chapter 27. the electric fieldthe electric field...
TRANSCRIPT
Chapter 27. The Electric FieldThe 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
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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.
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
Chapter 27. The Electric FieldChapter 27. The Electric Field
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• 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
Chapter 27. Reading QuizzesChapter 27. Reading Quizzes
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Chapter 27. Reading QuizzesChapter 27. Reading Quizzes
What device provides a practical
way to produce a uniform electric
field?
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A. A long thin resistor
B. A Faraday cage
C. A parallel plate capacitor
D. A toroidal inductor
E. An electric field uniformizer
What device provides a practical
way to produce a uniform electric
field?
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A. A long thin resistor
B. A Faraday cage
C. A parallel plate capacitor
D. A toroidal inductor
E. An electric field uniformizer
For charged particles, what is the
quantity q/m called?
A. Linear charge density
B. Charge-to-mass ratio
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B. Charge-to-mass ratio
C. Charged mass density
D. Massive electric dipole
E. Quadrupole moment
For charged particles, what is the
quantity q/m called?
A. Linear charge density
B. Charge-to-mass ratio
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B. Charge-to-mass ratio
C. Charged mass density
D. Massive electric dipole
E. Quadrupole moment
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
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B. A parallel-plate capacitor
C. A ring of charge
D. A plane of charge
E. They were all determined
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
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B. A parallel-plate capacitor
C. A ring of charge
D. A plane of charge
E. They were all determined
The worked examples of charged-
particle motion are relevant to
A. a transistor.
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A. a transistor.
B. a cathode ray tube.
C. magnetic resonance imaging.
D. cosmic rays.
E. lasers.
The worked examples of charged-
particle motion are relevant to
A. a transistor.
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A. a transistor.
B. a cathode ray tube.
C. magnetic resonance imaging.
D. cosmic rays.
E. lasers.
Chapter 27. Basic Content and ExamplesChapter 27. Basic Content and Examples
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Chapter 27. Basic Content and ExamplesChapter 27. Basic Content and Examples
Electric Field ModelsElectric Field Models
The electric field of a point charge q at the origin, r = 0, is
where є0 = 8.85 × 10–12 C2/N m2 is the permittivity
constant.
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constant.
The net electric field due to a group of point charges is
Electric Field ModelsElectric Field Models
where Ei is the field from point charge i.
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ProblemProblem--Solving Strategy: The electric field of Solving Strategy: The electric field of
multiple point chargesmultiple point charges
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ProblemProblem--Solving Strategy: The electric field of Solving Strategy: The electric field of
multiple point chargesmultiple point charges
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ProblemProblem--Solving Strategy: The electric field of Solving Strategy: The electric field of
multiple point chargesmultiple point charges
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Electric Dipole: Two Equal but Opposite Charges
Separated by a Small Distance
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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
The Electric Field of a DipoleThe Electric Field of a Dipole
The dipole-moment magnitude p = qs determines the
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The dipole-moment magnitude p = qs determines the
electric field strength. The SI units of the dipole moment
are C m.
The Electric Field of a Dipole
+
−+
−=+= −+ 2
2/2
2/0 )(
)(
)(4
1)()()(
sy
q
sy
qEEE yyydipole
πε
⋅+⋅−=
22
0 )5.0()5.0(
2
4)(
sysy
ysqE ydipole
πε
On the axis of an electric dipole
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y>>s
3
0
2
4
1)(
y
qsE ydipole
πε≈
The electric field at a point on the axis of a dipole is
The Electric Field of a DipoleThe Electric Field of a Dipole
where r is the distance measured from the center of the
dipole.
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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 EXAMPLE 27.2 The electric field of a water
moleculemolecule
QUESTION:
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EXAMPLE 27.2 The electric field of a water EXAMPLE 27.2 The electric field of a water
moleculemolecule
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EXAMPLE 27.2 The electric field of a water EXAMPLE 27.2 The electric field of a water
moleculemolecule
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Tactics: Drawing and using electric field linesTactics: Drawing and using electric field lines
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Tactics: Drawing and using electric field linesTactics: Drawing and using electric field lines
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Continuous Charge Distribution: Charge Density
The total electric charge is Q.
Q
Q
Linear, length L
Volume V
Amount of charge in a
small volume dl:
Qdq dl dl
Lλλλλ= == == == =
Q
Lλλλλ ====
Linear charge density
Amount of charge in a
small volume dV: Q
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Q
Surface,
area A
Linear charge density
Amount of charge in a
small volume dA:
Qdq dA dA
Aσσσσ= == == == =
Q
Aσσσσ ====
Surface charge density
small volume dV:
Qdq dV dV
Vρρρρ= == == == =
Q
Vρρρρ ====
Volume charge density
Electric Field: Continuous Charge Distribution
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
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contributions of all the charge elements
Symmetry: take advantage of any symmetry to
simplify calculations
2ˆ
e
qk
r
∆∆ =E r
For the individual
charge elements
2 20ˆ ˆlim
i
ie i e
qi i
q dqk k
r r∆ →
∆= =∑ ∫E r r
Because the charge distribution
is continuous
Electric field due to three equal point charges?
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Electric Field: Symmetry
1Errrr
2Errrr
Electric field
1 10q µC= 2 10q µC=
5m
5m
1 2E E E= += += += +r r rr r rr r rr r r2
ˆe
qE k r
r====
rrrr rrrr
Errrr 2
Errrr
ϕϕϕϕ
ϕϕϕϕ
12 cosE E φ=
1, 2,2 2z zE E E= =
1,zErrrr
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++++
1 2
++++
6m
5m
++++++++
++++++++
1Errrr
1,zErrrr
Errrr
1, 24 4 coszE E E φ= =
1, 1 coszE E φ=
ϕϕϕϕ
ϕϕϕϕ
The symmetry gives us the direction
of resultant electric field
The Electric Field of a Continuous
Charge Distribution at a point in the plane that
bisects the rode
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The Electric Field of a Continuous
Charge Distribution
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i
i
ixixir
QEE θ
πεθ cos
4
1cos)()(
2
0
∆==
The Electric Field of a Continuous
Charge Distribution on a Thin Rod
i
i
ixixir
QEE θ
πεθ cos
4
1cos)()(
2
0
∆==
2222
04
1)(
ry
r
ry
QE
ii
xi
++
∆=
πε
1 Qr∆
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2/322
0 )(4
1)(
ry
QrE
i
xi+
∆=
πε
∑+
∆=
i i
xry
yrLQE
2/322
0 )(4
/
πε
yLQyQ ∆=∆=∆ )/(λ
∫− +
=2/
2/
2/3220 )(4
/L
L
xry
rdyLQE
πε
How to evaluate this integral? ∫
− +=
2/
2/
2/3220 )(4
/L
L
xry
rdyLQE
πε-Trigonometric Substitution
http://en.wikipedia.org/wiki/Trigonometric_substitution
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Electric Field due to a Thin Rod
∫− +
=2/
2/
2/3220 )(4
/L
L
xry
rdyLQE
πεhttp://integrals.wolfram.com/index.jsp
(to evaluate, x=a tanθ, dx=a sec2θ dθ
sinθ = x/(x2+a2)1/2
Use:
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220 )2/(4
1
rLr
QEx
+=
πε
+−
−−
+=
+= −
22220
2/
2/22
0 )2/(
2/
)2/(
2/
4
/
4
/
rLr
L
rLr
LLQ
ryr
yLQE
L
Lxπεπε
sinθ = x/(x +a )
An infinite Line of Charge
220 )2/(4
1
rLr
QEx
+=
πε
220 )2/(4
1lim
rLr
QE
L
line
+=
∞→ πε
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2/4
1
0 rL
QE line
πε=
rEline
λ
πε
2
4
1
0
=
negligible
An infinite Line of Charge
rEline
λ
πε
2
4
1
0
=
Unlike a point charge, for which the field decreases as
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which the field decreases as 1/r2, the field of an infinitely long charged wire decreases more slowly – as only 1/r
Find electric field
Parallel-Plate Capacitor
0σ >
0σ− <
+ + + + ++ + + + +
+
− − − − −− − − − −
−
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0σ >
E−
r
E+
r
02
σE
ε+ =
0σ− <
E+
r
E−
r
02
σE
ε− =
Find electric field
Er
Er
Parallel-Plate Capacitor
0σ >
0σ− <
+ + + + ++ + + + +
+
− − − − −− − − − −
−02
σE
ε+ =
02
σE
ε− =
E E E+ −= +r r r
0E E E= − =
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0σ >E−
rE+
r
0σ− <E−
r
E+
r
E+
rE−
r
0E E E+ −= − =
0E E E+ −= − =
0
σE E E
ε+ −= + =
Parallel-Plate Capacitor
0σ >
0σ− <
+ + + + ++ + + + +
+
− − − − −− − − − −
−
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0σ >
0σ− <
Er
0E =
0
σE
ε=
0E =
EXAMPLE 27.7 The electric field inside a EXAMPLE 27.7 The electric field inside a
capacitorcapacitor
QUESTIONS:
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EXAMPLE 27.7 The electric field inside a EXAMPLE 27.7 The electric field inside a
capacitorcapacitor
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EXAMPLE 27.7 The electric field inside a EXAMPLE 27.7 The electric field inside a
capacitorcapacitor
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EXAMPLE 27.7 The electric field inside a EXAMPLE 27.7 The electric field inside a
capacitorcapacitor
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Electric Field: Continuous Charge Distribution on a ring
P
h
What is the electric field at point P ?
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OR
o90
λλλλ - linear charge density
λλλλ - linear charge densityWhat is the electric field at point P ?
1. Symmetry determines the direction of the electric field.
Errrr
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OR
o90
P
h
λλλλ - linear charge densityWhat is the electric field at point P ?
2. Divide the charge distribution into small elements, each
of which contains ∆q
Errrr
z
E∆∆∆∆rrrr
z
all elements all elements
E E E= ∆ = ∆= ∆ = ∆= ∆ = ∆= ∆ = ∆∑ ∑∑ ∑∑ ∑∑ ∑r r rr r rr r rr r r
2cos cos
z e
qE E k
rϕ ϕϕ ϕϕ ϕϕ ϕ
∆∆∆∆∆ = ∆ =∆ = ∆ =∆ = ∆ =∆ = ∆ =
z
all elements
E E= ∆= ∆= ∆= ∆∑∑∑∑
cosq
E k ϕϕϕϕ∆∆∆∆
==== ∑∑∑∑
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OR
o90
P
h
q lλλλλ∆ = ∆∆ = ∆∆ = ∆∆ = ∆
l∆∆∆∆
ϕϕϕϕ
ϕϕϕϕ
E∆∆∆∆rrrr z
E∆∆∆∆rrrr
r
2cos
e
all elements
qE k
rϕϕϕϕ
∆∆∆∆==== ∑∑∑∑
λλλλ - linear charge densityWhat is the electric field at point P ?
3. Evaluate the total field by summing the contributions of
all the charge elements ∆q
Errrr
q l Rλ λ ψλ λ ψλ λ ψλ λ ψ∆ = ∆ = ∆∆ = ∆ = ∆∆ = ∆ = ∆∆ = ∆ = ∆
z
E∆∆∆∆rrrr
2cos
e
all elements
qE k
rϕϕϕϕ
∆∆∆∆==== ∑∑∑∑
l R ψψψψ∆ = ∆∆ = ∆∆ = ∆∆ = ∆
Rλ ψλ ψλ ψλ ψ∆∆∆∆∑∑∑∑
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OR
o90
P
hl∆∆∆∆
ϕϕϕϕ
ϕϕϕϕ
E∆∆∆∆rrrr z
E∆∆∆∆rrrr
r
x
ψψψψψψψψ∆∆∆∆
2cos
e
all elements
RE k
r
λ ψλ ψλ ψλ ψϕϕϕϕ
∆∆∆∆==== ∑∑∑∑
Replace Sum by Integral
2
2
0
cose
RE d k
r
ππππ λλλλψ ϕψ ϕψ ϕψ ϕ==== ∫∫∫∫
λλλλ - linear charge densityWhat is the electric field at point P ?
4. Evaluate the integral
Errrr
z
E∆∆∆∆rrrr
2 2
2 2 2
0 0
cos cos 2 cose e e
R R RE d k k d k
r r r
π ππ ππ ππ πλ λ λλ λ λλ λ λλ λ λ
ψ ϕ ϕ ψ π ϕψ ϕ ϕ ψ π ϕψ ϕ ϕ ψ π ϕψ ϕ ϕ ψ π ϕ= = == = == = == = =∫ ∫∫ ∫∫ ∫∫ ∫
2 2cos
h h
r h Rϕϕϕϕ = == == == =
++++
2 2r h R= += += += +
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OR
o90
P
hl∆∆∆∆
ϕϕϕϕ
ϕϕϕϕ
E∆∆∆∆rrrr z
E∆∆∆∆rrrr
r
x
ψψψψψψψψ∆∆∆∆
r h R++++
(((( ))))3/ 2
2 22
e
hRE k
h Rπλπλπλπλ====
++++
2/322
0 )(4
1
Rh
hQE
+=
πε
A Circular Disk of Charge
2/322
0 )(4
1
Rh
hQE
+=
πεElectric field due to a ring:
2R
Q
A
Q
πη ==
2/3220 )(4
1)(
i
izi
rz
QzE
+
∆=
πε
∑∑== +
∆==
N
i i
iN
i
zizdiskrz
QzEE
12/322
01 )(4)()(
πε
∑+
∆=
Ni
zdiskrz
rrzE
2/322 )(2)(
ε
η
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∑= +
=i i
zdiskrz
E1
2/3220 )(2
)(ε
∫ +=
R
zdiskrz
rdrzE
0
2/322
0 )(2)(
ε
η
+−=
220
12
)(Rz
zE zdisk
ε
η
Quiz: If z>>R, then E = ?
Motion of a Charged Particle
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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
F qE====r rr rr rr r
- Coulomb’s law
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F qE====r rr rr rr r
F ma====rrrr rrrr
- Newton’s second law
- Coulomb’s law
qa E
m====
rrrrrrrr
Motion of Charged Particle
What is the final velocity?
F qE====r rr rr rr r
F ma====rrrr rrrr
- Newton’s second law
- Coulomb’s law
| |y
qa E
m= −= −= −= −
fv
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m
0
lt
v==== - travel time
Motion in x – with constant acceleration
0v
| |y
qa E
m= −= −= −= −
Motion in x – with constant velocity
After time t the velocity in y direction becomes
| |y y
qv a t Et
m= = −= = −= = −= = −
fv
0xv v====
yv
then
2
2
0f
qv v Et
m
= += += += +
Chapter 27. Summary SlidesChapter 27. Summary Slides
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Chapter 27. Summary SlidesChapter 27. Summary Slides
General PrinciplesGeneral Principles
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General PrinciplesGeneral Principles
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ApplicationsApplications
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ApplicationsApplications
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Chapter 27. QuestionsChapter 27. Questions
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Chapter 27. QuestionsChapter 27. Questions
At the
position of
the dot, the
electric field
points
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A. Up.
B. Down.
C. Left.
D. Right.
E. The electric field is zero.
points
At the
position of
the dot, the
electric field
points
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A. Up.
B. Down.
C. Left.
D. Right.
E. The electric field is zero.
points
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 . Rank in order, from largest
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with surface charge density 3. Rank in order, from largest
to smallest, the surface charge densities 1 to 3.
A. η2 = η3 > η1
B. η1 > η2 > η3
C. η1 > η2 = η3
D. η3 > η2 > η1
E. η1 = η2 = η3
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 . Rank in order, from largest
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with surface charge density 3. Rank in order, from largest
to smallest, the surface charge densities 1 to 3.
A. η2 = η3 > η1
B. η1 > η2 > η3
C. η1 > η2 = η3
D. η3 > η2 > η1
E. η1 = η2 = η3
Which of the following
actions will increase the
electric field strength at
the position of the dot?
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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.
Which of the following
actions will increase the
electric field strength at
the position of the dot?
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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.
Rank in order,
from largest to
smallest, the
electric field
strengths Ea to Ee
at these five points
near a plane of
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near a plane of
charge.
A. Ea > Ec > Eb > Ee > Ed
B. Ea = Eb = Ec = Ed = Ee
C. Ea > Eb = Ec > Ed = Ee
D. Eb = Ec = Ed = Ee > Ea
E. Ee > Ed > Ec > Eb > Ea
Rank in order,
from largest to
smallest, the
electric field
strengths Ea to Ee
at these five points
near a plane of
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A. Ea > Ec > Eb > Ee > Ed
B. Ea = Eb = Ec = Ed = Ee
C. Ea > Eb = Ec > Ed = Ee
D. Eb = Ec = Ed = Ee > Ea
E. Ee > Ed > Ec > Eb > Ea
near a plane of
charge.
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
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this parallel-plate
capacitor.
A. Fa = Fb = Fd = Fe > Fc
B. Fa = Fb > Fc > Fd = Fe
C. Fa = Fb = Fc = Fd = Fe
D. Fe = Fd > Fc > Fa = Fb
E. Fe > Fd > Fc > Fb > Fa
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
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A. Fa = Fb = Fd = Fe > Fc
B. Fa = Fb > Fc > Fd = Fe
C. Fa = Fb = Fc = Fd = Fe
D. Fe = Fd > Fc > Fa = Fb
E. Fe > Fd > Fc > Fb > Fa
this parallel-plate
capacitor.
Which electric field
is responsible for
the trajectory of the
proton?
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Which electric field
is responsible for
the trajectory of the
proton?
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