Download 21-6 The Electric Field

Chapter 21
Electric Charge and
Electric Field
Copyright © 2009 Pearson Education, Inc.
ConcepTest 21.4b Electric Force II
Two balls with charges +Q and +4Q are separated by 3R. Where
should you place another charged ball Q0 on the line between
the two charges such that the net force on Q0 will be zero?
+4Q
+Q
1
2
3
4
2R
R
3R
5
ConcepTest 21.4b Electric Force II
Two balls with charges +Q and +4Q are separated by 3R. Where
should you place another charged ball Q0 on the line between
the two charges such that the net force on Q0 will be zero?
+4Q
+Q
1
2
3
4
5
2R
R
3R
The force on Q0 due to +Q is:
F = k(Q0)(Q)/R2
The force on Q0 due to +4Q is:
F = k(Q0)(4Q)/(2R)2
Since +4Q is 4 times bigger than +Q, Q0 needs to be farther
from +4Q. In fact, Q0 must be twice as far from +4Q, since
the distance is squared in Coulomb’s law.
ConcepTest 21.5c Proton and Electron III
A proton and an electron
are held apart a distance
of 1 m and then let go.
Where would they meet?
1) in the middle
2) closer to the electron’s side
3) closer to the proton’s side
p
e
ConcepTest 21.5c Proton and Electron III
A proton and an electron
are held apart a distance
of 1 m and then let go.
Where would they meet?
1) in the middle
2) closer to the electron’s side
3) closer to the proton’s side
By Newton’s 3rd law, the electron and proton
feel the same force. But, since F = ma, and
since the proton’s mass is much greater, the
proton’s acceleration will be much smaller!
p
Thus, they will meet closer to the proton’s
original position.
Follow-up: Which particle will be moving faster when they meet?
e
21-6 The Electric Field
For a point charge:
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21-6 The Electric Field
An electric field surrounds every charge.
Copyright © 2009 Pearson Education, Inc.
21-6 The Electric Field
Force on a point
charge in an
electric field:
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21-6 The Electric Field
Example 21-6: Electric field of a single point
charge.
Calculate the magnitude and direction of the
electric field at a point P which is 30 cm to the
right of a point charge Q = -3.0 x 10-6 C.
Copyright © 2009 Pearson Education, Inc.
21-6 The Electric Field
Example 21-7: E at a point between two charges.
Two point charges are separated by a distance of 10.0 cm.
One has a charge of -25 μC and the other +50 μC. (a)
Determine the direction and magnitude of the electric field
at a point P between the two charges that is 2.0 cm from
the negative charge. (b) If an electron (mass = 9.11 x 10-31
kg) is placed at rest at P and then released, what will be its
initial acceleration (direction and magnitude)?
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ConcepTest 21.9b Superposition II
-2 C
What is the electric field at
2
-2 C
1
the center of the square?
3
-2 C
1)
2)
3)
4)
5)
E
E
E
E
E
=
=
=
=
=
0
1
2
3
4
4
-2 C
ConcepTest 21.9b Superposition II
What is the electric field at
-2 C
2
-2 C
1
the center of the square?
3
5) E = 0
-2 C
The four E field vectors all point outward
from the center of the square toward their
respective charges. Because they are all
equal, the net E field is zero at the center!!
Follow-up: What if the upper two charges were +2 C?
What if the right-hand charges were +2 C?
4
-2 C
21-7 Electric Field Calculations for
Continuous Charge Distributions
A continuous distribution of charge may be
treated as a succession of infinitesimal (point)
charges. The total field is then the integral of
the infinitesimal fields due to each bit of
charge:
Remember that the electric field is a vector;
you will need a separate integral for each
component.
Copyright © 2009 Pearson Education, Inc.
21-7 Electric Field Calculations for
Continuous Charge Distributions
Example 21-9: A ring of charge.
A thin, ring-shaped object of radius a holds a
total charge +Q distributed uniformly around
it. Determine the electric field at a point P on
its axis, a distance x from the center. Let λ be
the charge per unit length (C/m).
Copyright © 2009 Pearson Education, Inc.
21-7 Electric Field Calculations for
Continuous Charge Distributions
Example 21-9: A ring of charge.
A thin, ring-shaped object of radius a holds a
total charge +Q distributed uniformly around
it. Determine the electric field at a point P on
its axis, a distance x from the center. Let λ be
the charge per unit length (C/m).
SYMMETRY!!
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21-7 Electric Field Calculations for
Continuous Charge Distributions
Conceptual Example 21-10: Charge at the
center of a ring.
Imagine a small positive charge placed at
the center of a nonconducting ring carrying
a uniformly distributed negative charge. Is
the positive charge in equilibrium if it is
displaced slightly from the center along the
axis of the ring, and if so is it stable? What
if the small charge is negative? Neglect
gravity, as it is much smaller than the
electrostatic forces.
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21-7 Electric Field Calculations for
Continuous Charge Distributions
Example 21-11: Long line of
charge.
Determine the magnitude
of the electric field at any
point P a distance x from a
very long line (a wire, say)
of uniformly distributed
charge. Assume x is much
smaller than the length of
the wire, and let λ be the
charge per unit length
(C/m).
Copyright © 2009 Pearson Education, Inc.
21-7 Electric Field Calculations for
Continuous Charge Distributions
Example 21-11: Long line of
charge.
Determine the magnitude
of the electric field at any
point P a distance x from a
very long line (a wire, say)
of uniformly distributed
charge. Assume x is much
smaller than the length of
the wire, and let λ be the
charge per unit length
(C/m).
Copyright © 2009 Pearson Education, Inc.
SYMMETRY!!
21-7 Electric Field Calculations for
Continuous Charge Distributions
Example 21-12: Uniformly charged disk.
Charge is distributed uniformly over a thin
circular disk of radius R. The charge per unit
area (C/m2) is σ. Calculate the electric field at
a point P on the axis of the disk, a distance z
above its center.
Copyright © 2009 Pearson Education, Inc.
21-7 Electric Field Calculations for
Continuous Charge Distributions
Example 21-12: Uniformly charged disk.
Charge is distributed uniformly over a thin
circular disk of radius R. The charge per unit
area (C/m2) is σ. Calculate the electric field at
a point P on the axis of the disk, a distance z
above its center.
SYMMETRY!!
Copyright © 2009 Pearson Education, Inc.
21-7 Electric Field Calculations for
Continuous Charge Distributions
In the previous example, if we are very close
to the disk (that is, if z << R), the electric field
is:
This is the field due to an infinite plane
of charge.
Copyright © 2009 Pearson Education, Inc.
21-7 Electric Field Calculations for
Continuous Charge Distributions
Example 21-13: Two parallel plates.
Determine the electric field between two large parallel plates or
sheets, which are very thin and are separated by a distance d
which is small compared to their height and width. One plate
carries a uniform surface charge density σ and the other carries a
uniform surface charge density -σ as shown (the plates extend
upward and downward beyond the part shown).
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21-8 Field Lines
The electric field can be represented by field
lines. These lines start on a positive charge
and end on a negative charge.
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21-8 Field Lines
The number of field lines starting (ending)
on a positive (negative) charge is
proportional to the magnitude of the charge.
The electric field is stronger where the field
lines are closer together.
Schematic only – fields in 3-d, not 2-d
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21-8 Field Lines
Electric dipole: two equal charges, opposite in
sign:
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ConcepTest 21.12b Electric Field Lines II
Which of the charges has
the greater magnitude?
1)
2)
3) both the same
ConcepTest 21.12b Electric Field Lines II
Which of the charges has
the greater magnitude?
1)
2)
3) both the same
The field lines are denser around
the red charge, so the red one
has the greater magnitude.
Follow-up: What is the red/green ratio
of magnitudes for the two charges?
21-8 Field Lines
The electric field between
two closely spaced,
oppositely charged parallel
plates is constant.
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21-8 Field Lines
Summary of field lines:
1. Field lines indicate the direction of the
field; the field is tangent to the line.
2. The magnitude of the field is proportional
to the density of the lines.
3. Field lines start on positive charges and
end on negative charges; the number is
proportional to the magnitude of the
charge.
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21-9 Electric Fields and Conductors
The static electric field inside a conductor is
zero – if it were not, the charges would move.
The net charge on a conductor resides on its
outer surface.
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21-9 Electric Fields and Conductors
The electric field is perpendicular to the
surface of a conductor – again, if it were not,
charges would move.
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21-9 Electric Fields and Conductors
Conceptual Example
21-14: Shielding, and
safety in a storm.
A neutral hollow
metal box is placed
between two parallel
charged plates as
shown. What is the
field like inside the
box?
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21-10 Motion of a Charged Particle in
an Electric Field
The force on an object of charge q in
an electric field E is given by:
F = qE
Therefore, if we know the mass and
charge of a particle, we can describe
its subsequent motion in an electric
field.
Copyright © 2009 Pearson Education, Inc.
1
ConcepTest 21.6 Forces in 2D
2
3
Which of the arrows best
4
represents the direction
of the net force on charge
d
+2Q
+Q
+Q due to the other two
charges?
d
+4Q
5
1
ConcepTest 21.6 Forces in 2D
2
3
Which of the arrows best
4
represents the direction
of the net force on charge
d
+2Q
+Q
+Q due to the other two
d
charges?
+4Q
The charge +2Q repels +Q toward the
right. The charge +4Q repels +Q
upward, but with a stronger force.
Therefore, the net force is up and to
+2Q
the right, but mostly up.
Follow-up: What would happen if
the yellow charge were +3Q?
+4Q
5
21-10 Motion of a Charged Particle in
an Electric Field
Example 21-15: Electron
accelerated by electric field.
An electron (mass m = 9.11 x 10-31 kg)
is accelerated in the uniform field E
(E = 2.0 x 104 N/C) between two parallel
charged plates. The separation of the
plates is 1.5 cm. The electron is
accelerated from rest near the negative
plate and passes through a tiny hole in
the positive plate. (a) With what speed
does it leave the hole? (b) Show that
the gravitational force can be ignored.
Assume the hole is so small that it
does not affect the uniform field
between the plates.
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21-10 Motion of a Charged Particle in
an Electric Field
Example 21-16: Electron moving perpendicular to E.
Suppose an electron traveling with speed v0 = 1.0 x 107 m/s
enters a uniform electric field E, which is at right angles to
v0 as shown. Describe its motion by giving the equation of
its path while in the electric field. Ignore gravity.
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21-11 Electric Dipoles
An electric dipole consists of two charges
Q, equal in magnitude and opposite in
sign, separated by a distance . The
dipole moment, p = Q , points from the
negative to the positive charge.
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21-11 Electric Dipoles
An electric dipole in a uniform electric field
will experience no net force, but it will, in
general, experience a torque:
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21-11 Electric Dipoles
The electric field created by a dipole is the sum
of the fields created by the two charges; far from
the dipole, the field shows a 1/r3 dependence:
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21-11 Electric Dipoles
Example 21-17: Dipole in a field.
The dipole moment of a water molecule is
6.1 x 10-30 C·m. A water molecule is placed
in a uniform electric field with magnitude
2.0 x 105 N/C. (a) What is the magnitude of
the maximum torque that the field can
exert on the molecule? (b) What is the
potential energy when the torque is at its
maximum? (c) In what position will the
potential energy take on its greatest
value? Why is this different than the
position where the torque is maximum?
Copyright © 2009 Pearson Education, Inc.
21-12 Electric Forces in Molecular
Biology; DNA
Molecular biology is the
study of the structure and
functioning of the living cell
at the molecular level.
The DNA molecule is a
double helix:
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21-12 Electric Forces in Molecular
Biology; DNA
The A-T and G-C
nucleotide bases
attract each other
through electrostatic
forces.
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21-12 Electric Forces in Molecular
Biology; DNA
Replication: DNA is in a “soup” of A, C, G,
and T in the cell. During random collisions, A
and T will be attracted to each other, as will
G and C; other combinations will not.
Copyright © 2009 Pearson Education, Inc.
Summary of Chapter 21
• Two kinds of electric charge – positive and
negative.
• Charge is conserved.
• Charge on electron:
e = 1.602 x 10-19 C.
• Conductors: electrons free to move.
• Insulators: nonconductors.
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Summary of Chapter 21
• Charge is quantized in units of e.
• Objects can be charged by conduction or
induction.
• Coulomb’s law:
•Electric field is force per unit charge:
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Summary of Chapter 21
• Electric field of a point charge:
• Electric field can be represented by electric
field lines.
• Static electric field inside conductor is zero;
surface field is perpendicular to surface.
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Chapter 22
Gauss’s Law
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Units of Chapter 22
• Electric Flux
• Gauss’s Law
• Applications of Gauss’s Law
• Experimental Basis of Gauss’s and
Coulomb’s Laws
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22-1 Electric Flux
Electric flux:
Electric flux through an
area is proportional to the
total number of field lines
crossing the area.
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22-1 Electric Flux
Example 22-1: Electric flux.
Calculate the electric flux through the
rectangle shown. The rectangle is 10 cm by
20 cm, the electric field is uniform at 200
N/C, and the angle θ is 30°.
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22-1 Electric Flux
Flux through a closed surface:
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Think in terms of flux lines:
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22-2 Gauss’s Law
The net number of field lines through the surface
is proportional to the charge enclosed, and also
to the flux, giving Gauss’s law:
This can be used to find the electric field in
situations with a high degree of symmetry.
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22-2 Gauss’s Law
For a point charge,
Spherical
Therefore,
Solving for E gives the result
we expect from Coulomb’s
law:
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22-2 Gauss’s Law
Using Coulomb’s law to
evaluate the integral of
the field of a point charge
over the surface of a
sphere surrounding the
charge gives:
Looking at the arbitrarily shaped surface A2, we
see that the same flux passes through it as
passes through A1. Therefore, this result should
be valid for any closed surface.
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22-2 Gauss’s Law
Finally, if a gaussian surface encloses several
point charges, the superposition principle shows
that:
Therefore, Gauss’s law is valid for any charge
distribution. Note, however, that it only refers to
the field due to charges within the gaussian
surface – charges outside the surface will also
create fields.
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22-2 Gauss’s Law
Conceptual Example 22-2: Flux from Gauss’s law.
Consider the two gaussian surfaces, A1 and A2, as
shown. The only charge present is the charge Q at
the center of surface A1. What is the net flux through
each surface, A1 and A2?
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22-3 Applications of Gauss’s Law
Example 22-3: Spherical
conductor.
A thin spherical shell of
radius r0 possesses a
total net charge Q that is
uniformly distributed on it.
Determine the electric
field at points (a) outside
the shell, and (b) within
the shell. (c) What if the
conductor were a solid
sphere?
Copyright © 2009 Pearson Education, Inc.
Spherical
22-3 Applications of Gauss’s Law
Example 22-4: Solid
sphere of charge.
An electric charge Q
is distributed uniformly
throughout a
nonconducting sphere
of radius r0.
Determine the electric
field (a) outside the
sphere
(r > r0) and (b) inside
the sphere (r < r0).
Copyright © 2009 Pearson Education, Inc.
Spherical/radial
22-3 Applications of Gauss’s Law
Example 22-5:
Nonuniformly charged
solid sphere.
Suppose the charge
density of a solid sphere is
given by ρE = αr2, where α
is a constant. (a) Find α in
terms of the total charge Q
on the sphere and its
radius r0. (b) Find the
electric field as a function
of r inside the sphere.
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22-3 Applications of Gauss’s Law
Example 22-6: Long uniform line of charge.
A very long straight wire possesses a uniform
positive charge per unit length, λ. Calculate the
electric field at points near (but outside) the
wire, far from the ends.
Linear,
Cylindrical
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22-3 Applications of Gauss’s Law
Example 22-7: Infinite
plane of charge.
Charge is distributed
uniformly, with a surface
charge density σ (σ =
charge per unit area =
dQ/dA) over a very large
but very thin nonconducting
flat plane surface.
Determine the electric field
at points near the plane.
Planar & cylindrical
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22-3 Applications of Gauss’s Law
Example 22-8: Electric
field near any
conducting surface.
Show that the electric
field just outside the
surface of any good
conductor of arbitrary
shape is given by
E = σ/ε0
where σ is the surface
charge density on the conductor’s
surface at that point.
Copyright © 2009 Pearson Education, Inc.
22-3 Applications of Gauss’s Law
The difference between the electric field outside
a conducting plane of charge and outside a
nonconducting plane of charge can be thought
of in two ways:
1. The field inside the conductor is zero, so the
flux is all through one end of the cylinder.
2. The nonconducting plane has a total charge
density σ, whereas the conducting plane has a
charge density σ on each side, effectively giving
it twice the charge density.
Copyright © 2009 Pearson Education, Inc.
22-3 Applications of Gauss’s Law
Conceptual Example
22-9: Conductor with
charge inside a cavity.
Suppose a conductor
carries a net charge
+Q and contains a
cavity, inside of
which resides a point
charge +q. What can
you say about the
charges on the inner
and outer surfaces of
the conductor?
Copyright © 2009 Pearson Education, Inc.
22-3 Applications of Gauss’s Law
Procedure for Gauss’s law problems:
1. Identify the symmetry, and choose a gaussian
surface that takes advantage of it (with surfaces
along surfaces of constant field).
2. Draw the surface.
3. Use the symmetry to find the direction of E.
4. Evaluate the flux by integrating.
5. Calculate the enclosed charge.
6. Solve for the field.
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E
22-4 Experimental Basis of Gauss’s
and Coulomb’s Laws
In the experiment shown,
Gauss’s law predicts that the
charge on the ball flows onto
the surface of the cylinder when
they are in contact. This can be
tested by measuring the charge
on the ball after it is removed –
it should be zero.
Copyright © 2009 Pearson Education, Inc.
Summary of Chapter 22
• Electric flux:
• Gauss’s law:
• Gauss’s law can be used to calculate the field in
situations with a high degree of symmetry.
• Gauss’s law applies in all situations, and
therefore is more general than Coulomb’s law.
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Homework Asignment # 2
Chapter 21 – 40, 58, 62
Chapter 22 – 10, 22, 34
Homework Asignment # 3
Chapter 23 – 4, 14, 30, 36, 52, 62
Copyright © 2009 Pearson Education, Inc.