain
JG JG
Q
2
E
w
∫∫ ⋅ d A = Er A = Er ( 4π r ) =
ε0
S
(5.2.8)
or
Er =
1
Q
4πε o r 2
(5.2.9)
Therefore, the potential difference between the two conducting shells is:
b
Q
a
4πε 0
∆ V = Vb − Va = − ∫ Er dr = −
∫
dr
Q ⎛1 1⎞
Q ⎛ b−a ⎞
=−
⎜ − ⎟=−
⎜
⎟
2
r
4πε 0 ⎝ a b ⎠
4πε 0 ⎝ ab ⎠
(5.2.10)
Q
⎛ ab ⎞
= 4πε 0 ⎜
⎟
| ∆V |
⎝ b−a ⎠
(5.2.11)
b
a
which yields
C=
Again, the capacitance C depends only on the physical dimensions, a and b.
An “isolated” conductor (with the second conductor placed at infinity) also has a
capacitance. In the limit where b → ∞ , the above equation becomes a
⎛ ab ⎞
lim C = lim 4πε 0 ⎜
4πε 0
⎟ = lim
b →∞
b →∞
b
→∞
⎛ a
⎝ b−a ⎠
⎜ 1−
⎝ b
⎞
⎟
⎠
= 4πε 0 a
(5.2.12)
Thus, for a single isolated spherical conductor of radius R, the capacitance is
C = 4πε 0 R
(5.2.13)
The above expression can also be obtained by noting that a conducting sphere of radius R
with a charge Q uniformly distributed over its surface has V = Q / 4πε 0 R , using infinity
as the reference point having zero potential, V (∞ ) = 0 . This gives
C=
Q
Q
=
= 4πε 0 R
| ∆ V | Q / 4πε 0 R
(5.2.14)
As expected, the capacitance of an isolated charged sphere only depends on its geometry,
namely, the radius R.
5.3 Capacitors in Electric Circuits
A capacitor can be charged by connecting the plates to the terminals of a battery, which
are maintained at a potential difference ∆V called the terminal voltage.
Figure 5.3.1 Charging a capacitor.
The connection results in sharing the charges between the terminals and the plates. For
example, the plate that is connected to the (positive) negative terminal will acquire some
(positive) negative charge. The sharing causes a momentary reduction of charges on the
terminals, and a decrease in the terminal voltage. Chemical reactions are then triggered to
transfer more charge from one terminal to the other to compensate for the loss of charge
to the capacitor plates, and maintain the terminal voltage at its initial level. The battery
could thus be thought of as a charge pump that brings a charge Q from one plate to the
other. 5.3.1 Parallel Connection
Suppose we have two capacitors C1 with charge Q1 and C2 with charge Q2 that are
connected in parallel, as shown in Figure 5.3.2.
Figure 5.3.2 Capacitors in parallel and an equivalent capacitor.
The left plates of both capacitors C1 and C2 are connected to the positive terminal of the
battery and have the same electric potential as the positive terminal. Similarly, both right
plates are negatively charged and have the same potential as the negative terminal. Thus,
the potential difference | ∆V | is the same across each capacitor. This gives
C1 =
Q1
,
| ∆V |
C2 =
Q2
| ∆V |
(5.3.1)
These two capacitors can be replaced by a single equivalent capacitor Ceq with a total
charge Q supplied by the battery. However, since Q is shared by the two capacitors, we
must have
Q = Q1 + Q2 = C1 | ∆V | +C2 | ∆V |= ( C1 + C2 ) | ∆V |
(5.3.2)
The equivalent capacitance is then seen to be given by
Ceq =
Q
= C1 + C2
| ∆V |
(5.3.3)
Thus, capacitors that are connected in parallel add. The generalization to any number of
capacitors is
N
Ceq = C1 + C2 + C3 + " + C N = ∑ Ci
i =1
(parallel)
(5.3.4) 5.3.2 Series Connection
Suppose two initially uncharged capacitors C1 and C2 are connected in series, as shown
in Figure 5.3.3. A potential difference | ∆V | is then applied across both capacitors. The
left plate of capacitor 1 is connected to the positive terminal of the battery and becomes
positively charged with a charge +Q, while the right plate of capacitor 2 is connected to
the negative terminal and becomes negatively charged with charge –Q as electrons flow
in. What about the inner plates? They were initially uncharged; now the outside plates
each attract an equal and opposite charge. So the right plate of capacitor 1 will acquire a
charge –Q and the left plate of capacitor +Q.
Figure 5.3.3 Capacitors in series and an equivalent capacitor
The potential differences across capacitors C1 and C2 are
| ∆ V1 | =
Q
Q
, | ∆ V2 | =
C1
C2
(5.3.5)
respectively. From Figure 5.3.3, we see that the total potential difference is simply the
sum of the two individual potential differences:
| ∆ V | = | ∆ V1 | + | ∆ V2 |
(5.3.6)
In fact, the total potential difference across any number of capacitors in series connection
is equal to the sum of potential differences across the individual capacitors. These two
capacitors can be replaced by a single equivalent capacitor Ceq = Q / | ∆ V | . Using the fact
that the potentials add in series,
Q
Q Q
= +
Ceq C1 C2
and so the equivalent capacitance for two capacitors in series becomes
1
1
1
= +
Ceq C1 C2
(5.3.7) The generalization to any number of capacitors connected in series is
N
1
1
1
1
1
= +
+" +
=∑
Ceq C1 C2
CN i =1 Ci
( series )
(5.3.8)
Example 5.4: Equivalent Capacitance
Find the equivalent capacitance for the combination of capacitors shown in Figure 5.3.4(a)
Figure 5.3.4 (a) Capacitors connected in series and in parallel
Solution:
Since C1 and C2 are connected in parallel, their equivalent capacitance C12 is given by
C12 = C1 + C2
Figure 5.3.4 (b) and (c) Equivalent circuits.
Now capacitor C12 is in series with C3, as seen from Figure 5.3.4(b). So, the equivalent
capacitance C123 is given by
1
1
1
=
+
C123 C12 C3
or
C123 =
( C + C2 ) C3
C12C3
= 1
C12 + C3 C1 + C2 + C3 5.4 Storing Energy in a Capacitor
As discussed in the introduction, capacitors can be used to stored electrical energy. The
amount of energy stored is equal to the work done to charge it. During the charging
process, the battery does work to remove charges from one plate and deposit them onto
the other.
Figure 5.4.1 Work is done by an external agent in bringing +dq from the negative plate and
depositing the charge on the positive plate.
Let the capacitor be initially uncharged. In each plate of the capacitor, there are many
negative and positive charges, but the number of negative charges balances the number of
positive charges, so that there is no net charge, and therefore no electric field between the
plates. We have a magic bucket and a set of stairs from the bottom plate to the top plate
(Figure 5.4.1).
We start out at the bottom plate, fill our magic bucket with a charge + dq , carry the
bucket up the stairs and dump the contents of the bucket on the top plate, charging it up
positive to charge + dq . However, in doing so, the bottom plate is now charged to − dq .
Having emptied the bucket of charge, we now descend the stairs, get another bucketful of
charge +dq, go back up the stairs and dump that charge on the top plate. We then repeat
this process over and over. In this way we build up charge on the capacitor, and create
electric field where there was none initially.
Suppose the amount of charge on the top plate at some instant is + q , and the potential
difference between the two plates is | ∆V |= q / C . To dump another bucket of charge
+ dq on the top plate, the amount of work done to overcome electrical repulsion is
dW =| ∆V | dq . If at the end of the charging process, the charge on the top plate is +Q ,
then the total amount of work done in this process is
Q
Q
0
0
W = ∫ dq | ∆ V |= ∫ dq
q 1 Q2
=
C 2 C
(5.4.1)
This is equal to the electrical potential energy U E of the system:
UE =
1 Q2 1
1
= Q | ∆ V |= C | ∆ V |2
2 C 2
2
(5.4.2) 5.4.1 Energy Density of the Electric Field
One can think of the energy stored in the capacitor as being stored in the electric field
itself. In the case of a parallel-plate capacitor, with C = ε 0 A / d and | ∆V |= Ed , we have
UE =
1
1 ε0 A
1
2
C | ∆ V |2 =
( Ed ) = ε 0 E 2 ( Ad )
2
2 d
2
(5.4.3)
Since the quantity Ad represents the volume between the plates, we can define the electric
energy density as
uE =
UE
1
= ε0E2
Volume 2
(5.4.4)
Note that uE is proportional to the square of the electric field. Alternatively, one may
obtain the energy stored in the capacitor from the point of view of external work. Since
the plates are oppositely charged, force must be applied to maintain a constant separation
between them. From Eq. (4.4.7), we see that a small patch of charge
∆q = σ (∆A) experiences an attractive force ∆F = σ 2 (∆A) / 2ε 0 . If the total area of the
plate is A, then an external agent must exert a force Fext = σ 2 A / 2ε 0 to pull the two plates
apart. Since the electric field strength in the region between the plates is given by
E = σ / ε 0 , the external force can be rewritten as
Fext =
ε0
2
E2 A
(5.4.5)
Note that Fext is independent of d . The total amount of work done externally to separate
the plates by a distance d is then
G
⎛ ε E2 A ⎞
G
Wext = ∫ Fext ⋅ d s = Fext d = ⎜ 0
⎟d
⎝ 2 ⎠
(5.4.6)
consistent with Eq. (5.4.3). Since the potential energy of the system is equal to the work
done by the external agent, we have u E = Wext / Ad = ε 0 E 2 / 2 . In addition, we note that the
expression for uE is identical to Eq. (4.4.8) in Chapter 4. Therefore, the electric energy
density uE can also be interpreted as electrostatic pressure P.
Interactive Simulation 5.2: Charge Placed between Capacitor Plates
This applet shown in Figure 5.4.2 is a simulation of an experiment in which an aluminum
sphere sitting on the bottom plate of a capacitor is lifted to the top plate by the
electrostatic force generated as the capacitor is charged. We have placed a non- conducting barrier just below the upper plate to prevent the sphere from touching it and
discharging.
Figure 5.4.2 Electrostatic force experienced by an aluminum sphere placed between the
plates of a parallel-plate capacitor.
While the sphere is in contact with the bottom plate, the charge density of the bottom of
the sphere is the same as that of the lower plate. Thus, as the capacitor is charged, the
charge density on the sphere increases proportional to the potential difference between
the plates. In addition, energy flows in to the region between the plates as the electric
field builds up. This can be seen in the motion of the electric field lines as they move
from the edge to the center of the capacitor.
As the potential difference between the plates increases, the sphere feels an increasing
attraction towards the top plate, indicated by the increasing tension in the field as more
field lines "attach" to it. Eventually this tension is enough to overcome the downward
force of gravity, and the sphere is lifted. Once separated from the lower plate, the sphere
charge density no longer increases, and it feels both an attractive force towards the upper
plate (whose charge is roughly opposite that of the sphere) and a repulsive force from the
lower one (whose charge is roughly equal to that of the sphere). The result is a net force
upwards.
Example 5.5: Electric Energy Density of Dry Air
The breakdown field strength at which dry air loses its insulating ability and allows a
discharge to pass through is Eb = 3 × 106 V/m . At this field strength, the electric energy
density is:
(
1
1
u E = ε 0 E 2 = 8.85 × 10−12 C 2 /N ⋅ m 2
2
2
)( 3 ×10 V/m )
6
2
= 40 J/m3
Example 5.6: Energy Stored in a Spherical Shell
Find the energy stored in a metallic spherical shell of radius a and charge Q.
Solution:
(5.4.7) The electric field associated of a spherical shell of radius a is (Example 4.3)
⎧ Q ˆ
r, r > a
JG ⎪
2
E = ⎨ 4πε 0 r
⎪ 0G ,
r 1 . Note that every dielectric material has a characteristic dielectric strength
which is the maximum value of electric field before breakdown occurs and charges begin
to flow.
Material
κe
Dielectric strength (106 V / m )
Air
1.00059
3
Paper
3.7
16
Glass
4−6
9
Water
80
−
The fact that capacitance increases in the presence of a dielectric can be explained from a
molecular point of view. We shall show that κ e is a measure of the dielectric response to
an external electric field. There are two types of dielectrics. The first type is polar
dielectrics, which are dielectrics that have permanent electric dipole moments. An
example of this type of dielectric is water.
G
G
G
Figure 5.5.1 Orientations of polar molecules when (a) E0 = 0 and (b) E0 ≠ 0 .
As depicted in Figure 5.5.1, the orientation of polar molecules is random in the absence
JG
of an external field. When an external electric field E0 is present, a torque is set up and
G
causes the molecules to align with E0 . However, the alignment is not complete due to
random thermal motion. The aligned molecules then generate an electric field that is
opposite to the applied field but smaller in magnitude.
The second type of dielectrics is the non-polar dielectrics, which are dielectrics that do
not possess permanent electric dipole moment. Electric dipole moments can be induced
by placing the materials in an externally applied electric field. G
G
G
G
Figure 5.5.2 Orientations of non-polar molecules when (a) E0 = 0 and (b) E0 ≠ 0 .
Figure 5.5.2 illustrates the orientation of non-polar molecules with and without an
JG
G
external field E0 . The induced surface charges on the faces produces an electric field E P
JG G
G
G
G
G
in the direction opposite to E0 , leading to E = E0 + E P , with | E | < | E0 | . Below we show
JG
how the induced electric field E P is calculated.
5.5.1 Polarization
We have shown that dielectric materials consist of many permanent or induced electric
dipoles. One of the concepts crucial to the understanding of dielectric materials is the
average electric field produced by many little electric dipoles which are all aligned.
Suppose we have a piece of material in the form of a cylinder with area A and height h,
as shown in Figure 5.5.3, and that it consists of N electric dipoles, each with electric
G
dipole moment p spread uniformly throughout the volume of the cylinder.
Figure 5.5.3 A cylinder with uniform dipole distribution.
G
We furthermore assume for the moment that all of the electric dipole moments p are
aligned with the axis of the cylinder. Since each electric dipole has its own electric field
associated with it, in the absence of any external electric field, if we average over all the
individual fields produced by the dipole, what is the average electric field just due to the
presence of the aligned dipoles?
G
To answer this question, let us define the polarization vector P to be the net electric
dipole moment vector per unit volume: G
P=
1
Volume
N
G
∑p
i
(5.5.2)
i=1
In the case of our cylinder, where all the dipoles are perfectly aligned, the magnitude of
G
P is equal to
P=
Np
Ah
(5.5.3)
G
and the direction of P is parallel to the aligned dipoles.
Now, what is the average electric field these dipoles produce? The key to figuring this
out is realizing that the situation shown in Figure 5.5.4(a) is equivalent that shown in
Figure 5.5.4(b), where all the little ± charges associated with the electric dipoles in the
interior of the cylinder are replaced with two equivalent charges, ±QP , on the top and
bottom of the cylinder, respectively.
Figure 5.5.4 (a) A cylinder with uniform dipole distribution. (b) Equivalent charge
distribution.
The equivalence can be seen by noting that in the interior of the cylinder, positive charge
at the top of any one of the electric dipoles is canceled on average by the negative charge
of the dipole just above it. The only place where cancellation does not take place is for
electric dipoles at the top of the cylinder, since there are no adjacent dipoles further up.
Thus the interior of the cylinder appears uncharged in an average sense (averaging over
many dipoles), whereas the top surface of the cylinder appears to carry a net positive
charge. Similarly, the bottom surface of the cylinder will appear to carry a net negative
charge.
How do we find an expression for the equivalent charge QP in terms of quantities we
know? The simplest way is to require that the electric dipole moment QP produces,
QP h , is equal to the total electric dipole moment of all the little electric dipoles. This
gives QP h = Np , or
Np
(5.5.4)
QP =
h To compute the electric field produced by QP , we note that the equivalent charge
distribution resembles that of a parallel-plate capacitor, with an equivalent surface charge
density σ P that is equal to the magnitude of the polarization:
σP =
QP Np
=
=P
A
Ah
(5.5.5)
Note that the SI units of P are (C ⋅ m)/m3 , or C/m 2 , which is the same as the surface
charge density. In general if the polarization vector makes an angle θ with n̂ , the
outward normal vector of the surface, the surface charge density would be
G
σ P = P ⋅ nˆ = P cos θ
(5.5.6)
Thus, our equivalent charge system will produce an average electric field of magnitude
G
EP = P / ε 0 . Since the direction of this electric field is opposite to the direction of P , in
vector notation, we have
G
G
E P = −P / ε 0
(5.5.7)
Thus, the average electric field of all these dipoles is opposite to the direction of the
dipoles themselves. It is important to realize that this is just the average field due to all
the dipoles. If we go close to any individual dipole, we will see a very different field.
We have assumed here that all our electric dipoles are aligned. In general, if these
G
dipoles are randomly oriented, then the polarization P given in Eq. (5.5.2) will be zero,
and there will be no average field due to their presence. If the dipoles have some
G G
tendency toward a preferred orientation, then P ≠ 0 , leading to a non-vanishing average
G
field E P .
Let us now examine the effects of introducing dielectric material into a system. We shall
first assume that the atoms or molecules comprising the dielectric material have a
permanent electric dipole moment. If left to themselves, these permanent electric dipoles
in a dielectric material never line up spontaneously, so that in the absence of any applied
G G
external electric field, P = 0 due to the random alignment of dipoles, and the average
G
electric field E P is zero as well. However, when we place the dielectric material in an
G
G G G
external field E0 , the dipoles will experience a torque τ = p × E0 that tends to align the
G
G
G
G
dipole vectors p with E0 . The effect is a net polarization P parallel to E0 , and therefore
G
G
an average electric field of the dipoles E P anti-parallel to E0 , i.e., that will tend to
G
G
reduce the total electric field strength below E0 . The total electric field E is the sum of
these two fields: G G
G
G G
E = E0 + E P = E0 − P / ε 0
(5.5.8)
G
G
In most cases, the polarization P is not only in the same direction as E0 , but also linearly
G
G
proportional to E0 (and hence E .) This is reasonable because without the external field
G
G
E0 there would be no alignment of dipoles and no polarization P . We write the linear
G
G
relation between P and E as
G
G
P = ε 0 χeE
(5.5.9)
where χ e is called the electric susceptibility. Materials they obey this relation are linear
dielectrics. Combing Eqs. (5.5.8) and (5.5.7) gives
G
G
G
E0 = (1 + χ e )E = κ e E
(5.5.10)
κ e = (1 + χ e )
(5.5.11)
where
is the dielectric constant. The dielectric constant κ e is always greater than one since
χ e > 0 . This implies
E=
E0
κe
< E0
(5.5.12)
Thus, we see that the effect of dielectric materials is always to decrease the electric field
below what it would otherwise be.
In the case of dielectric material where there are no permanent electric dipoles, a similar
G
effect is observed because the presence of an external field E0 induces electric dipole
G
moments in the atoms or molecules. These induced electric dipoles are parallel to E0 ,
G
G
again leading to a polarization P parallel to E0 , and a reduction of the total electric field
strength.
5.5.2 Dielectrics without Battery
As shown in Figure 5.5.5, a battery with a potential difference | ∆V0 | across its terminals
is first connected to a capacitor C0, which holds a charge Q0 = C0 | ∆V0 | . We then
disconnect the battery, leaving Q0 = const. Figure 5.5.5 Inserting a dielectric material between the capacitor plates while keeping the
charge Q0 constant
If we then insert a dielectric between the plates, while keeping the charge constant,
experimentally it is found that the potential difference decreases by a factor of κ e :
| ∆ V |=
| ∆ V0 |
κe
(5.5.13)
This implies that the capacitance is changed to
C=
Q0
Q0
Q
=
= κe
= κ e C0
| ∆ V | | ∆ V0 | / κ e
| ∆ V0 |
(5.5.14)
Thus, we see that the capacitance has increased by a factor of κ e .The electric field within
the dielectric is now
E=
| ∆ V | | ∆ V0 | / κ e 1 ⎛ | ∆ V0 | ⎞ E0
=
= ⎜
=
d
d
κ e ⎝ d ⎟⎠ κ e
(5.5.15)
We see that in the presence of a dielectric, the electric field decreases by a factor of κ e .
5.5.3 Dielectrics with Battery
Consider a second case where a battery supplying a potential difference | ∆V0 | remains
connected as the dielectric is inserted. Experimentally, it is found (first by Faraday) that
the charge on the plates is increased by a factor κ e :
Q = κ eQ0
where Q0 is the charge on the plates in the absence of any dielectric.
Figure 5.5.6 Inserting a dielectric material between the capacitor plates while
maintaining a constant potential difference | V0 | .
(5.5.16) The capacitance becomes
C=
κQ
Q
= e 0 = κ e C0
| ∆ V0 | | ∆ V0 |
(5.5.17)
which is the same as the first case where the charge Q0 is kept constant, but now the
charge has increased.
5.5.4 Gauss’s Law for Dielectrics
Consider again a parallel-plate capacitor shown in Figure 5.5.7:
Figure 5.5.7 Gaussian surface in the absence of a dielectric.
G
When no dielectric is present, the electric field E0 in the region between the plates can be
found by using Gauss’s law:
w
∫∫
S
JG G
Q
E ⋅ d A = E0 A =
,
ε0
⇒ E0 =
σ
ε0
We have see that when a dielectric is inserted (Figure 5.5.8), there is an induced
charge QP of opposite sign on the surface, and the net charge enclosed by the Gaussian
surface is Q − QP .
Figure 5.5.8 Gaussian surface in the presence of a dielectric. Gauss’s law becomes
JG G
Q − QP
E
w
∫∫ ⋅ d A = EA =
ε0
S
(5.5.18)
or
E=
Q − QP
ε0 A
(5.5.19)
However, we have just seen that the effect of the dielectric is to weaken the original field
E0 by a factor κ e . Therefore,
E=
E0
κe
=
Q
κ eε 0 A
=
Q − QP
ε0 A
(5.5.20)
from which the induced charge QP can be obtained as
⎛
1 ⎞
QP = Q ⎜ 1 − ⎟
⎝ κe ⎠
(5.5.21)
In terms of the surface charge density, we have
⎛
σ P = σ ⎜ 1−
⎝
1 ⎞
⎟
κe ⎠
(5.5.22)
Note that in the limit κ e = 1 , QP = 0 which corresponds to the case of no dielectric
material.
Substituting Eq. (5.5.21) into Eq. (5.5.18), we see that Gauss’s law with dielectric can be
rewritten as
G G
Q
Q
E
=
w
∫∫ ⋅ d A =
κ eε 0
S
ε
(5.5.23)
where ε = κ eε 0 is called the dielectric permittivity. Alternatively, we may also write
w
∫∫
G G
D⋅ dA = Q
S
JG
G
where D = ε 0κ E is called the electric displacement vector.
(5.5.24) Example 5.7: Capacitance with Dielectrics
A non-conducting slab of thickness t , area A and dielectric constant κ e is inserted into
the space between the plates of a parallel-plate capacitor with spacing d, charge Q and
area A, as shown in Figure 5.5.9(a). The slab is not necessarily halfway between the
capacitor plates. What is the capacitance of the system?
Figure 5.5.9 (a) Capacitor with a dielectric. (b) Electric field between the plates.
Solution:
To find the capacitance C, we first calculate the potential difference ∆V . We have
already seen that in the absence of a dielectric, the electric field between the plates is
given by E0 = Q / ε 0 A , and ED = E0 / κ e when a dielectric of dielectric constant κ e is
present, as shown in Figure 5.5.9(b). The potential can be found by integrating the
electric field along a straight line from the top to the bottom plates:
−
∆ V = − ∫ Edl = − ∆ V0 − ∆ VD = − E0 ( d − t ) − ED t = −
+
⎛
Q ⎡
1 ⎞⎤
=−
⎢ d − t ⎜ 1− ⎟ ⎥
Aε 0 ⎣⎢
⎝ κ e ⎠ ⎦⎥
Q
Q
t
(d −t )−
Aε 0
Aε 0κ e
(5.5.25)
where ∆VD = ED t is the potential difference between the two faces of the dielectric. This
gives
ε0 A
Q
C=
=
(5.5.26)
| ∆V |
⎛
1 ⎞
d − t ⎜ 1−
⎟
κe ⎠
⎝
It is useful to check the following limits:
(i) As t → 0, i.e., the thickness of the dielectric approaches zero, we have
C = ε 0 A / d = C0 , which is the expected result for no dielectric.
(ii) As κ e → 1 , we again have C → ε 0 A / d = C0 , and the situation also correspond to the
case where the dielectric is absent. (iii) In the limit where t → d ,
have C → κ eε 0 A / d = κ eC0 .
the
space
is
filled
with
dielectric,
we
We also comment that the configuration is equivalent to two capacitors connected in
series, as shown in Figure 5.5.10.
Figure 5.5.10 Equivalent configuration.
Using Eq. (5.3.8) for capacitors connected in series, the equivalent capacitance is
1 d −t
t
=
+
C ε 0 A κ eε 0 A
(5.5.28)
5.6 Creating Electric Fields
Animation 5.1: Creating an Electric Dipole
Electric fields are created by electric charge. If there is no electric charge present, and
there never has been any electric charge present in the past, then there would be no
electric field anywhere is space. How is electric field created and how does it come to fill
up space? To answer this, consider the following scenario in which we go from the
electric field being zero everywhere in space to an electric field existing everywhere in
space.
Figure 5.6.1 Creating an electric dipole. (a) Before any charge separation. (b) Just
after the charges are separated. (c) A long time after the charges are separated. Suppose we have a positive point charge sitting right on top of a negative electric charge,
so that the total charge exactly cancels, and there is no electric field anywhere in space.
Now let us pull these two charges apart slightly, so that they are separated by a small
distance. If we allow them to sit at that distance for a long time, there will now be a
charge imbalance – an electric dipole. The dipole will create an electric field.
Let us see how this creation of electric field takes place in detail. Figure 5.6.1 shows
three frames of an animation of the process of separating the charges. In Figure 5.6.1(a),
there is no charge separation, and the electric field is zero everywhere in space. Figure
5.6.1(b) shows what happens just after the charges are first separated. An expanding
sphere of electric fields is observed. Figure 5.6.1(c) is a long time after the charges are
separated (that is, they have been at a constant distance from another for a long time). An
electric dipole has been created.
What does this sequence tell us? The following conclusions can be drawn:
(1) It is electric charge that generates electric field — no charge, no field.
(2) The electric field does not appear instantaneously in space everywhere as soon as
there is unbalanced charge — the electric field propagates outward from its source at
some finite speed. This speed will turn out to be the speed of light, as we shall see later.
(3) After the charge distribution settles down and becomes stationary, so does the field
configuration. The initial field pattern associated with the time dependent separation of
the charge is actually a burst of “electric dipole radiation.” We return to the subject of
radiation at the end of this course. Until then, we will neglect radiation fields. The field
configuration left behind after a long time is just the electric dipole pattern discussed
above.
We note that the external agent who pulls the charges apart has to do work to keep them
separate, since they attract each other as soon as they start to separate. Therefore, the
external work done is to overcome the electrostatic attraction. In addition, the work also
goes into providing the energy carried off by radiation, as well as the energy needed to
set up the final stationary electric field that we see in Figure 5.6.1(c).
Figure 5.6.2 Creating the electric fields of two point charges by pulling apart two
opposite charges initially on top of one another. We artificially terminate the field lines
at a fixed distance from the charges to avoid visual confusion. Finally, we ignore radiation and complete the process of separating our opposite point
charges that we began in Figure 5.6.1. Figure 5.6.2 shows the complete sequence. When
we finish and have moved the charges far apart, we see the characteristic radial field in
the vicinity of a point charge.
Animation 5.2: Creating and Destroying Electric Energy
Let us look at the process of creating electric energy in a different context. We ignore
energy losses due to radiation in this discussion. Figure 5.6.3 shows one frame of an
animation that illustrates the following process.
Figure 5.6.3 Creating and destroying electric energy.
We start out with five negative electric charges and five positive charges, all at the same
point in space. Sine there is no net charge, there is no electric field. Now we move one
of the positive charges at constant velocity from its initial position to a distance L away
along the horizontal axis. After doing that, we move the second positive charge in the
same manner to the position where the first positive charge sits. After doing that, we
continue on with the rest of the positive charges in the same manner, until all the positive
charges are sitting a distance L from their initial position along the horizontal axis.
Figure 5.6.3 shows the field configuration during this process. We have color coded the
“grass seeds” representation to represent the strength of the electric field. Very strong
fields are white, very weak fields are black, and fields of intermediate strength are
yellow.
Over the course of the “create” animation associated with Figure 5.6.3, the strength of the
electric field grows as each positive charge is moved into place. The electric energy
flows out from the path along which the charges move, and is being provided by the
agent moving the charge against the electric field of the other charges. The work that this
agent does to separate the charges against their electric attraction appears as energy in the
electric field. We also have an animation of the opposite process linked to Figure 5.6.3.
That is, we return in sequence each of the five positive charges to their original positions.
At the end of this process we no longer have an electric field, because we no longer have
an unbalanced electric charge.
On the other hand, over the course of the “destroy” animation associated with Figure
5.6.3, the strength of the electric field decreases as each positive charge is returned to its
original position. The energy flows from the field back to the path along which the charges move, and is now being provided to the agent moving the charge at constant
speed along the electric field of the other charges. The energy provided to that agent as
we destroy the electric field is exactly the amount of energy that the agent put into
creating the electric field in the first place, neglecting radiative losses (such losses are
small if we move the charges at speeds small compared to the speed of light). This is a
totally reversible process if we neglect such losses. That is, the amount of energy the
agent puts into creating the electric field is exactly returned to that agent as the field is
destroyed.
There is one final point to be made. Whenever electromagnetic energy is being created,
G G
an electric charge is moving (or being moved) against an electric field ( q v ⋅ E < 0 ).
Whenever electromagnetic energy is being destroyed, an electric charge is moving (or
G G
being moved) along an electric field ( q v ⋅ E > 0 ). When we return to the creation and
destruction of magnetic energy, we will find this rule holds there as well.
5.7 Summary
•
A capacitor is a device that stores electric charge and potential energy. The
capacitance C of a capacitor is the ratio of the charge stored on the capacitor
plates to the the potential difference between them:
C=
Q
| ∆V |
System
Capacitance
C = 4πε 0 R
Isolated charged sphere of radius R
Parallel-plate capacitor of plate area A and plate separation d
Cylindrical capacitor of length L , inner radius a and outer radius b
Spherical capacitor with inner radius a and outer radius b
•
C = ε0
C=
A
d
2πε 0 L
ln(b / a)
C = 4πε 0
ab
(b − a )
The equivalent capacitance of capacitors connected in parallel and in series are
Ceq = C1 + C2 + C3 +"
(parallel)
1
1
1
1
= +
+ + " (series)
Ceq C1 C2 C3 •
The work done in charging a capacitor to a charge Q is
U=
Q2 1
1
= Q | ∆V |= C | ∆V |2
2C 2
2
This is equal to the amount of energy stored in the capacitor.
•
G
The electric energy can also be thought of as stored in the electric field E . The
energy density (energy per unit volume) is
1
uE = ε 0 E 2
2
The energy density uE is equal to the electrostatic pressure on a surface.
•
•
When a dielectric material with dielectric constant κ e is inserted into a
capacitor, the capacitance increases by a factor κ e :
C = κ eC0
G
The polarization vector P is the magnetic dipole moment per unit volume:
G 1
P=
V
N
G
∑p
i =1
i
The induced electric field due to polarization is
G
G
E P = −P / ε 0
•
In the presence of a dielectric with dielectric constant κ e , the electric field
becomes
G G
G
G
E = E0 + E P = E0 / κ e
G
where E0 is the electric field without dielectric.
5.8 Appendix: Electric Fields Hold Atoms Together
In this Appendix, we illustrate how electric fields are responsible for holding atoms
together. “…As our mental eye penetrates into smaller and smaller distances and
shorter and shorter times, we find nature behaving so entirely differently
from what we observe in visible and palpable bodies of our surroundings
that no model shaped after our large-scale experiences can ever be "true".
A completely satisfactory model of this type is not only practically
inaccessible, but not even thinkable. Or, to be precise, we can, of course,
think of it, but however we think it, it is wrong.”
Erwin Schroedinger
5.8.1 Ionic and van der Waals Forces
Electromagnetic forces provide the “glue” that holds atoms together—that is, that keep
electrons near protons and bind atoms together in solids. We present here a brief and
very idealized model of how that happens from a semi-classical point of view.
(a)
(b)
Figure 5.8.1 (a) A negative charge and (b) a positive charge moves past a massive
positive particle at the origin and is deflected from its path by the stresses transmitted by
the electric fields surrounding the charges.
Figure 5.8.1(a) illustrates the examples of the stresses transmitted by fields, as we have
seen before. In Figure 5.8.1(a) we have a negative charge moving past a massive positive
charge and being deflected toward that charge due to the attraction that the two charges
feel. This attraction is mediated by the stresses transmitted by the electromagnetic field,
and the simple interpretation of the interaction shown in Figure 5.8.1(b) is that the
attraction is primarily due to a tension transmitted by the electric fields surrounding the
charges.
In Figure 5.8.1(b) we have a positive charge moving past a massive positive charge and
being deflected away from that charge due to the repulsion that the two charges feel.
This repulsion is mediated by the stresses transmitted by the electromagnetic field, as we
have discussed above, and the simple interpretation of the interaction shown in Figure
5.8.1(b) is that the repulsion is primarily due to a pressure transmitted by the electric
fields surrounding the charges.
Consider the interaction of four charges of equal mass shown in Figure 5.8.2. Two of the
charges are positively charged and two of the charges are negatively charged, and all
have the same magnitude of charge. The particles interact via the Coulomb force. We also introduce a quantum-mechanical “Pauli” force, which is always repulsive and
becomes very important at small distances, but is negligible at large distances. The
critical distance at which this repulsive force begins to dominate is about the radius of the
spheres shown in Figure 5.8.2. This Pauli force is quantum mechanical in origin, and
keeps the charges from collapsing into a point (i.e., it keeps a negative particle and a
positive particle from sitting exactly on top of one another).
Additionally, the motion of the particles is damped by a term proportional to their
velocity, allowing them to "settle down" into stable (or meta-stable) states.
Figure 5.8.2 Four charges interacting via the Coulomb force, a repulsive Pauli force at
close distances, with dynamic damping.
When these charges are allowed to evolve from the initial state, the first thing that
happens (very quickly) is that the charges pair off into dipoles. This is a rapid process
because the Coulomb attraction between unbalanced charges is very large. This process is
called "ionic binding", and is responsible for the inter-atomic forces in ordinary table salt,
NaCl. After the dipoles form, there is still an interaction between neighboring dipoles, but
this is a much weaker interaction because the electric field of the dipoles falls off much
faster than that of a single charge. This is because the net charge of the dipole is zero.
When two opposite charges are close to one another, their electric fields “almost” cancel
each other out.
Although in principle the dipole-dipole interaction can be either repulsive or attractive, in
practice there is a torque that rotates the dipoles so that the dipole-dipole force is
attractive. After a long time, this dipole-dipole attraction brings the two dipoles together
in a bound state. The force of attraction between two dipoles is termed a “van der Waals”
force, and it is responsible for intermolecular forces that bind some substances together
into a solid.
Interactive Simulation 5.3: Collection of Charges in Two Dimensions
Figure 5.8.3 is an interactive two-dimensional ShockWave display that shows the same
dynamical situation as in Figure 5.8.2 except that we have included a number of positive
and negative charges, and we have eliminated the representation of the field so that we can interact with this simulation in real time. We start the charges at rest in random
positions in space, and then let them evolve according to the forces that act on them
(electrostatic attraction/repulsion, Pauli repulsion at very short distances, and a dynamic
drag term proportional to velocity). The particles will eventually end up in a
configuration in which the net force on any given particle is essentially zero. As we saw
in the animation in Figure 5.8.3, generally the individual particles first pair off into
dipoles and then slowly combine into larger structures. Rings and straight lines are the
most common configurations, but by clicking and dragging particles around, the user can
coax them into more complex meta-stable formations.
Figure 5.8.3 A two dimensional interactive simulation of a collection of positive and
negative charges affected by the Coulomb force and the Pauli repulsive force, with
dynamic damping.
In particular, try this sequence of actions with the display. Start it and wait until the
simulation has evolved to the point where you have a line of particles made up of seven
or eight particles. Left click on one of the end charges of this line and drag it with the
mouse. If you do this slowly enough, the entire line of chares will follow along with the
charge you are virtually “touching”. When you move that charge, you are putting
“energy” into the charge you have selected on one end of the line. This “energy” is going
into moving that charge, but it is also being supplied to the rest of the charges via their
electromagnetic fields. The “energy” that the charge on the opposite end of the line
receives a little while after you start moving the first charge is delivered to it entirely by
energy flowing through space in the electromagnetic field, from the site where you create
that energy.
This is a microcosm of how you interact with the world. A physical object lying on the
floor in front is held together by electrostatic forces. Quantum mechanics keeps it from
collapsing; electrostatic forces keep it from flying apart. When you reach down and pick
that object up by one end, energy is transferred from where you grasp the object to the
rest of it by energy flow in the electromagnetic field. When you raise it above the floor,
the “tail end” of the object never “touches” the point where you grasp it. All of the
energy provided to the “tail end” of the object to move it upward against gravity is
provided by energy flow via electromagnetic fields, through the complicated web of
electromagnetic fields that hold the object together. Interactive Simulation 5.4: Collection of Charges in Three Dimensions
Figure 5.8.4 is an interactive three-dimensional ShockWave display that shows the same
dynamical situation as in Figure 5.8.3 except that we are looking at the scene in three
dimensions. This display can be rotated to view from different angles by right-clicking
and dragging in the display. We start the charges at rest in random positions in space,
and then let them evolve according to the forces that act on them (electrostatic
attraction/repulsion, Pauli repulsion at very short distances, and a dynamic drag term
proportional to velocity). Here the configurations are more complex because of the
availability of the third dimension. In particular, one can hit the “w” key to toggle a force
that pushes the charges together on and off. Toggling this force on and letting the
charges settle down in a “clump”, and then toggling it off to let them expand, allows the
construction of complicated three dimension structures that are “meta-stable”.
An
example of one of these is given in Figure 5.8.4.
Figure 5.8.4 An three-dimensional interactive simulation of a collection of positive and
negative charges affected by the Coulomb force and the Pauli repulsive force, with
dynamic damping.
Interactive Simulation 5.5: Collection of Dipoles in Two Dimensions
Figure 5.8.5 shows an interactive ShockWave simulation that allows one to interact in
two dimensions with a group of electric dipoles.
Figure 5.8.5 An interactive simulation of a collection of electric dipoles affected by the
Coulomb force and the Pauli repulsive force, with dynamic damping. The dipoles are created with random positions and orientations, with all the electric
dipole vectors in the plane of the display. As we noted above, although in principle the
dipole-dipole interaction can be either repulsive or attractive, in practice there is a torque
that rotates the dipoles so that the dipole-dipole force is attractive. In the ShockWave
simulation we see this behavior—that is, the dipoles orient themselves so as to attract,
and then the attraction gathers them together into bound structures.
Interactive Simulation 5.6: Charged Particle Trap
Figure 5.8.6 shows an interactive simulation of a charged particle trap.
Figure 5.8.6 An interactive simulation of a particle trap.
Particles interact as before, but in addition each particle feels a force that pushes them
toward the origin, regardless of the sign of their charge. That “trapping” force increases
linearly with distance from the origin. The charges initially are randomly distributed in
space, but as time increases the dynamic damping “cools” the particles and they
“crystallize” into a number of highly symmetric structures, depending on the number of
particles. This mimics the highly ordered structures that we see in nature (e.g.,
snowflakes).
Exercise:
Start the simulation. The simulation initially introduces 12 positive charges in random
positions (you can of course add more particles of either sign, but for the moment we deal
with only the initial 12). About half the time, the 12 charges will settle down into an
equilibrium in which there is a charge in the center of a sphere on which the other 11
charges are arranged. The other half of the time all 12 particles will be arranged on the
surface of a sphere, with no charge in the middle. Whichever arrangement you initially
find, see if you can move one of the particles into position so that you get to the other
stable configuration. To move a charge, push shift and left click, and use the arrow
buttons to move it up, down, left, and right. You may have to select several different
charges in turn to find one that you can move into the center, if you initial equilibrium
does not have a center charge. Here is another exercise. Put an additional 8 positive charges into the display (by
pressing “p” eight times) for a total of 20 charges. By moving charges around as above,
you can get two charges in inside a spherical distribution of the other 18. Is this the
lowest number of charges for which you can get equilibrium with two charges inside?
That is, can you do this with 18 charges? Note that if you push the “s” key you will get
generate a surface based on the positions of the charges in the sphere, which will make its
symmetries more apparent.
Interactive Simulation 5.6: Lattice 3D
Lattice 3D, shown in Figure 5.8.7, simulates the interaction of charged particles in three
dimensions. The particles interact via the classical Coulomb force, as well as the
repulsive quantum-mechanical Pauli force, which acts at close distances (accounting for
the “collisions” between them). Additionally, the motion of the particles is damped by a
term proportional to their velocity, allowing them to “settle down” into stable (or metastable) states.
Figure 5.8.7 Lattice 3D simulating the interaction of charged particles in three
dimensions.
In this simulation, the proportionality of the Coulomb and Pauli forces has been adjusted
to allow for lattice formation, as one might see in a crystal. The “preferred” stable state is
a rectangular (cubic) lattice, although other formations are possible depending on the
number of particles and their initial positions.
Selecting a particle and pressing “f” will toggle field lines illustrating the local field
around that particle. Performance varies depending on the number of particles / field lines
in the simulation.
Interactive Simulation 5.7: 2D Electrostatic Suspension Bridge
To connect electrostatic forces to one more example of the real world, Figure 5.8.8 is a
simulation of a 2D “electrostatic suspension bridge.” The bridge is created by attaching a
series of positive and negatively charged particles to two fixed endpoints, and adding a
downward gravitational force. The tension in the “bridge” is supplied simply by the Coulomb interaction of its constituent parts and the Pauli force keeps the charges from
collapsing in on each other. Initially, the bridge only sags slightly under the weight of
gravity. However the user can introduce additional “neutral” particles (by pressing “o”)
to stress the bridge more, until the electrostatic bonds “break” under the stress and the
bridge collapses.
Figure 5.8.8 A ShockWave simulation of a 2D electrostatic suspension bridge.
Interactive Simulation 5.8: 3D Electrostatic Suspension Bridge
In the simulation shown in Figure 5.8.9, a 3D “electrostatic suspension bridge” is created
by attaching a lattice of positive and negatively charged particles between four fixed
corners, and adding a downward gravitational force. The tension in the “bridge” is
supplied simply by the Coulomb interaction of its constituent parts and the Pauli force
keeping them from collapsing in on each other. Initially, the bridge only sags slightly
under the weight of gravity, but what would happen to it under a rain of massive neutral
particles? Press “o” to find out.
Figure 5.8.9 A ShockWave simulation of a 3D electrostatic suspension bridge.
5.9 Problem-Solving Strategy: Calculating Capacitance
In this chapter, we have seen how capacitance C can be calculated for various systems.
The procedure is summarized below: (1) Identify the direction of the electric field using symmetry.
(2) Calculate the electric field everywhere.
(3) Compute the electric potential difference ∆V.
(4) Calculate the capacitance C using C = Q / | ∆V | .
In the Table below, we illustrate how the above steps are used to calculate the
capacitance of a parallel-plate capacitor, cylindrical capacitor and a spherical capacitor.
Capacitors
Parallel-plate
Cylindrical
Spherical
Figure
(1) Identify the
direction of the
electric
field
using symmetry
(2)
Calculate
electric
field
everywhere
(3) Compute the
electric potential
difference ∆V
JG JG
Q
w
∫∫ E ⋅ d A = EA =
ε0
S
E=
Q
σ
=
Aε 0 ε 0
∆V = V− − V+ = − ∫
−
+
= − Ed
G G
E ⋅ ds
JG
JG
Q
w
∫∫ E ⋅ d A = E ( 2π rl ) = ε
S
E=
0
λ
2πε 0 r
JG
JG
w
∫∫ E ⋅ d A = E ( 4π r ) = ε
2
Q
r
S
1
Q
Er =
4πε o r 2
b
∆ V = Vb − Va = − ∫ Er dr
a
λ
⎛b⎞
=−
ln ⎜ ⎟
2πε 0 ⎝ a ⎠
b
∆ V = Vb − Va = − ∫ Er dr
a
Q ⎛ b−a ⎞
=−
⎜
⎟
4πε 0 ⎝ ab ⎠
0 (4) Calculate C
using
C = Q / | ∆V |
C=
ε0 A
d
C=
2πε 0l
ln(b / a )
⎛ ab ⎞
C = 4πε 0 ⎜
⎟
⎝ b−a ⎠
5.10 Solved Problems
5.10.1 Equivalent Capacitance
Consider the configuration shown in Figure 5.10.1. Find the equivalent capacitance,
assuming that all the capacitors have the same capacitance C.
Figure 5.10.1 Combination of Capacitors
Solution:
For capacitors that are connected in series, the equivalent capacitance is
1
1
1
1
= +
+" = ∑
Ceq C1 C2
i Ci
(series)
On the other hand, for capacitors that are connected in parallel, the equivalent
capacitance is
Ceq = C1 + C2 + " = ∑ Ci
(parallel)
i
Using the above formula for series connection, the equivalent configuration is shown in
Figure 5.10.2.
Figure 5.10.2 Now we have three capacitors connected in parallel. The equivalent capacitance is given
by
⎛ 1 1 ⎞ 11
Ceq = C ⎜1 + + ⎟ = C
⎝ 2 3⎠ 6
5.10.2 Capacitor Filled with Two Different Dielectrics
Two dielectrics with dielectric constants κ1 and κ 2 each fill half the space between the
plates of a parallel-plate capacitor as shown in Figure 5.10.3.
Figure 5.10.3 Capacitor filled with two different dielectrics.
Each plate has an area A and the plates are separated by a distance d. Compute the
capacitance of the system.
Solution:
Since the potential difference on each half of the capacitor is the same, we may treat the
system as being composed of two capacitors connected in parallel. Thus, the capacitance
of the system is
C = C1 + C2
With
Ci =
κ iε 0 ( A / 2)
d
,
i = 1, 2
we obtain
C=
κ 1ε 0 ( A / 2) κ 2ε 0 ( A / 2)
d
+
d
=
ε0 A
2d
(κ 1 + κ 2 )
5.10.3 Capacitor with Dielectrics
Consider a conducting spherical shell with an inner radius a and outer radius c. Let the
space between two surfaces be filed with two different dielectric materials so that the dielectric constant is κ1 between a and b, and κ 2 between b and c, as shown in Figure
5.10.4. Determine the capacitance of this system.
Figure 5.10.4 Spherical capacitor filled with dielectrics.
Solution:
The system can be treated as two capacitors connected in series, since the total potential
difference across the capacitors is the sum of potential differences across individual
capacitors. The equivalent capacitance for a spherical capacitor of inner radius r1 and
outer radius r2 filled with dielectric with dielectric constant κ e is given by
⎛ rr ⎞
C = 4πε 0κ e ⎜ 1 2 ⎟
⎝ r2 − r1 ⎠
Thus, the equivalent capacitance of this system is
1
1
1
κ c(b − a ) + κ1a(c − b)
=
+
= 2
4πε 0κ1κ 2 abc
C 4πε 0κ1ab 4πε 0κ 2bc
−
−
b
a
c
b
(
)
( )
or
C=
4πε 0κ1κ 2 abc
κ 2 c (b − a ) + κ 1 a ( c − b )
It is instructive to check the limit where κ1 , κ 2 → 1 . In this case, the above expression
reduces to
C=
4πε 0 abc
4πε 0 abc 4πε 0 ac
=
=
c(b − a) + a (c − b) b(c − a) (c − a)
which agrees with Eq. (5.2.11) for a spherical capacitor of inner radius a and outer radius
c. 5.10.4 Capacitor Connected to a Spring
Consider an air-filled parallel-plate capacitor with one plate connected to a spring having
a force constant k, and another plate held fixed. The system rests on a table top as shown
in Figure 5.10.5.
Figure 5.10.5 Capacitor connected to a spring.
If the charges placed on plates a and b are +Q and −Q , respectively, how much does the
spring expand?
Solution:
G
The spring force Fs acting on plate a is given by
G
Fs = −kx ˆi
G
Similarly, the electrostatic force Fe due to the electric field created by plate b is
G
⎛ σ ⎞ ˆ Q2 ˆ
ˆ
Fe = QE i = Q ⎜
i
⎟i =
2 Aε 0
⎝ 2ε 0 ⎠
where A is the area of the plate . Notice that charges on plate a cannot exert a force on
itself, as required by Newton’s third law. Thus, only the electric field due to plate b is
considered. At equilibrium the two forces cancel and we have
⎛ Q ⎞
kx = Q ⎜
⎟
⎝ 2 Aε 0 ⎠
which gives
Q2
x=
2kAε 0
5.11 Conceptual Questions
1. The charges on the plates of a parallel-plate capacitor are of opposite sign, and they
attract each other. To increase the plate separation, is the external work done positive or
negative? What happens to the external work done in this process? 2. How does the stored energy change if the potential difference across a capacitor is
tripled?
3. Does the presence of a dielectric increase or decrease the maximum operating voltage
of a capacitor? Explain.
4. If a dielectric-filled capacitor is cooled down, what happens to its capacitance?
5.12 Additional Problems
5.12.1 Capacitors in Series and in Parallel
A 12-Volt battery charges the four capacitors shown in Figure 5.12.1.
Figure 5.12.1
Let C1 = 1 µF, C2 = 2 µF, C3 = 3 µF, and C4 = 4 µF.
(a) What is the equivalent capacitance of the group C1 and C2 if switch S is open (as
shown)?
(b) What is the charge on each of the four capacitors if switch S is open?
(c) What is the charge on each of the four capacitors if switch S is closed?
5.12.2 Capacitors and Dielectrics
(a) A parallel-plate capacitor of area A and spacing d is filled with three dielectrics as
shown in Figure 5.12.2. Each occupies 1/3 of the volume. What is the capacitance of this
system? [Hint: Consider an equivalent system to be three parallel capacitors, and justify
this assumption.] Show that you obtain the proper limits as the dielectric constants
approach unity, κi → 1.]
Figure 5.12.2 (b) This capacitor is now filled as shown in Figure 5.12.3. What is its capacitance? Use
Gauss's law to find the field in each dielectric, and then calculate ∆V across the entire
capacitor. Again, check your answer as the dielectric constants approach unity, κi → 1.
Could you have assumed that this system is equivalent to three capacitors in series?
Figure 5.12.3
5.12.3 Gauss’s Law in the Presence of a Dielectric
A solid conducting sphere with a radius R1 carries a free charge Q and is surrounded by a
concentric dielectric spherical shell with an outer radius R2 and a dielectric constant κ e .
This system is isolated from other conductors and resides in air ( κ e ≈ 1 ), as shown in
Figure 5.12.4.
Figure 5.12.4
G
(a) Determine the displacement vector D everywhere, i.e. its magnitude and direction in
the regions r < R1 , R1 < r < R2 and r > R2 .
G
(b) Determine the electric field E everywhere.
5.12.4 Gauss’s Law and Dielectrics
A cylindrical shell of dielectric material has inner radius a and outer radius b, as shown in
Figure 5.12.5. Figure 5.12.5
The material has a dielectric constant κ e = 10 . At the center of the shell there is a line
charge running parallel to the axis of the cylindrical shell, with free charge per unit length
λ.
(a) Find the electric field for: r < a , a < r < b and r > b .
(b) What is the induced surface charge per unit length on the inner surface of the
spherical shell? [Ans: −9λ /10 .]
(c) What is the induced surface charge per unit length on the outer surface of the
spherical shell? [Ans: +9λ /10 .]
5.12.5 A Capacitor with a Dielectric
A parallel plate capacitor has a capacitance of 112 pF, a plate area of 96.5 cm2, and a
mica dielectric ( κ e = 5.40 ). At a 55 V potential difference, calculate
(a) the electric field strength in the mica; [Ans: 13.4 kV/m.]
(b) the magnitude of the free charge on the plates; [Ans: 6.16 nC.]
(c) the magnitude of the induced surface charge; [Ans: 5.02 nC.]
G
(d) the magnitude of the polarization P [Ans: 520 nC/m2.]
5.12.6 Force on the Plates of a Capacitor
The plates of a parallel-plate capacitor have area A and carry total charge ±Q (see Figure
5.12.6). We would like to show that these plates attract each other with a force given by
F = Q2/(2εoA). Figure 5.12.6
(a) Calculate the total force on the left plate due to the electric field of the right plate,
using Coulomb's Law. Ignore fringing fields.
(b) If you pull the plates apart, against their attraction, you are doing work and that work
goes directly into creating additional electrostatic energy. Calculate the force necessary
G G
to increase the plate separation from x to x+dx by equating the work you do, F ⋅ dx , to the
increase in electrostatic energy, assuming that the electric energy density is εoE2/2, and
that the charge Q remains constant.
(c) Using this expression for the force, show that the force per unit area (the electrostatic
stress) acting on either capacitor plate is given by εoE2/2. This result is true for a
G
conductor of any shape with an electric field E at its surface.
(d) Atmospheric pressure is 14.7 lb/in2, or 101,341 N/m2. How large would E have to be
to produce this force per unit area? [Ans: 151 MV/m. Note that Van de Graff
accelerators can reach fields of 100 MV/m maximum before breakdown, so that
electrostatic stresses are on the same order as atmospheric pressures in this extreme
situation, but not much greater].
5.12.7 Energy Density in a Capacitor with a Dielectric
Consider the case in which a dielectric material with dielectric constant κ e completely
fills the space between the plates of a parallel-plate capacitor. Show that the energy
G G
density of the field between the plates is uE = E ⋅ D / 2 by the following procedure:
G G
G
(a) Write the expression uE = E ⋅ D / 2 as a function of E and κ e (i.e. eliminate D ).
(b) Given the electric field and potential of such a capacitor with free charge q on it
(problem 4-1a above), calculate the work done to charge up the capacitor from q = 0 to
q = Q , the final charge.
(c) Find the energy density uE .
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