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May Term in Guatemala
GDS 3559/STS 3500: Engineering Public Health:
An Interdisciplinary Exploration of Community
Development in Guatemala
Interested in Public
Health? Global
Health? Medicine?
Want to Learn
how engineering
affects
community
health?
Contact Kent Wayland (kaw6r@virginia.edu), Eric Anderson
(ewa3a@virginia.edu), or David R. Burt, MD, at drb5p@virginia.edu
for more information!
Copyright © 2009 Pearson Education, Inc.
Chapter 24
Capacitance, Dielectrics,
Electric Energy Storage
Copyright © 2009 Pearson Education, Inc.
Recap:
Capacitor is any pair of conductors which
a) have a potential difference between them and
b) have on them charges of equal magnitude and opposite sign.
Q  CV
C   Q V ;
1F  1C 1V ; 1 pF  1012 F  1"puff"
Simplest version/model is parallel-plate capacitor
A
C pp   0
d
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.
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:
Va 
 Vb
Va 
 Vb
 Va
Vb 
Qtot  CeqV  Q1  Q2  Q3  C1V  C2V  C3V   C1  C2  C3 V
 Ceq  C1  C2  C3
Copyright © 2009 Pearson Education, Inc.
[Parallel]
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!
Q  CV  V 
Q
C
V  V1  V2  V3
Q
Q Q Q
1
1
1
1
 


 

Ceq C1 C2 C3
Ceq C1 C2 C3
Copyright © 2009 Pearson Education, Inc.
[Series]
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.
C1  C2  C3  C
Copyright © 2009 Pearson Education, Inc.
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.
C1  C 2  C 3  C
C 23  C 2  C 3  2C
1
1
1


C123 C 23 C1
 C123
C1C 23

C1  C 23

C  2C 2
 C
C  2C 3
Copyright © 2009 Pearson Education, Inc.
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.
C 23  2C
Copyright © 2009 Pearson Education, Inc.
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.
Q
Q
Q
Q
V  0

V  Vbatt
Copyright © 2009 Pearson Education, Inc.
C 23  2C
2
C123  C
3
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.
2
Q  C123Vbatt  CVbatt  8 C  Q1
3
 Q2  Q3  4 C
  1  Vbatt 4
Q  2
V23 
  C Vbatt  
  3  3V
C 23  3
2
C


4
 V2  V3  V
3
8
 V1  Vbatt  V2  V  V1
3
Copyright © 2009 Pearson Education, Inc.
Q
Q
Q
Q
V  0

V  Vbatt
C 23  2C
2
C123  C
3
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 F
10 V
C1 = 1.0 F
C3 = 1.0 F
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 F
10 V
C1 = 1.0 F
C3 = 1.0 F
Follow-up: What is the current in this
circuit?
24-4 Electric Energy Storage
A charged capacitor stores electric energy;
the energy stored is equal to the work done
to charge the capacitor. Consider two sheets
of charge, one positive and one negative on
the surface of conductor b.
Work required to pull the negatively
charged sheet to conductor a is:
Wtot  K  0  Wext  WE  Wext  U E
Q A
Q2
 Wext  U E   FE  d    Q 
d
2 0
2C
Q 2 CV 2 QV
Q  CV  U E 


2C
2
2
Copyright © 2009 Pearson Education, Inc.
24-4 Electric Energy Storage
National Ignition Facility (NIF)
Lawrence Livermore National Laboratory
Copyright © 2009 Pearson Education, Inc.
NIF
Laser system driven by 4000 300 μF capacitors
which store a total of 422 MJ. They take 60 s to
charge and are discharged in 400 μs.
1) What is the potential difference across each
capacitor?
2) What is the power delivered during the
discharge?
Copyright © 2009 Pearson Education, Inc.
NIF
Laser system driven by 4000 300 μF capacitors
which store a total of 422 MJ. They take 60 s to
charge and are discharged in 400 μs.
1) What is the potential difference across each
capacitor?
2) What is the power delivered during the
discharge?
Solution:
1) U = CV2/2 →V = (2U/C)1/2
→V = [2(422x106)/4000/300x10-6] ½ = 26.5 kV
2) P = W/t = U/t = 422x106 /400x10-6 ~ 1012 W
= 1000 GW! cf. 1.0-1.5 GW for power plant
Copyright © 2009 Pearson Education, Inc.
24-4 Electric Energy Storage
Conceptual Example: 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. a) How has
the energy stored in this capacitor changed?
b) Would the answer change if the battery
were not disconnected?
Copyright © 2009 Pearson Education, Inc.
24-4 Electric Energy Storage
0 A
Q2
a) U E 
; C
2C
d
Q2 d

2 0A
2
Q
2d
 U E' 
 2U E
2 0A

Work done pulling plates apart
2
0A
1
V
2
b) U E  CV 
2
2 d
2
0 A UE
V
'
 UE 

2 2d
2
 Work done "pushing" charge back to battery
Copyright © 2009 Pearson Education, Inc.
24-4 Electric Energy Storage
Heart defibrillators
use electric
discharge to “jumpstart” the heart, and
can save lives.
Copyright © 2009 Pearson Education, Inc.
24-4 Electric Energy Storage
The energy density, u, defined as the energy per
unit volume, is the same no matter the origin of
the electric field:
2
U E U E Q 2C
Q
d 0
1  
E2
u



 
u
  
Vol Ad
Ad
2 0
2 A d  0 A  0 2 0   0 
2
2
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!
Copyright © 2009 Pearson Education, Inc.
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:
Copyright © 2009 Pearson Education, Inc.
24-5 Dielectrics
Dielectric strength is the
maximum field a
dielectric can experience
without breaking down.
Copyright © 2009 Pearson Education, Inc.
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.
Copyright © 2009 Pearson Education, Inc.
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.
Copyright © 2009 Pearson Education, Inc.
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.
Copyright © 2009 Pearson Education, Inc.
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.
Copyright © 2009 Pearson Education, Inc.
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:
Copyright © 2009 Pearson Education, Inc.
Summary of Chapter 24
• Capacitor: nontouching conductors carrying
equal and opposite charge.
• Capacitance:
• Capacitance of a parallel-plate capacitor:
Copyright © 2009 Pearson Education, Inc.
Summary of Chapter 24
• Capacitors in parallel:
• Capacitors in series:
Copyright © 2009 Pearson Education, Inc.
Summary of Chapter 24
• Energy density in electric field:
• A dielectric is an insulator.
• Dielectric constant gives ratio of total field to
external field.
• For a parallel-plate capacitor:
Copyright © 2009 Pearson Education, Inc.
Questions?
Copyright © 2009 Pearson Education, Inc.