Download Chapter 27. The Electric Field

Chapter 27. The Electric Field
Electric fields are
responsible for the electric
currents that flow through
your computer and the
nerves in your body. Electric
fields also line up polymer
molecules to form the
images in a liquid crystal
display (LCD).
Chapter Goal: To learn how
to calculate and use the
electric field.
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Chapter 27. The Electric Field
Topics:
• Electric Field Models
• The Electric Field of Multiple Point Charges
• The Electric Field of a Continuous Charge
Distribution
• The Electric Fields of Rings, Disks, Planes, and
Spheres
• The Parallel-Plate Capacitor
• Motion of a Charged Particle in an Electric
Field
• Motion of a Dipole in an Electric Field
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What device provides a practical
way to produce a uniform electric
field?
Chapter 27. Reading Quizzes
A. A long thin resistor
B. A Faraday cage
C. A parallel plate capacitor
D. A toroidal inductor
E. An electric field uniformizer
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What device provides a practical
way to produce a uniform electric
field?
A. A long thin resistor
B. A Faraday cage
C. A parallel plate capacitor
D. A toroidal inductor
E. An electric field uniformizer
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For charged particles, what is the
quantity q/m called?
A. Linear charge density
B. Charge-to-mass ratio
C. Charged mass density
D. Massive electric dipole
E. Quadrupole moment
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For charged particles, what is the
quantity q/m called?
Which of these charge distributions did
not have its electric field determined in
Chapter 27?
A. Linear charge density
B. Charge-to-mass ratio
C. Charged mass density
D. Massive electric dipole
E. Quadrupole moment
A. A line of charge
B. A parallel-plate capacitor
C. A ring of charge
D. A plane of charge
E. They were all determined
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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Which of these charge distributions did
not have its electric field determined in
Chapter 27?
A. A line of charge
B. A parallel-plate capacitor
C. A ring of charge
D. A plane of charge
E. They were all determined
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The worked examples of chargedparticle motion are relevant to
A. a transistor.
B. a cathode ray tube.
C. magnetic resonance imaging.
D. cosmic rays.
E. lasers.
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The worked examples of chargedparticle motion are relevant to
A. a transistor.
B. a cathode ray tube.
C. magnetic resonance imaging.
D. cosmic rays.
E. lasers.
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Chapter 27. Basic Content and Examples
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Electric Field Models
Electric Field Models
The electric field of a point charge q at the origin, r = 0, is
The net electric field due to a group of point charges is
where є0 = 8.85 × 10–12 C2/N m2 is the permittivity
constant.
where Ei is the field from point charge i.
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Problem--Solving Strategy: The electric field of
Problem
multiple point charges
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Problem--Solving Strategy: The electric field of
Problem
multiple point charges
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Problem--Solving Strategy: The electric field of
Problem
multiple point charges
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The Electric Field of a Dipole
We can represent an electric dipole by two opposite charges
±q separated by the small distance s.
The dipole moment is defined as the vector
Electric Dipole: Two Equal but Opposite Charges
Separated by a Small Distance
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The Electric Field of a Dipole
On the axis of an electric dipole
( Edipole ) y = ( E+ ) y + ( E− ) y =
q
1 
(−q) 
+

2
4πε 0  ( y − s / 2) ( y + s / 2) 2 
( Edipole ) y =
The dipole-moment magnitude p = qs determines the
electric field strength. The SI units of the dipole moment
are C m.
q
4πε 0


2 ys
 ( y − 0.5 ⋅ s) 2 ( y + 0.5 ⋅ s ) 2 


y>>s
( Edipole ) y ≈
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1
2qs
4πε 0 y 3
The Electric Field of a Dipole
The electric field at a point on the axis of a dipole is
where r is the distance measured from the center of the
dipole.
The electric field in the plane that bisects and is
perpendicular to the dipole is
This field is opposite to the dipole direction, and it is only
half the strength of the on-axis field at the same distance.
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EXAMPLE 27.2 The electric field of a water
molecule
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EXAMPLE 27.2 The electric field of a water
molecule
QUESTION:
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Tactics: Drawing and using electric field lines
EXAMPLE 27.2 The electric field of a water
molecule
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Continuous Charge Distribution: Charge Density
Tactics: Drawing and using electric field lines
The total electric charge is Q.
Volume V
Linear, length L
Q
Amount of charge in a
small volume dl:
Q
Q
λ=
dq = dl = λ dl
L
L
Linear charge density
Surface,
area A
Q
Q
Amount of charge in a
small volume dV:
Q
ρ=
Q
dq = dV = ρ dV
V
V
Volume charge density
Amount of charge in a
small volume dA:
dq =
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Q
dA = σ dA
A
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Q
A
Surface charge density
σ=
Electric Field: Continuous Charge Distribution
Electric field due to three equal point charges?
Procedure:
– Divide the charge distribution into small
elements, each of which contains ∆q
– Calculate the electric field due to one of
these elements at point P
– Evaluate the total field by summing the
contributions of all the charge elements
Symmetry: take advantage of any symmetry to
simplify calculations
For the individual
charge elements
Because the charge distribution
is continuous
E = ke lim
∆qi →0
∆q i
∑r
i
2
rˆi = ke ∫
i
∆E = ke
dq
rˆ
r2
∆q
rˆ
r2
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The Electric Field of a Continuous
Charge Distribution at a point in the plane that
bisects the rode
Electric Field: Symmetry
r
q r
E = ke 2 rˆ
r
q1 = 10 µC q2 = 10 µC
r
E2
5m
r
E
r
E1,z
r
E2 Electric field
r r
r
E = E1 + E2
ϕ
r
E1
E = 2E1 cos φ
E = 2E1,z = 2E2,z
ϕ
5m
5
m
+
r
E1,z
+
1
r
E
ϕ
6m
E1,z = E1 cos φ
r
E1
E = 4E1,z = 4E2 cos φ
ϕ
+
2
+
The symmetry gives us the direction
field
+
+
resultant
electric
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Pearson
Addison-Wesley.
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The Electric Field of a Continuous
Charge Distribution
The Electric Field of a Continuous
Charge Distribution on a Thin Rod
( Ei ) x = ( Ei ) x cos θ i =
( Ei ) x =
∆Q
4πε 0 y + r 2
( Ei ) x =
r∆Q
r∆
4πε 0 ( y + r 2 ) 3 / 2
Ex =
( Ei ) x = ( Ei ) x cos θ i =
∆Q
cos θ i
4πε 0 ri 2
1
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How to evaluate this integral?
-Trigonometric Substitution
Ex =
Q/L
4πε 0
∆Q
cos θ i
4πε 0 ri 2
1
1
r
2
i
2
i
y + r2
1
∆Q = λ∆y = (Q / L)∆y
2
i
Q/L
r∆y
∑
2
4πε 0 i ( yi + r 2 )3 / 2
L/2
Q/L
rdy
Ex =
∫
2
4πε 0 − L / 2 ( y + r 2 )3 / 2
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L/2
rdy
2 3/ 2
−L /2 ( y + r )
∫
2
http://en.wikipedia.org/wiki/Trigonometric_substitution
Electric Field due to a Thin Rod
L/ 2
Q/ L
rdy
Ex =
4πε0 −L∫/ 2 ( y2 + r 2 )3/ 2
Use:
http://integrals.wolfram.com/index.jsp
(to evaluate, x=a tanθ, dx=a sec2θ dθ
sinθ = x/(x2+a2)1/2
Ex =
Q/ L
y
4πε0 r y 2 + r 2
Ex =
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1
L/ 2
−L / 2
=


Q/ L 
L/2
− L/ 2

−
4πε0  r (L / 2)2 + r 2 r (−L / 2)2 + r 2 


Q
4πε 0 r ( L / 2) 2 + r 2
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An infinite Line of Charge
Ex =
1
Q
4πε 0 r ( L / 2) 2 + r 2
E line =
lim
L→ ∞
E line =
1
Q
4 πε 0 r ( L / 2 ) 2 + r
1
4 πε 0
Eline =
Q
rL / 2
An infinite Line of Charge
Eline =
2
2λ
4πε 0 r
1
Unlike a point charge, for
which the field decreases as
1/r2, the field of an infinitely
long charged wire decreases
more slowly – as only 1/r
negligible
2λ
4πε 0 r
1
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Parallel-Plate Capacitor
Parallel-Plate Capacitor
Find electric field
+ + + + +
+ + + + + +
− − − − −
− − − − − −
r
E+
r
E+
σ >0
E+ =
σ
2ε0
− − − − −
− − − − − −
−σ < 0
r r
r
E = E+ + E−
σ >0
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E− =
−σ < 0
r
E−
σ
E+ =
2ε0
σ
2ε0
σ >0
+ + + + +
+ + + + + +
Find electric field
−σ < 0
r
E−
σ
E− =
2ε0
r
E+
r
E−
r
E+
r
E−
r
E+
r
E−
σ >0
−σ < 0
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E = E+ − E− = 0
E = E+ + E− =
σ
ε0
E = E+ − E− = 0
Parallel-Plate Capacitor
+ + + + +
+ + + + + +
− − − − −
− − − − − −
E =0
E=
σ
ε0
σ >0
−σ < 0
EXAMPLE 27.7 The electric field inside a
capacitor
QUESTIONS:
σ >0
r
E
−σ < 0
E =0
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EXAMPLE 27.7 The electric field inside a
capacitor
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EXAMPLE 27.7 The electric field inside a
capacitor
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Electric Field: Continuous Charge Distribution on a ring
EXAMPLE 27.7 The electric field inside a
capacitor
What is the electric field at point P ?
P
h
90o
R
λ
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What is the electric field at point P ?
λ
- linear charge density
1. Symmetry determines the direction of the electric field.
O
- linear charge density
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λ - linear charge density
2. Divide the charge distribution into small elements, each
of which contains ∆q
What is the electric field at point P ?
r
E=
r
∆E =
∑
all elements
z
r
E
∑
E=
r
E
all elements
∆E z
all elements
∆E z = ∆E cos ϕ = ke
P
r
∆E
ϕ
r
∆E z
E=
ϕ
h
r
∆ q = λ ∆l
∆l
90o
R
O
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R
∑
all elements
P
h
r
∆E z
∑
90o
O
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ke
∆q
cos ϕ
r2
∆q
cos ϕ
r2
What is the electric field at point P ?
λ
- linear charge density
What is the electric field at point P ?
3. Evaluate the total field by summing the contributions of
all the charge elements ∆q
z
r
∆E
ϕ
E=
P
λR
e
0
r
2
cos ϕ = ke
r
∑
ke
λ R ∆ψ
r2
2π
E=
∫ dψ ke
0
∆ψ
ψ
r
2
z
r
∆E
cos ϕ
ϕ
λR
r
2
cos ϕ =
r
∆E z
h
x
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90o
r2
cos ϕ
∆l
ψ
h
h
=
2
r
h + R2
E = 2πλ ke
r
cos ϕ
O
0
λR
r = h2 + R 2
ϕ
R
2π
cos ϕ ∫ dψ = 2π ke
P
Replace Sum by Integral
∆l
λR
r
E
all elements
h
90o
∫ dψ k
∆ q = λ ∆l = λ R ∆ ψ
ϕ
R
2π
E=
∆l = R∆ψ
r
∆E z
- linear charge density
4. Evaluate the integral
∆q
E = ∑ ke 2 cos ϕ
r
all elements
r
E
λ
∆ψ
E=
hR
(h
2
+ R2 )
1
hQ
4πε 0 (h + R 2 )3 / 2
O
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A Circular Disk of Charge
Electric field due to a ring:
η=
Q
Q
=
A π R2
E =
(Ei )z =
1
4 πε 0
1
4 πε
N
0
hQ
(h + R 2 )3/ 2
2
z∆ Qi
2
( z 2 + ri ) 3 / 2
( Edisk ) z = ∑ ( Ei ) z =
z
N
∑ (z
4πε 0 i =1
ri ∆r
ηz N
( Edisk ) z =
∑
2
2ε 0 i =1 ( z + ri 2 ) 3 / 2
i =1
2
∆Qi
2
+ ri ) 3 / 2
Motion of a Charged Particle
ηz R
rdr
2
∫
2ε 0 0 ( z + r 2 ) 3 / 2

z
η 
)z =
1 −

2
2
2ε 0 
z +R 
( Edisk ) z =
( E disk
Quiz: If z>>R, then E = ?
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3/ 2
2
x
Motion of Charged Particle
• When a charged particle is placed in an electric field, it experiences an
electrical force
• If this is the only force on the particle, it must be the net force
• The net force will cause the particle to accelerate according to Newton’s
second law
r
r
F = qE
r
r
F = ma
- Coulomb’s law
Motion of Charged Particle
What is the final velocity?
r
r
F = qE
r
r
F = ma
- Coulomb’s law
- Newton’s second law
vf
|q|
ay = −
E
m
Motion in x – with constant velocity
- Newton’s second law
Motion in x – with constant acceleration
t=
r q r
a= E
m
v0
l
v0
ay = −
|q|
E
m
- travel time
v x = v0
After time t the velocity in y direction becomes
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2
vy
|q|
Et then v f = v02 +  q Et 
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m 
v y = ayt = −
General Principles
Chapter 27. Summary Slides
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vf
General Principles
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Applications
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Applications
Chapter 27. Questions
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At the
position of
the dot, the
electric field
points
At the
position of
the dot, the
electric field
points
A. Up.
B. Down.
C. Left.
D. Right.
E. The electric field is zero.
A. Up.
B. Down.
C. Left.
D. Right.
E. The electric field is zero.
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A piece of plastic is uniformly charged with surface
charge density 1. The plastic is then broken into a large
piece with surface charge density 2 and a small piece
with surface charge density 3. Rank in order, from largest
to smallest, the surface charge densities 1 to 3.
A.
B.
C.
D.
E.
η2
η1
η1
η3
η1
= η3 > η1
> η2 > η3
> η2 = η3
> η2 > η1
= η2 = η3
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A piece of plastic is uniformly charged with surface
charge density 1. The plastic is then broken into a large
piece with surface charge density 2 and a small piece
with surface charge density 3. Rank in order, from largest
to smallest, the surface charge densities 1 to 3.
A.
B.
C.
D.
E.
η2 = η3 > η1
η1 > η2 > η3
η1 > η2 = η3
η3 > η2 > η1
η1 = η2 = η3
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Which of the following
actions will increase the
electric field strength at
the position of the dot?
A. Make the rod longer without changing the charge.
B. Make the rod fatter without changing the charge.
C. Make the rod shorter without changing the charge.
D. Remove charge from the rod.
E. Make the rod narrower without changing the charge.
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Rank in order,
from largest to
smallest, the
electric field
strengths Ea to Ee
at these five points
near a plane of
charge.
A.
B.
C.
D.
E.
Ea > Ec > Eb > Ee > Ed
Ea = Eb = Ec = Ed = Ee
Ea > Eb = Ec > Ed = Ee
Eb = Ec = Ed = Ee > Ea
Ee > Ed > Ec > Eb > Ea
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Which of the following
actions will increase the
electric field strength at
the position of the dot?
A. Make the rod longer without changing the charge.
B. Make the rod fatter without changing the charge.
C. Make the rod shorter without changing the charge.
D. Remove charge from the rod.
E. Make the rod narrower without changing the charge.
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Rank in order,
from largest to
smallest, the
electric field
strengths Ea to Ee
at these five points
near a plane of
charge.
A.
B.
C.
D.
E.
Ea > Ec > Eb > Ee > Ed
Ea = Eb = Ec = Ed = Ee
Ea > Eb = Ec > Ed = Ee
Eb = Ec = Ed = Ee > Ea
Ee > Ed > Ec > Eb > Ea
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Rank in order, from
largest to smallest, the
forces Fa to Fe a proton
would experience if
placed at points a – e in
this parallel-plate
capacitor.
A.
B.
C.
D.
E.
Fa = Fb = Fd = Fe > Fc
Fa = Fb > Fc > Fd = Fe
Fa = Fb = Fc = Fd = Fe
Fe = Fd > Fc > Fa = Fb
Fe > Fd > Fc > Fb > Fa
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Which electric field
is responsible for
the trajectory of the
proton?
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Rank in order, from
largest to smallest, the
forces Fa to Fe a proton
would experience if
placed at points a – e in
this parallel-plate
capacitor.
A.
B.
C.
D.
E.
Fa = Fb = Fd = Fe > Fc
Fa = Fb > Fc > Fd = Fe
Fa = Fb = Fc = Fd = Fe
Fe = Fd > Fc > Fa = Fb
Fe > Fd > Fc > Fb > Fa
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Which electric field
is responsible for
the trajectory of the
proton?
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