Transcript Lecture_4
Chapter 23
Electric Potential
Copyright © 2009 Pearson Education, Inc.
23-4 Potential Due to Any Charge
Distribution
The potential due to an arbitrary charge
distribution can be expressed as a sum or
integral (if the distribution is continuous):
or
Copyright © 2009 Pearson Education, Inc.
23-4 Potential Due to Any Charge
Distribution
Example 23-8:
Potential due to a
ring of charge.
A thin circular ring of
radius R has a
uniformly distributed
charge Q. Determine
the electric potential
at a point P on the
axis of the ring a
distance x from its
center.
Copyright © 2009 Pearson Education, Inc.
23-4 Potential Due to Any Charge
Distribution
Example 23-9:
Potential due to a
charged disk.
A thin flat disk, of
radius R0, has a
uniformly
distributed charge
Q. Determine the
potential at a point
P on the axis of the
disk, a distance x
from its center.
Copyright © 2009 Pearson Education, Inc.
ConcepTest 23.5 Equipotential Surfaces
Which of these configurations gives V = 0 at
+2mC
+2mC
+2mC
+1mC
-2mC
+1mC
all
points
on the
x axis?
x
-1mC
-2mC
x
1)
+1mC
-1mC
2)
4) all of the above
Copyright © 2009 Pearson Education, Inc.
-2mC
-1mC
x
3)
5) none of the above
ConcepTest 23.5 Equipotential Surfaces
Which of these configurations gives V = 0 at
+2mC
+2mC
+2mC
+1mC
-2mC
+1mC
all
points
on the
x axis?
x
-1mC
-2mC
x
-2mC
-1mC
1)
x
2)
4) all of the above
+1mC
-1mC
3)
5) none of the above
Only in case (1), where opposite charges lie
directly across the x axis from each other, do
the potentials from the two charges above the
x axis cancel the ones below the x axis.
Copyright © 2009 Pearson Education, Inc.
23-5 Equipotential Surfaces
An equipotential is a line
or surface over which the
potential is constant.
Electric field lines are
perpendicular to
equipotentials.
The surface of a conductor
is an equipotential
(E// =0 → ∂V/∂s// = 0)
Copyright © 2009 Pearson Education, Inc.
23-5 Equipotential Surfaces
Equipotential surfaces are always
perpendicular to field lines; they are
always closed surfaces (unlike field lines,
which begin and end on charges).
Copyright © 2009 Pearson Education, Inc.
23-5 Equipotential Surfaces
A gravitational analogy to equipotential surfaces
is the topographical map – the lines connect
points of equal gravitational potential (altitude).
Copyright © 2009 Pearson Education, Inc.
23-6 Electric Dipole Potential
The potential due to an
electric dipole is just the sum
of the potentials due to each
charge, and can be calculated
exactly. For distances large
compared to the charge
separation:
Copyright © 2009 Pearson Education, Inc.
23-7 E Determined from V
If we know the field, we can determine the
potential by integrating. Inverting this
process, if we know the potential, we can
find the field by differentiating:
This is a vector differential equation;
here it is in component form:
Copyright © 2009 Pearson Education, Inc.
23-7 E Determined from V
Example 23-11: E for ring and disk.
Use electric potential to determine the electric
field at point P on the axis of (a) a circular ring
of charge and (b) a uniformly charged disk.
Copyright © 2009 Pearson Education, Inc.
23-8 Electrostatic Potential Energy;
the Electron Volt
The potential energy of a charge in an
electric potential is U = qV. To find the electric
potential energy of two charges, imagine
bringing each in from infinitely far away. The
first one takes no work, as there is no field.
To bring in the second one, we must do work
due to the field of the first one; this means
the potential energy of the pair is:
Copyright © 2009 Pearson Education, Inc.
23-8 Electrostatic Potential Energy;
the Electron Volt
One electron volt (eV) is the energy gained by
an electron moving through a potential
difference of one volt:
1 eV = 1.6 × 10-19 J.
The electron volt is often a much more
convenient unit than the joule for measuring
the energy of individual particles.
Copyright © 2009 Pearson Education, Inc.
Summary of Chapter 23
• Electric potential is potential energy per
unit charge:
• Potential difference between two points:
• Potential of a point charge:
Copyright © 2009 Pearson Education, Inc.
Summary of Chapter 23
• Equipotential: line or surface along which
potential is the same.
• Electric dipole potential is proportional to 1/r2.
• To find the field from the potential:
Copyright © 2009 Pearson Education, Inc.
Chapter 24
Capacitance, Dielectrics,
Electric Energy Storage
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.
Copyright © 2009 Pearson Education, Inc.
24-1 Capacitors
Parallel-plate capacitor connected to battery. (b)
is a circuit diagram.
Copyright © 2009 Pearson Education, Inc.
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. It is
defined for ANY pair of separated conductors
at different potentials.
Unit of capacitance: the farad (F):
1 F = 1 C/V.
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.
24-2 Determination of Capacitance
For a parallel-plate capacitor
as shown, the field between
the plates is
E = σ/ε0 = Q/ε0A.
Integrating along a path
between the plates gives the
potential difference:
Vba = Ed = (Q/ε0A)d.
This gives the capacitance:
Copyright © 2009 Pearson Education, Inc.
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.
24-2 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.
Copyright © 2009 Pearson Education, Inc.
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.
Copyright © 2009 Pearson Education, Inc.
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.
Copyright © 2009 Pearson Education, Inc.
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).
Copyright © 2009 Pearson Education, Inc.
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:
Copyright © 2009 Pearson Education, Inc.
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!
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.
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.
Copyright © 2009 Pearson Education, Inc.
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.
Copyright © 2009 Pearson Education, Inc.
24-4 Electric Energy Storage
A charged capacitor stores electric energy;
the energy stored is equal to the work done
to charge the capacitor:
Copyright © 2009 Pearson Education, Inc.
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
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?
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!
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.
24-5 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.
Homework Assignment # 4
Chapter 23 – 36
Chapter 24 – 6, 16, 28, 46, 60, 82
Tentative HW # 5:
Chapter 25 – 10, 20, 34, 40, 54, 58
Copyright © 2009 Pearson Education, Inc.