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Chapter 24
Capacitance, Dielectrics,
Electric Energy Storage
Copyright © 2009 Pearson Education, Inc.
Units of Chapter 24
• Capacitors
• Determination of Capacitance
• Capacitors in Series and Parallel
• Electric Energy Storage
• Dielectrics
• Molecular Description of Dielectrics
Copyright © 2009 Pearson Education, Inc.
24-1 Capacitors
A capacitor consists of two conductors
that are close but not touching. A
capacitor has the ability to store electric
charge.
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24-1 Capacitors
Parallel-plate capacitor connected to battery. (b)
is a circuit diagram.
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24-1 Capacitors
When a capacitor is connected to a battery, the
charge on its plates is proportional to the
voltage:
The quantity C is called the capacitance.
Unit of capacitance: the farad (F):
1 F = 1 C/V.
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24-2 Determination of Capacitance
For a parallel-plate capacitor
as shown, the field between
the plates is
E = Q/ε0A.
Integrating along a path
between the plates gives the
potential difference:
Vba = Qd/ε0A.
This gives the capacitance:
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24-2 Determination of Capacitance
Example 24-1: Capacitor calculations.
(a) Calculate the capacitance of a parallel-plate
capacitor whose plates are 20 cm × 3.0 cm
and are separated by a 1.0-mm air gap. (b)
What is the charge on each plate if a 12-V
battery is connected across the two plates? (c)
What is the electric field between the plates?
(d) Estimate the area of the plates needed to
achieve a capacitance of 1 F, given the same
air gap d.
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24-2 Determination of Capacitance
Capacitors are now made with capacitances
of 1 farad or more, but they are not parallelplate capacitors. Instead, they are activated
carbon, which acts as a capacitor on a very
small scale. The capacitance of 0.1 g of
activated carbon is about 1 farad.
Some computer keyboards
use capacitors;
depressing the
key changes the
capacitance, which
is detected in a
circuit.
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24-2 Determination of Capacitance
Example 24-2: Cylindrical
capacitor.
A cylindrical capacitor consists of a
cylinder (or wire) of radius Rb
surrounded by a coaxial cylindrical
shell of inner radius Ra. Both
cylinders have length l which we
assume is much greater than the
separation of the cylinders, so we can
neglect end effects. The capacitor is
charged (by connecting it to a battery)
so that one cylinder has a charge +Q
(say, the inner one) and the other one
a charge –Q. Determine a formula for
the capacitance.
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24-2 Determination of Capacitance
Example 24-3: Spherical capacitor.
A spherical capacitor
consists of two thin
concentric spherical
conducting shells of radius
ra and rb as shown. The
inner shell carries a
uniformly distributed
charge Q on its surface,
and the outer shell an
equal but opposite charge –Q.
Determine the capacitance of the
two shells.
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24-2 Determination of Capacitance
Example 24-4: Capacitance of two long parallel
wires.
Estimate the capacitance per unit length of two
very long straight parallel wires, each of radius
R, carrying uniform charges +Q and –Q, and
separated by a distance d which is large
compared to R (d >> R).
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24-3 Capacitors in Series and Parallel
Capacitors in parallel
have the same voltage
across each one. The
equivalent capacitor is
one that stores the
same charge when
connected to the same
battery:
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24-3 Capacitors in Series and Parallel
Capacitors in series have the same charge. In
this case, the equivalent capacitor has the
same charge across the total voltage drop.
Note that the formula is for the inverse of the
capacitance and not the capacitance itself!
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24-3 Capacitors in Series and Parallel
Example 24-5: Equivalent capacitance.
Determine the capacitance of a single
capacitor that will have the same effect as
the combination shown.
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24-3 Capacitors in Series and Parallel
Example 24-6: Charge and voltage on
capacitors.
Determine the charge on each capacitor and
the voltage across each, assuming C = 3.0 μF
and the battery voltage is V = 4.0 V.
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24-3 Capacitors in Series and Parallel
Example 24-7: Capacitors
reconnected.
Two capacitors, C1 = 2.2 μF and
C2 = 1.2 μF, are connected in
parallel to a 24-V source as
shown. After they are charged
they are disconnected from the
source and from each other and
then reconnected directly to
each other, with plates of
opposite sign connected
together. Find the charge on
each capacitor and the potential
across each after equilibrium is
established.
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24-4 Electric Energy Storage
A charged capacitor stores electric energy;
the energy stored is equal to the work done
to charge the capacitor:
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24-4 Electric Energy Storage
Example 24-8: Energy stored in a capacitor.
A camera flash unit stores energy in a 150-μF
capacitor at 200 V. (a) How much electric energy
can be stored? (b) What is the power output if
nearly all this energy is released in 1.0 ms?
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24-4 Electric Energy Storage
Conceptual Example 24-9: Capacitor plate
separation increased.
A parallel-plate capacitor carries charge Q
and is then disconnected from a battery. The
two plates are initially separated by a
distance d. Suppose the plates are pulled
apart until the separation is 2d. How has the
energy stored in this capacitor changed?
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24-4 Electric Energy Storage
Example 24-10: Moving parallel capacitor
plates.
The plates of a parallel-plate capacitor have
area A, separation x, and are connected to a
battery with voltage V. While connected to
the battery, the plates are pulled apart until
they are separated by 3x. (a) What are the
initial and final energies stored in the
capacitor? (b) How much work is required to
pull the plates apart (assume constant
speed)? (c) How much energy is exchanged
with the battery?
Copyright © 2009 Pearson Education, Inc.
24-4 Electric Energy Storage
The energy density, defined as the energy per
unit volume, is the same no matter the origin of
the electric field:
The sudden discharge of electric energy can be
harmful or fatal. Capacitors can retain their
charge indefinitely even when disconnected
from a voltage source – be careful!
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24-4 Electric Energy Storage
Heart defibrillators
use electric
discharge to “jumpstart” the heart, and
can save lives.
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24-5 Dielectrics
A dielectric is an insulator, and is
characterized by a dielectric constant K.
Capacitance of a parallel-plate capacitor filled
with dielectric:
Using the dielectric constant, we define the
permittivity:
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24-5 Dielectrics
Dielectric strength is the
maximum field a
dielectric can experience
without breaking down.
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24-5 Dielectrics
Here are two experiments where we insert and
remove a dielectric from a capacitor. In the
first, the capacitor is connected to a battery,
so the voltage remains constant. The
capacitance increases, and therefore the
charge on the plates increases as well.
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24-5 Dielectrics
In this second experiment, we charge a
capacitor, disconnect it, and then insert the
dielectric. In this case, the charge remains
constant. Since the dielectric increases the
capacitance, the potential across the
capacitor drops.
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24-5 Dielectrics
Example 24-11: Dielectric removal.
A parallel-plate capacitor, filled with a
dielectric with K = 3.4, is connected to
a 100-V battery. After the capacitor is
fully charged, the battery is
disconnected. The plates have area A
= 4.0 m2 and are separated by d = 4.0
mm. (a) Find the capacitance, the
charge on the capacitor, the electric
field strength, and the energy stored
in the capacitor. (b) The dielectric is
carefully removed, without changing
the plate separation nor does any
charge leave the capacitor. Find the
new values of capacitance, electric
field strength, voltage between the plates,
and the energy stored in the capacitor.
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24-6 Molecular Description of
Dielectrics
The molecules in a dielectric, when in an
external electric field, tend to become oriented
in a way that reduces the external field.
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24-6 Molecular Description of
Dielectrics
This means that the electric field within the
dielectric is less than it would be in air, allowing
more charge to be stored for the same potential.
This reorientation of the molecules results in an
induced charge – there is no net charge on the
dielectric, but the charge is asymmetrically
distributed.
The magnitude of the induced charge depends on
the dielectric constant:
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Review Questions
Copyright © 2009 Pearson Education, Inc.
ConcepTest 24.1 Capacitors
Capacitor C1 is connected across
1) C1
a battery of 5 V. An identical
2) C2
capacitor C2 is connected across
a battery of 10 V. Which one has
more charge?
3) both have the same charge
4) it depends on other factors
ConcepTest 24.1 Capacitors
Capacitor C1 is connected across
1) C1
a battery of 5 V. An identical
2) C2
capacitor C2 is connected across
a battery of 10 V. Which one has
more charge?
3) both have the same charge
4) it depends on other factors
Since Q = CV and the two capacitors are
identical, the one that is connected to the
greater voltage has more charge, which
is C2 in this case.
ConcepTest 24.2a Varying Capacitance I
What must be done to
1) increase the area of the plates
a capacitor in order to
2) decrease separation between the plates
increase the amount of
3) decrease the area of the plates
charge it can hold (for
a constant voltage)?
4) either (1) or (2)
5) either (2) or (3)
+Q – Q
ConcepTest 24.2a Varying Capacitance I
What must be done to
1) increase the area of the plates
a capacitor in order to
2) decrease separation between the plates
increase the amount of
3) decrease the area of the plates
charge it can hold (for
a constant voltage)?
4) either (1) or (2)
5) either (2) or (3)
+Q – Q
Since Q = CV, in order to increase the charge
that a capacitor can hold at constant voltage,
one has to increase its capacitance. Since the
capacitance is given by C   0 A , that can be
d
done by either increasing A or decreasing d.
ConcepTest 24.2b Varying Capacitance II
A parallel-plate capacitor
1) the voltage decreases
initially has a voltage of 400 V
2) the voltage increases
and stays connected to the
3) the charge decreases
battery. If the plate spacing is
now doubled, what happens?
4) the charge increases
5) both voltage and charge change
+Q – Q
ConcepTest 24.2b Varying Capacitance II
A parallel-plate capacitor
1) the voltage decreases
initially has a voltage of 400 V
2) the voltage increases
and stays connected to the
3) the charge decreases
battery. If the plate spacing is
now doubled, what happens?
4) the charge increases
5) both voltage and charge change
Since the battery stays connected, the
voltage must remain constant! Since
C   0 A , when the spacing d is
d
doubled, the capacitance C is halved.
And since Q = CV, that means the
charge must decrease.
Follow-up: How do you increase the charge?
+Q – Q
ConcepTest 24.2c Varying Capacitance III
A parallel-plate capacitor initially has
a potential difference of 400 V and is
then disconnected from the charging
battery. If the plate spacing is now
doubled (without changing Q), what
is the new value of the voltage?
+Q – Q
1) 100 V
2) 200 V
3) 400 V
4) 800 V
5) 1600 V
ConcepTest 24.2c Varying Capacitance III
A parallel-plate capacitor initially has
a potential difference of 400 V and is
then disconnected from the charging
battery. If the plate spacing is now
doubled (without changing Q), what
is the new value of the voltage?
Once the battery is disconnected, Q has to
remain constant, since no charge can flow
either to or from the battery. Since
C   0 A , when the spacing d is doubled, the
d
capacitance C is halved. And since Q = CV,
that means the voltage must double.
1) 100 V
2) 200 V
3) 400 V
4) 800 V
5) 1600 V
+Q – Q
ConcepTest 24.3a
Capacitors I
1) Ceq = 3/2C
What is the equivalent capacitance,
2) Ceq = 2/3C
Ceq , of the combination below?
3) Ceq = 3C
4) Ceq = 1/3C
5) Ceq = 1/2C
o
Ceq
o
C
C
C
ConcepTest 24.3a
Capacitors I
1) Ceq = 3/2C
What is the equivalent capacitance,
2) Ceq = 2/3C
Ceq , of the combination below?
3) Ceq = 3C
4) Ceq = 1/3C
5) Ceq = 1/2C
The 2 equal capacitors in series add
o
up as inverses, giving 1/2C. These
are parallel to the first one, which
Ceq
add up directly. Thus, the total
equivalent capacitance is 3/2C.
o
C
C
C
ConcepTest 24.3b
Capacitors II
How does the voltage V1 across
1) V1 = V2
the first capacitor (C1) compare
2) V1 > V2
to the voltage V2 across the
3) V1 < V2
second capacitor (C2)?
4) all voltages are zero
C2 = 1.0 mF
10 V
C1 = 1.0 mF
C3 = 1.0 m
ConcepTest 24.3b
Capacitors II
How does the voltage V1 across
1) V1 = V2
the first capacitor (C1) compare
2) V1 > V2
to the voltage V2 across the
3) V1 < V2
second capacitor (C2)?
4) all voltages are zero
The voltage across C1 is 10 V.
The combined capacitors C2 +
C3 are parallel to C1. The
voltage across C2 + C3 is also
10 V. Since C2 and C3 are in
series, their voltages add.
Thus the voltage across C2
and C3 each has to be 5 V,
which is less than V1.
C2 = 1.0 mF
10 V
C1 = 1.0 mF
C3 = 1.0 m
Follow-up: What is the current in this
circuit?
ConcepTest 24.3c
How does the charge Q1 on the first
Capacitors III
1) Q1 = Q2
2) Q1 > Q2
capacitor (C1) compare to the
charge Q2 on the second capacitor
3) Q1 < Q2
(C2)?
4) all charges are zero
C2 = 1.0 mF
10 V
C1 = 1.0 mF
C3 = 1.0 mF
ConcepTest 24.3c
How does the charge Q1 on the first
Capacitors III
1) Q1 = Q2
2) Q1 > Q2
capacitor (C1) compare to the
charge Q2 on the second capacitor
3) Q1 < Q2
(C2)?
4) all charges are zero
We already know that the
C2 = 1.0 mF
voltage across C1 is 10 V
and the voltage across both
C2 and C3 is 5 V each. Since
Q = CV and C is the same for
all the capacitors, we have
V1 > V2 and therefore Q1 > Q2.
10 V
C1 = 1.0 mF
C3 = 1.0 mF
Ch. 24 Homework
• Read Ch. 24 and begin looking over Ch.
25.
• Group problems #’s 2, 12, 22
• HW problems #’s 3, 5, 7, 11, 13, 17, 23,
25, 29, 35, 41, 47, 59
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