Transcript Calculate the electric potential

Chapter 23
Electric Potential
Copyright © 2009 Pearson Education, Inc.
23-1 Electrostatic Potential Energy
and Potential Difference
The electrostatic force is
conservative – potential
energy can be defined.
Change in electric potential
energy is negative of work
done by electric force:
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23-1 Electrostatic Potential Energy
and Potential Difference
Electric potential is defined as potential
energy per unit charge:
Unit of electric potential: the volt (V):
1 V = 1 J/C.
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23-1 Electrostatic Potential Energy
and Potential Difference
Only changes in potential can be measured,
allowing free assignment of V = 0:
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23-1 Electrostatic Potential Energy
and Potential Difference
Conceptual Example 23-1: A negative charge.
Suppose a negative charge, such as an electron, is
placed near the negative plate at point b, as shown
here. If the electron is free to move, will its electric
potential energy increase or decrease? How will
the electric potential change?
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23-1 Electrostatic Potential Energy
and Potential Difference
Analogy between gravitational and electrical
potential energy:
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23-1 Electrostatic Potential Energy
and Potential Difference
Electrical sources
such as batteries and
generators supply a
constant potential
difference. Here are
some typical potential
differences, both
natural and
manufactured:
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23-1 Electrostatic Potential Energy
and Potential Difference
Example 23-2: Electron in CRT.
Suppose an electron in a
cathode ray tube is
accelerated from rest
through a potential
difference Vb – Va = Vba =
+5000 V. (a) What is the
change in electric potential
energy of the electron? (b)
What is the speed of the
electron (m = 9.1 × 10-31 kg)
as a result of this acceleration?
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23-2 Relation between Electric
Potential and Electric Field
The general relationship
between a conservative force
and potential energy:
Substituting the
potential difference
and the electric field:
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23-2 Relation between Electric
Potential and Electric Field
The simplest case is a uniform field:
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23-2 Relation between Electric
Potential and Electric Field
Example 23-3: Electric field
obtained from voltage.
Two parallel plates are
charged to produce a potential
difference of 50 V. If the
separation between the plates
is 0.050 m, calculate the
magnitude of the electric field
in the space between the
plates.
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23-2 Relation between Electric
Potential and Electric Field
Example 23-4:
Charged conducting
sphere.
Determine the
potential at a distance
r from the center of a
uniformly charged
conducting sphere of
radius r0 for (a) r > r0,
(b) r = r0, (c) r < r0. The
total charge on the
sphere is Q.
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23-2 Relation between Electric
Potential and Electric Field
The previous example
gives the electric
potential as a function of
distance from the surface
of a charged conducting
sphere, which is plotted
here, and compared with
the electric field:
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23-2 Relation between Electric
Potential and Electric Field
Example 23-5: Breakdown voltage.
In many kinds of equipment, very high voltages are used.
A problem with high voltage is that the air can become
ionized due to the high electric fields: free electrons in
the air (produced by cosmic rays, for example) can be
accelerated by such high fields to speeds sufficient to
ionize O2 and N2 molecules by collision, knocking out one
or more of their electrons. The air then becomes
conducting and the high voltage cannot be maintained as
charge flows. The breakdown of air occurs for electric
fields of about 3.0 × 106 V/m. (a) Show that the
breakdown voltage for a spherical conductor in air is
proportional to the radius of the sphere, and (b) estimate
the breakdown voltage in air for a sphere of diameter 1.0
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23-3 Electric Potential Due to Point
Charges
To find the electric potential due to a point
charge, we integrate the field along a field line:
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23-3 Electric Potential Due to Point
Charges
Setting the potential to zero at r = ∞ gives the
general form of the potential due to a point
charge:
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23-3 Electric Potential Due to Point
Charges
Example 23-6: Work required to bring two
positive charges close together.
What minimum work must be done by an
external force to bring a charge q = 3.00 μC
from a great distance away (take r = ∞) to a
point 0.500 m from a charge Q = 20.0 µC?
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23-3 Electric Potential Due to Point
Charges
Example 23-7: Potential above two charges.
Calculate the electric potential (a) at point A in
the figure due to the two charges shown, and
(b) at point B.
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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
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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.
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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.
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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.
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23-5 Equipotential Surfaces
Example 23-10: Point
charge equipotential
surfaces.
For a single point charge
with Q = 4.0 × 10-9 C,
sketch the equipotential
surfaces (or lines in a
plane containing the
charge) corresponding
to V1 = 10 V, V2 = 20 V,
and V3 = 30 V.
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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).
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23-5 Equipotential Surfaces
A gravitational analogy to equipotential surfaces
is the topographical map – the lines connect
points of equal gravitational potential (altitude).
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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:
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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:
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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.
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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:
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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.
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23-8 Electrostatic Potential Energy;
the Electron Volt
Example 23-12: Disassembling a hydrogen
atom.
Calculate the work needed to “disassemble” a
hydrogen atom. Assume that the proton and
electron are initially separated by a distance
equal to the “average” radius of the hydrogen
atom in its ground state, 0.529 × 10-10 m, and
that they end up an infinite distance apart from
each other.
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23-9 Cathode Ray Tube: TV and
Computer Monitors, Oscilloscope
A cathode ray tube
contains a wire cathode
that, when heated, emits
electrons. A voltage
source causes the
electrons to travel to the
anode.
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23-9 Cathode Ray Tube: TV and
Computer Monitors, Oscilloscope
The electrons can be steered using electric or
magnetic fields.
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23-9 Cathode Ray Tube: TV and
Computer Monitors, Oscilloscope
Televisions and computer monitors (except for
LCD and plasma models) have a large
cathode ray tube
as their display.
Variations in the
field steer the
electrons on their
way to the screen.
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23-9 Cathode Ray Tube: TV and
Computer Monitors, Oscilloscope
An oscilloscope displays an electrical signal on
a screen, using it to deflect the beam vertically
while it sweeps horizontally.
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Review Questions
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ConcepTest 23.1a Electric Potential Energy I
A) proton
A proton and an electron are in
a constant electric field created
by oppositely charged plates.
You release the proton from
the positive side and the
electron from the negative side.
Which feels the larger electric
force?
B) electron
C) both feel the same force
D) neither – there is no force
E) they feel the same magnitude
force but opposite direction
Electron
electron
-
+
Proton
proton

E
ConcepTest 23.1a Electric Potential Energy I
A proton and an electron are in
a constant electric field created
by oppositely charged plates.
You release the proton from
the positive side and the
electron from the negative side.
Which feels the larger electric
force?
A) proton
B) electron
C) both feel the same force
D) neither – there is no force
E) they feel the same magnitude
force but opposite direction
Since F = qE and the proton and electron
have the same charge in magnitude, they
both experience the same force. However,
Electron
electron
-
the forces point in opposite directions
because the proton and electron are
oppositely charged.
+
Proton
proton

E
ConcepTest 23.1b Electric Potential Energy II
A proton and an electron are in
a constant electric field created
by oppositely charged plates.
You release the proton from
the positive side and the
electron from the negative side.
Which has the larger
acceleration?
A) proton
B) electron
C) both feel the same acceleration
D) neither – there is no acceleration
E) they feel the same magnitude
acceleration but opposite
direction
Electron
electron
-
+
Proton
proton

E
ConcepTest 23.1b Electric Potential Energy II
A proton and an electron are in
a constant electric field created
by oppositely charged plates.
You release the proton from
the positive side and the
electron from the negative side.
Which has the larger
acceleration?
A) proton
B) electron
C) both feel the same acceleration
D) neither – there is no acceleration
E) they feel the same magnitude
acceleration but opposite
direction
Since F = ma and the electron is much less
massive than the proton, the electron
experiences the larger acceleration.
Electron
electron
-
+
Proton
proton

E
ConcepTest 23.1c Electric Potential Energy III
A) proton
A proton and an electron are in
a constant electric field created
by oppositely charged plates.
You release the proton from
the positive side and the
electron from the negative side.
When it strikes the opposite
plate, which one has more KE?
B) electron
C) both acquire the same KE
D) neither – there is no change of
KE
E) they both acquire the same KE
but with opposite signs
Electron
electron
-
+
Proton
proton

E
ConcepTest 23.1c Electric Potential Energy III
A proton and an electron are in
a constant electric field created
by oppositely charged plates.
You release the proton from
the positive side and the
electron from the negative side.
When it strikes the opposite
plate, which one has more KE?
A) proton
B) electron
C) both acquire the same KE
D) neither – there is no change of
KE
E) they both acquire the same KE
but with opposite signs
Since PE = qV and the proton and electron
have the same charge in magnitude, they
both have the same electric potential energy
Electron
electron
-
initially. Because energy is conserved, they
both must have the same kinetic energy after
they reach the opposite plate.
+
Proton
proton

E
ConcepTest 23.2 Work and Potential Energy
Which group of charges took more work to bring together
from a very large initial distance apart?
+2
d
+1
+1
d
+1
Both took the same amount of work.
d
d
+1
ConcepTest 23.2 Work and Potential Energy
Which group of charges took more work to bring together
from a very large initial distance apart?
d
+2
+1
d
+1
+1
Both took the same amount of work.
The work needed to assemble
a collection of charges is the
same as the total PE of those
charges:
Q1Q 2
PE  k
r
added over
all pairs
For case 1:
d
d
+1
only 1 pair
(2)(1)
2
PE  k
k
d
d
For case 2:
there are 3
(1)(1)
1
PE  3k
 3k
d
d
pairs
ConcepTest 23.3a Electric Potential I
A) V > 0
What is the electric
potential at point A?
B) V = 0
C) V < 0
A
B
ConcepTest 23.3a Electric Potential I
A) V > 0
What is the electric
potential at point A?
B) V = 0
C) V < 0
Since Q2 (which is positive) is closer
to point A than Q1 (which is negative)
and since the total potential is equal
to V1 + V2, the total potential is
positive.
A
B
ConcepTest 23.3b Electric Potential II
A) V > 0
What is the electric
potential at point B?
B) V = 0
C) V < 0
A
B
ConcepTest 23.3b Electric Potential II
A) V > 0
What is the electric
potential at point B?
B) V = 0
C) V < 0
Since Q2 and Q1 are equidistant
from point B, and since they have
equal and opposite charges, the
total potential is zero.
Follow-up: What is the potential
at the origin of the x y axes?
A
B
ConcepTest 23.4 Hollywood Square
Four point charges are
arranged at the corners of a
square. Find the electric
field E and the potential V at
the center of the square.
A) E = 0
V=0
B) E = 0
V0
C) E  0
V0
D) E  0
V=0
E) E = V regardless of the value
-Q
+Q
-Q
+Q
ConcepTest 23.4 Hollywood Square
Four point charges are
arranged at the corners of a
square. Find the electric
field E and the potential V at
the center of the square.
A) E = 0
V=0
B) E = 0
V0
C) E  0
V0
D) E  0
V=0
E) E = V regardless of the value
The potential is zero: the scalar
contributions from the two positive
charges cancel the two minus charges.
However, the contributions from the
electric field add up as vectors, and
they do not cancel (so it is non-zero).
Follow-up: What is the direction
of the electric field at the center?
-Q
+Q
-Q
+Q
ConcepTest 23.5a Equipotential Surfaces I
1
E) all of them
At which point
does V = 0?
2
+Q
3
4
–Q
ConcepTest 23.5a Equipotential Surfaces I
1
E) all of them
At which point
does V = 0?
2
+Q
3
–Q
4
All of the points are equidistant from both charges. Since
the charges are equal and opposite, their contributions to
the potential cancel out everywhere along the mid-plane
between the charges.
Follow-up: What is the direction of the electric field at all 4 points?
ConcepTest 23.5b Equipotential Surfaces II
Which of these configurations gives V = 0 at all points on the x axis?
+2mC
+1mC
+2mC
+1mC
x
-1mC
-2mC
+2mC
-2mC
x
-2mC
-1mC
A)
B)
D) all of the above
x
+1mC
-1mC
C)
E) none of the above
ConcepTest 23.5b Equipotential Surfaces II
Which of these configurations gives V = 0 at all points on the x axis?
+2mC
+1mC
+2mC
+1mC
x
-1mC
-2mC
+2mC
-2mC
x
-2mC
-1mC
A)
x
B)
D) all of the above
+1mC
-1mC
C)
E) none of the above
Only in case (A), 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.
ConcepTest 23.5c Equipotential Surfaces III
Which of these configurations gives V = 0 at all points on the y axis?
+2mC
+1mC
+2mC
+1mC
x
-1mC
-2mC
+2mC
-2mC
x
-2mC
-1mC
A)
B)
D) all of the above
x
+1mC
-1mC
C)
E) none of the above
ConcepTest 23.5c Equipotential Surfaces III
Which of these configurations gives V = 0 at all points on the y axis?
+2mC
+1mC
+2mC
+1mC
x
-1mC
-2mC
+2mC
-2mC
x
-2mC
-1mC
A)
x
B)
D) all of the above
+1mC
-1mC
C)
E) none of the above
Only in case (C), where opposite charges lie
directly across the y axis from each other, do
the potentials from the two charges above the
y axis cancel the ones below the y axis.
Follow-up: Where is V = 0 for configuration #2?
ConcepTest 23.6 Equipotential of Point Charge
A) A and C
Which two points have
the same potential?
B) B and E
C) B and D
D) C and E
E) no pair
A
C
B
E
Q
D
ConcepTest 23.6 Equipotential of Point Charge
A) A and C
Which two points have
the same potential?
B) B and E
C) B and D
D) C and E
E) no pair
Since the potential of a point charge is:
A
Q
V k
r
only points that are at the same distance
from charge Q are at the same potential.
This is true for points C and E.
C
B
They lie on an equipotential surface.
Follow-up: Which point has the smallest potential?
E
Q
D
ConcepTest 23.7a Work and Electric Potential I
A) P  1
Which requires the most work,
to move a positive charge from
P to points 1, 2, 3 or 4 ? All
points are the same distance
from P.
B) P  2
C) P  3
D) P  4
E) all require the same
amount of work
3
2
1
P

E
4
ConcepTest 23.7a Work and Electric Potential I
Which requires the most work,
to move a positive charge from
P to points 1, 2, 3 or 4 ? All
points are the same distance
from P.
For path #1, you have to push the
positive charge against the E field,
which is hard to do. By contrast,
path #4 is the easiest, since the
field does all the work.
A) P  1
B) P  2
C) P  3
D) P  4
E) all require the same
amount of work
3
2
1
P

E
4
ConcepTest 23.7b Work and Electric Potential II
A) P  1
Which requires zero work, to
move a positive charge from
P to points 1, 2, 3 or 4 ? All
points are the same distance
from P.
B) P  2
C) P  3
D) P  4
E) all require the same
amount of work
3
2
1
P

E
4
ConcepTest 23.7b Work and Electric Potential II
Which requires zero work, to
move a positive charge from
P to points 1, 2, 3 or 4 ? All
points are the same distance
from P.
A) P  1
B) P  2
C) P  3
D) P  4
E) all require the same
amount of work
For path #3, you are moving in a
direction perpendicular to the field
lines. This means you are moving
along an equipotential, which
requires no work (by definition).
Follow-up: Which path requires the least work?
3
2
1
P

E
4
Ch. 23 Assignments
• Group problems #’s 4, 10, 12, 30, 34, 36,
42
• Homework problems #’s 1, 9, 11, 13, 15,
27, 29, 33, 49, 51, 57
• Finish reading Ch. 23 and begin looking
over Ch. 24.
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