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Transcript
Chapter 21
Electric Charge and
Electric Field
Copyright © 2009 Pearson Education, Inc.
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
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
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:
1 dQ
dE  r  
rˆ
2
4 0 r
Remember that the electric field is a vector;
you will need a separate integral for each
component.
Copyright © 2009 Pearson Education, Inc.
Charge districutions
dQ can take various forms:
dQ   d ;     C m
  dA;    C
m2
  dV ;     C
m3
 ,  , and  can be functions of position
but we will confine our calculations to
cases where they are constant
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
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.
Copyright © 2009 Pearson Education, Inc.
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.
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:
1/2



2

  
 
z

z
z

1 
E
1
1 2   

1/2
2 0   z 2  R 2   2 0 
R 
R   2 0




This is the field due to an infinite plane of
charge. It’s uniform and doesn’t depend or z
because if R   then z/R always 1 .
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).
Copyright © 2009 Pearson Education, Inc.
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.
Copyright © 2009 Pearson Education, Inc.
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!
Copyright © 2009 Pearson Education, Inc.
21-8 Field Lines
Electric dipole: two equal charges, opposite in
sign:
Copyright © 2009 Pearson Education, Inc.
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.
Copyright © 2009 Pearson Education, Inc.
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.
4. Schematic only.
Copyright © 2009 Pearson Education, Inc.
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.
Copyright © 2009 Pearson Education, Inc.
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.

Copyright © 2009 Pearson Education, Inc.
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?
Copyright © 2009 Pearson Education, Inc.
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
Electron gun.
An electron (mass m = 9.11 x 10-31 kg)
is accelerated in the uniform field
(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.
Copyright © 2009 Pearson Education, Inc.
E
Example
A proton (m=1.67x10-27kg, q=+e=1.60x10-19 C)
7
v

2.5

10
m / s xˆ
with an initial velocity of
enters a region where there is a uniform
electric field E  8  106 N / C oriented at 30o to the
x-axis. Describe its trajectory.
Copyright © 2009 Pearson Education, Inc.
Copyright © 2009 Pearson Education, Inc.
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.
pH2O

Copyright © 2009 Pearson Education, Inc.
21-11 Electric Dipoles
Recall: 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:
E  r  rxˆ  
2kp
r
3
Copyright © 2009 Pearson Education, Inc.
[along axis of p]
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:
U   p E
(cf. Chapt. 10)
So how was I able to
move the 2x4 using the
electric field of the rod?
Copyright © 2009 Pearson Education, Inc.
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:
Copyright © 2009 Pearson Education, Inc.
21-11 Electric Dipoles
Example:
The dipole moment of a water molecule is
6.3 x 10-30 C·m. A sample contains 1021
water molecules, with their dipole
moments all oriented in the direction of an
electric field of 2.5 x 105 N/C. How much
work is required to rotate the dipoles from
this orientation ( = 0) to one in which all
moments are perpendicular to the field
( = 90)?
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:
Copyright © 2009 Pearson Education, Inc.
21-12 Electric Forces in Molecular
Biology; DNA
The A-T and G-C
nucleotide bases
attract each other
through electrostatic
forces.
Copyright © 2009 Pearson Education, Inc.
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.
21-13 Photocopy Machines and
Computer Printers Use
Electrostatics
Photocopy machine:
• drum is charged positively
• image is focused on drum
• only black areas stay charged and
therefore attract toner particles
• image is transferred to paper and sealed by
heat
Copyright © 2009 Pearson Education, Inc.
21-13 Photocopy Machines and
Computer Printers Use
Electrostatics
Copyright © 2009 Pearson Education, Inc.
21-13 Photocopy Machines and
Computer Printers Use
Electrostatics
Laser printer is similar, except a computer
controls the laser intensity to form the image
on the drum.
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.
Copyright © 2009 Pearson Education, Inc.
Summary of Chapter 21
• Charge is quantized in units of e.
• Objects can be charged by conduction or
induction.
• Coulomb’s law: F12  k
Q1Q2
1 Q1Q2
ˆ
r

rˆ12
12
2
2
r12
4 0 r12
where  0 
1
4 k
 8.85 10 12 C 2 / N  m 2
•Electric field is force per unit charge:
Copyright © 2009 Pearson Education, Inc.
Summary of Chapter 21
• Electric field of a point charge:
F
Qq / r 2
Q
ˆ
E k
r  k 2 rˆ for single point charge
q
q
r
• Electric field can be represented by electric
field lines.
• Static electric field inside conductor is zero;
surface field is perpendicular to surface.
Copyright © 2009 Pearson Education, Inc.
Questions?
Copyright © 2009 Pearson Education, Inc.