Download Module II Lesson 4 Capacitors Capacitors are important

Module II
Lesson 4
Capacitors
Capacitors are important components of electronic circuits and of electrical machinery
and power grids. You can find large oil-insulated capacitors on power line poles or
small ceramic-insulated capacitors in a radio. In each application, the capacitor is used
to store electrical charge and electrical energy; energy may be stored for a short time in
an alternating-current cycle, or for a longer time until the energy is needed, as in a
strobe light for a camera. Your body can be a capacitor, storing up enough charge and
energy to cause a painful spark when you “discharge” yourself to another conductor.
Practical capacitors are basically two conducting plates or sheets separated by an
insulator (dielectric) such as air, oil, paper, plastic film, or even the oxide layer on one of
the conducting surfaces. This lesson will treat the basic physics of capacitors.
4-1: Definition and Calculation of Capacitance
OBJECTIVES: Define the terms “capacitance” and use these definitions to relate
capacitance, voltage difference, and charge in a capacitor.
Derive and use expressions for the capacitance of capacitors that have
planar, cylindrical, or spherical symmetry.
PREREQUISITE: Determining electric fields and potentials for charge distributions with
planar, cylindrical, and spherical symmetries
Reading Assignment
Study in your textbook Chapter 24, Section 1.
Commentary
Once again we have an important physical quantity defined as a constant of
proportionality in a simple equation. In the equation
(1)
notice that the capacitance,
, is a quantity that is characteristic of a particular
capacitor- which we can think of as a pair of insulated conductors. We can evaluate
for any capacitor by placing equal but opposite charges of magnitude
conductors and measuring the voltage difference between them.
on the two
The charge
is
sometimes called the “charge on the capacitor” even though the capacitor as a whole,
of course, has a net charge of zero.
Obviously, if we know any two of the quantities in Eq. (1), it is possible to solve quickly
for the third.
EXAMPLE 4-1:
An initially uncharged
difference of
capacitor is charged to a potential
by transferring charge from one plate of the
capacitor to the other. How much charge was transferred?
Solution: Capacitance is defined as the ratio or the magnitude of the charge on one or
the other of the conductors to the potential difference between them:
(the larger the capacitance, the more charge the capacitor will hold for a given
potential difference). The answer is therefore
.
Capacitance, then does not depend on the charge applied to a capacitor or the resultant
potential difference; rather, it determines the relation between these quantities. For
now, we are concerned with capacitors whose plates are separated by a vacuum (or
air), and in this case the capacitance depends only on the geometry of the “plates” that
form the capacitor. Gauss’s law provides a means for us to calculate the capacitance of
certain geometric plate configurations. To do this we first imagine that the plates carry
charges of magnitude , and then use Gauss’s law (or some other method) to find the
electric field between the plates. Next, we must integrate to find the magnitude of the
potential difference
between the plates. (We are concerned only with
because
capacitance is always positive.) Finally, we use the definition
to find . This
procedure is illustrated in Section 24.1 of the text and the following additional examples.
(We recommend that you not try to memorize formulas for the capacitance of particular
capacitors; instead, you should concentrate on being able to derive these formulas.)
EXAMPLE 4-2: A spherical capacitor consists of two concentric spherical shells of radii
and (
). Determine its capacitance.
Solution: To derive the capacitance, we imagine that the spheres carry opposite
charges of magnitude . We know that the potential due to the outer sphere
is constant in the space between the conductors so that the total potential in
the region
depends on the distance from the center of the capacitor
as
. Therefore the potential difference between the conductors is
.
From the definition of capacitance,
, we obtain immediately
.
This problem is of a type you can solve without direct application of Gauss’s
law- to the extent that you can remember the forms of the potentials due to
particular charge distributions (e.g., Figure 23.17 in the text).
EXAMPLE 4-3: The cylindrical capacitor at right is
long and consists of two conducting
cylinders that are coaxial as shown.
Determine the capacitance of this capacitor
and compute the potential difference if the
inner conductor carries a charge of
.
(
.)
Solution: Before reading this solution, make sure for yourself that you can derive the
expression for the capacitance. In the space between the conductors, there
is no field due to the outer cylinder so that for a charge
on the inner
conductor, you can determine from Gauss’s law (do it!) that the electric field in
this region is given by
where
is the length of the cylinder. We integrate this expression to find the
potential difference and obtain
.
Therefore, using the given data,
/
and
.
If the gap between the capacitor plates is small compared to their other dimensions, as
in the preceding example, then it is reasonable to ignore edge effect (“fringing”); that is,
to treat the plates as infinite rather than correcting for the field curvature that occurs at
the boundaries of a finite capacitor.
As you reread the text assignment and study Sample Problems 1 through 4, take some
time to consider the relation of the unit of capacitance, the farad, to other SI units.
Sample Problem 1 should give you an idea of the magnitude of the farad, and help to
explain why
and
are more commonly seen in units. (The example may be
somewhat misleading, however. Since capacitance is inversely proportional to plate
gap, it is possible to package
of capacitance into a much smaller space by making
the gap between the plates very small and, as we shall see later, filling it with a
dielectric.)
Practice Exercise 4-1
Write your solutions to the following problems in your notebook.
1. A
capacitor (initially uncharged) is charged through the transfer of
electrons from one plate to the other. What is the resulting potential difference
between the plates?
2. A capacitor is made of two flat, circular, conducting plates of radius
separated in a vacuum by
,
. Determine the capacitance.
Check your answers with the Module 2 Answer Key and review Section 4-1 as
necessary before going on.
4-2: Connected Capacitors
OBJECTIVE: Determine the equivalent capacitance of a set of capacitors connected
together and determine the charge and voltage on each capacitor of the
set.
Reading Assignment
Study in your textbook Chapter 24, Section 2.
Commentary
This section provides a point of entry into the sizable topic of electrical devices as
elements in circuits. For one thing, it introduces a schematic notation that allows us to
easily represent a particular capacitor (or set of capacitors) charged to a given potential
difference. We also encounter the notion that a set of connected capacitors is
equivalent, as a circuit element, to a single capacitor of appropriately chosen
capacitance. (This idea will be extended to other circuit elements in later lessons.)
There are two basic ways that capacitors (or other circuit elements) can be connected
together- in series and in parallel- and these give rise to two separate cases in the
calculation of equivalent capacitance. The case of capacitors in series or parallel is
treated in the text. Example 4-4 below also shows how to derive the equivalent
capacitance for a parallel connection.
EXAMPLE 4-4: Diagram (a) below shows three capacitors-
and
- connected in
parallel (i.e., the plates of each capacitor are connected by wires to the
same terminals, and ).
Determine the equivalent capacitance for this arrangement. In other
words, determine the single capacitance [ in diagram
] that will
give the same potential difference
when a given charge
transferred from terminal
and distributes itself onto the
to terminal
is
plates.
Solution: Let charges on
charge
,
, and
be
,
, and
, respectively. The total
is simply the sum of the charges on the three capacitors:
. For this parallel connection the potential difference
is
the same across each capacitor (the upper plates are all at the same
potential and the potential of the lower plates differ from this by
). The
charges on the capacitor must then be
,
,
.
We can therefore immediately write
,
which says simply that we add the capacitances of capacitors connected in
parallel.
In most practical cases, combinations of capacitors can be consolidated into groups of
capacitors that are connected either in series or in parallel, which are then reduced
step-by-step to find the single equivalent capacitance. It is important to be systematic
as you work through a problem involving both series and parallel connections. Consider
the following examples.
EXAMPLE 4-5: Find the equivalent capacitance of the
combination of capacitors shown in the
figure at right, where
,
, and
Solution:
.
In this problem we first consolidate
and
into an
equivalent capacitor
to obtain the simplified
circuit at right. Now we have
and
in series, which
gives the following relation for the equivalent capacitance
:
.
Inserting the values given for
EXAMPLE 4-6: A potential difference of
an
capacitor (
,
, and
, we obtain
is applied to a
.
capacitor ( ) and
) connected in series.
(a) Determine the charge and the potential difference for each
capacitor.
(b) The charged capacitors are disconnected and then reconnected with
their positive plates together and negative plates together, no
external voltage being applied. What are the charge and the
potential difference for each capacitor after this procedure?
(c) The charged capacitors in part (a) are reconnected with plates of
opposite sign together. What are the charge and the potential
difference for each?
Solution: (a) Conservation of charge on the two
middle plates in the diagram at right lets
us conclude immediately that the two
capacitors carry equal charges. After
calculating the equivalent capacitance of
, you should be able to show that
,
,
.
(b) The important part of the problem is to be sure that you can picture what is
going on. For this purpose you may wish to sketch diagrams illustrating the
successive steps of the problem. Diagrams (i) and (ii) show the capacitors
disconnected and reconnected as described. Conservation of charge
requires that the total charge on the upper plates in diagram (ii) is
,
and that the charge on the lower plates is the negative of this. We can thus
draw diagram (iii) showing an equivalent capacitance
, with
charge
on it and potential difference
it.
across
Having calculated
charges
and
, we can go back to diagram (ii) and determine the
on
and
:
,
.
Sometimes you may find that pursuing a solution in terms of the algebraic
quantities gives a rather complicated expression, and it may be easier to
compute intermediate numbers (such as
in this case). Still, it is usually
worthwhile to try writing out an algebraic expression before you give up and
plug in numbers along the way. In this case, the direct method yields
and
These equations are not too complicated to evaluate in terms of the original
variables. The potential across both capacitors is then
(c) The sequence of steps in this case is similar to that in (b). The difference is
that now we have
.
Practice Exercise 4-2
Write your solutions to the following problems in your notebook.
1. In the circuit diagrammed at right, the
battery provides a constant potential
difference of
and
and
are initially
uncharged. Switch
the
capacitor
is closed, charging
.
Then
is
opened, disconnecting the battery from the
circuit. Following this
is closed. The value of the potential difference
is then measured to be
. Determine the capacitance of
across
.
2. In the circuit at right, the values of the
capacitors are
,
,
, and
.
(a) What is the equivalent capacitance
between terminals and ?
(b) What is the potential difference across
terminals
if a
-volt battery is connected to
and ?
Check your answers with the Module 2 Answer Key and review Section 4-2 as
necessary before going on.
4-3: Energy Storage and Energy Density
OBJECTIVE: Determine the energy stored in a capacitor or combination of capacitors,
and compute the energy stored per unit volume in a region where an
electric field exists.
Reading Assignment
Study in your textbook Chapter 24, Section 3.
Commentary
The concept of electrical energy “stored” in empty space may seem a bit strange to you
at first, but it stems rather naturally from our picture of an electric field that permeates
the space surrounding any charged object. The plates in a charged capacitor exert
forces on each other due to their opposite charges and this electrostatic force between
the plates has the potential to do work on either the plates or the charges themselves.
The amount of work that can be done by a capacitor (we will explore later how electrical
work is measured in circuits) is proportional to the charge on the capacitor times the
potential difference across it. Each electron (quantum of charge) that could be released
from one plate of the capacitor would gain
of kinetic energy per volt of potential
change as it crossed the gap distance ( ) to the oppositely charged plate. The transfer
of an increment of charge changes the potential, though, so we must integrate
to find the total work that can be done by a capacitance
carrying charge
:
(2)
We choose, for the present, to think of
as potential energy stored in the electric field
as opposed to potential energy due to the charge separation on the capacitor (the two
views are equivalent in our model).
The energy per unit volume stored in a (vector) electric field can be characterized by a
(scalar) field- the energy density - that is proportional to
. For fields in vacuum,
energy density is given by
.
As we will see in this next section, this result can be further generalized to electric fields
in other media by adding a correction to the permittivity constant . Still, it depends on
a detailed knowledge of
, which is generally available only for capacitors of simple
geometry (e.g., parallel-plate or radial shells). If we are interested merely in the total
energy stored in a capacitor with charge and potential difference , we may use
any of the relationships in Eq. (2).
Study Sample Problems 24.7, 24.8, and 24.9 in the text and the following additional
example.
EXAMPLE 4-7: Two capacitors,
and
initially by being connected to a
, are each charged
battery.
Then the two
capacitors are connected together. What is the total electric energy
stored in the capacitors, if they are connected such that
(a) plates of like charge are connected together?
(b) Plates of opposite charge are connected together?
(c) Account for the lost energy in part (b).
Solution: The first step here, as in previous problems, is to make sure you have a clear
picture of the situation- drawing a diagram is recommended. Since
,
we have
.
We note that the stored energy after the initial charging is
(a) When plates of like charge are connected
together, charge
is confined to the
other pair of plates. Since the like-charged
plates are already at the same potential ,
no charge is redistributed so that the
potential across and energy stored in the
capacitors remain the same as before they
were connected:
.
In fact, the situation would be identical if the capacitors had been
charged in parallel in the first place.
(b) When plates of opposite charge are connected, we may visualize
charge
on the upper plates in the diagram and charge
on
the
lower
plates.
The equivalent
and the potential difference is
.
Therefore, the stored energy is
capacitance
is
.
(c) The energy that “disappears” when plates of unlike potential
difference are connected goes into heating the wires and into
electromagnetic radiation as the charges redistribute themselves
through a flow of current from one plate to the other.
Practice Exercise 4-3
Write your solutions to the following problems in your notebook.
1. How much energy is required to charge a metal sphere
potential of
in diameter to a
?
2. Refer to Example 4-5 in this syllabus. How much energy is stored in the capacitors
if
volts?
3. A
to
capacitor ( ) is charged to
while an identical capacitor (
) is charged
. The oppositely charged plates are then connected together. Calculate the
stored energy before and after the connection is made and the energy dissipated by
the system.
Check your answer with the Module 2 Answer Key and review Section 4-3 as necessary
before going on to the next section.
4-4: Dielectrics
OBJECTIVE: Describe the effect on a capacitor’s capacitance, voltage, charge, and
stored energy, as well as the electric field in the capacitor, if the space
between the conductors of the capacitor contains dielectric material;
describe qualitatively the distribution of polarization charges that
accounts for these effects.
Reading Assignment
Study in your textbook Chapter 24, Sections 4, 5, and 6.
Commentary
Any insulating material can be characterized by a dimensionless number- its dielectric
constant, - which is a measure of the extent to which negative and positive charges in
the material will separate in the presence of an electric field. Charge separation in a
non-conducting material (dielectric) does not involve the movement of “free” charges,
but rather the alignment of molecular dipoles within the material to produce a surfacebound polarization charge (Figure 24.18 in the text). The dielectric constant of a
material is sometimes referred to as its relative permittivity; when we multiply
by
the permittivity of a vacuum, we obtain the absolute permittivity of the material
,
.
Dielectric constants of selected materials are listed in the left column of Table 24.2 in
the text.
The dielectric constant enables us to generalize Gauss’s law so that it applies to charge
distributions and resultant electric fields in media other than a vacuum. As you study
the derivation of this generalized form in Section 24.6 of the text, make sure that you
understand the significance of , the effective charge that appears on the surface of the
dielectric. The essential thing to recognize is that all matter consists of charged
particles, and that in the presence of an electric field these charges will tend to
redistribute themselves (to a greater or lesser extent, depending on the molecular
structure of the material) such that the field strength is reduced. Thus, the field strength
resulting from a particular distribution of free charges is greatest in vacuum and can
only be less in any material medium; the dielectric constant of the medium tells how
much less.
When you see a problem involving the insertion of a dielectric slab between the plates
of a parallel-plate capacitor, be sure to notice whether or not the capacitor remains
connected to a battery. If a battery is connected, you can assume that it will maintain a
constant potential difference between the capacitor plates so that the free charge on the
plates must increase as the dielectric is inserted. If the capacitor has been
disconnected from the charging battery, the charge on the plates remains fixed and the
potential between them must decrease as the dielectric is inserted.
Study Sample Problems 24.10, 24.11, and 24.12 in the text and the following example.
EXAMPLE 4-8: A parallel-plate capacitor is half filled with an
insulating material of dielectric constant , as
suggested by the diagram at right; a vacuum
exists in the remaining space. What is the capacitance in terms of ,
, and ?
Solution: We can solve this problem by treating the capacitor as two capacitors in
parallel, each having an area of
, with one capacitor filled with
dielectric and the other in vacuum. The capacitance of the dielectricfilled plates is then
and that of the plates in vacuum is
so that the total capacitance is
Notice that this reduces to the expected result in the limiting case
where
.
Practice Exercise 4-4
Write your solutions to the following problems in your notebook.
1. A Geiger counter is made of two long, concentric metal cylinders with a gas of
dielectric constant between them. Neglecting edge effects, use Gauss’s law to
calculate the capacitance of the configuration. The center rod has radius
, the
surrounding tube radius , and the length of the cylinders is .
2. A parallel-plate capacitor of plate area
and separation
is charged by a battery to
potential , and is then disconnected from the battery.
(a) Give expressions for the energy stored in, and charge on, the capacitor.
(b) A slab of dielectric with constant
is then inserted into the capacitor, completely
filling the space between the plates. Determine the capacitance, charge,
potential difference, energy stored, and electric field in the capacitor. Explain
quantitatively what happens to the “lost” energy, and describe the distribution of
polarization charge in the dielectric.
Check your answers with the Module 2 Answer Key and review Section 4-4 as
necessary before doing the self-check test. To see if you have achieved the objectives
for Lesson 4, try to solve the problems in Self-Check Test 4 without using any reference
materials.
Self-Check Test 4
Write your solutions to the following problems in your notebook.
1. Derive the expression for the capacitance of a capacitor that consists of concentric
cylindrical conducting shells of radius and (
) and of length .
2. A
capacitor is charged to
and an
capacitor is charged to
.
They are then connected in parallel (positive plate of one to the negative plate of the
other) with an initially uncharged
capacitor. Determine the energy stored in
the final configuration.
3. A parallel-plate capacitor with plate separation
(a) If a dielectric slab of thickness
has capacitance .
and dielectric constant
is inserted
between the plates and parallel to them, determine the ratio of the capacitance
with the dielectric in place to the capacitance without the dielectric.
(b) Make a sketch showing the location of free and polarization charges (with their
signs) when the capacitor is charged with the dielectric in place.
Check your answers with the Module 2 Answer Key and review Lesson 4 as necessary.
Assignment 4
When you have demonstrated mastery of the content of this lesson, log into the
Mastering Physics website and work Assignment 4.