Transcript ppt

From Last Time:


Electric field of a single charge
Electric field from multiple charges
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
Superposition
Ex: calculate electric field of dipole
Today:

Continuous charge distributions
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Electric field of line, ring, plane
Conductors : electrostatic equilibrium
Electrostatic potential
Oct. 1, 2009
Physics 208 Lecture 9
1
Exam 1 results
Physics 208 Exam 1
25
Mean = 66%
B
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Average = 66%
Curve
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Ave at B/BC
boundary
Contributes to
course grade as
curved score.
20
BC
AB
C
15
Count

Exams returned in
Monday
discussion
A
10
D
F
5
0
20
30
40
50
60
70
80
90 100
Exam % Score
Oct. 1, 2009
Physics 208 Lecture 9
2
From Last Time:


Electric field of a single charge
Electric field from multiple charges


Superposition
Ex: calculate electric field of dipole
Today:

Continuous charge distributions



Electric field of line, ring, plane
Conductors : electrostatic equilibrium
Electrostatic potential
Oct. 1, 2009
Physics 208 Lecture 9
3
Charge Densities

Volume charge density: when a charge is distributed
evenly throughout a volume
  = Q / V
dq =  dV

Surface charge density: when a charge is distributed
evenly over a surface area
  = Q / A
dq =  dA
Linear charge density: when a charge is distributed along
a line
  = Q /
dq =  d

Oct. 1, 2009
Physics 208 Lecture 9
4
Infinite line of charge
An infinite line of charge has a uniform charge density
 Coulombs / meter. What direction is the electric
field at point x?
B.
C.
x
A.
D.
E.
+ + + + + + + + + + + + + + + + + + +
Oct. 1, 2009
Physics 208 Lecture 9
5
E-field: infinite line of charge
dE y  cosdE
y


dE
2
r 2  x
dQ  dx

r
1
4 o

dQ
r2  x 2
dQ

4 o r 2  x 2

2
1

1
cos
r
r2  x 2
rdx
4 o r 2  x 2 3 / 2
+ + + + + + + + + + + + + + + + + + + +
x


Charge density Coulombs/meter
Oct. 1, 2009
Physics 208 Lecture 9
6
Add all these up
Ey 

 2 4
rdx
1
0
o
r
2
x

2 3/2


xdx




 2k
1/ 2
2
2
2o rr  x 
r
2o r
0

Units?


Oct. 1, 2009
[k]  N  m 2 /C 2
[  ]  C /m

[r]  m


Physics 208 Lecture 9
7
Non-infinite (finite) line of charge
What direction is the E-field
above the end of the line
charge?
C
B
A
D
E
+ + + + + + + + + + + + + + + + + + + +
Oct. 1, 2009
Physics 208 Lecture 9
8
A)
Ring of uniform positive charge
z
B)
y
Which is the graph of
E z on the z-axis?
x
C)

D)
z
Ez  k
E)
Oct. 1, 2009
Physics 208 Lecture 9
zQ
z
2
R

2 3/2
9
Quick quiz
We have an infinite sheet of charge of uniform
charge density . The electric field
A. increases with distance from plane
B. decreases with distance from plane
C. is independent of distance from plane
D. changes direction with distance from plane
Oct. 1, 2009
Physics 208 Lecture 9
10
E-field of plane of charge

Text does this as adding up rings of charge.

E plane 
2o
  Surface charge density


Oct. 1, 2009
Physics 208 Lecture 9
11
Electric fields and forces


Original definite of E-field was
(Coulomb Force) / charge
E-field produced force on charged particle.
F  qE
Oct. 1, 2009

Physics 208 Lecture 9
12
Force on charged particle
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Electric field produces force
qE on charged particle
Force produces an acceleration a = FE / m
Uniform E-field (direction & magnitude)
produces constant acceleration
Positive charge
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
accelerates in direction of the field
Negative charge

accelerates in direction opposite the electric field
Oct. 1, 2009
Physics 208 Lecture 9
13
Motion of charged particle

If no other forces, positive charge accelerates
in direction of E-field.
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But many systems have drag forces
(e.g. molecules in a liquid, etc)

Drag force is complex, but
usually depends on velocity.
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Particle reaches
terminal velocity,
determined by force balance
Oct. 1, 2009
Physics 208 Lecture 9
14
Application: Gel Electrophoresis
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Charged macromolecules in ‘gel’ with applied E-field
Electric force: FE = qE Steady state: F +F =0
E
D
Drag force: FD ~ -cv
v = qE / c
Speed depends on
charge and drag (molecule size)
Sometimes too complex to interpret
Protein electrophoresis:
soak in detergent to give proteins
all the same charge density.
Result, small proteins move faster
Oct. 1, 2009
Physics 208 Lecture 9
15
Conductors:
electrostatic equilibrium
Because charges are mobile in conductor :
 E-field inside conductor = 0
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Local net charge resides at surface
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Non-zero E-field causes charges to move
Stop at surface, generate canceling E-field
Repulsive force between charges pushes them away from
each other – stop at surface
E-field at locally perpendicular to surface


A parallel component would cause charge motion
Has magnitude E surf   /o
Surface charge density
Oct. 1, 2009
16
Charge distribution on conductor

What charge is
induced on
sphere to make
zero electric
field?
Oct. 1, 2009
+
17
Physics 208 Lecture 9
Electrogenic fish

Dipole +
nearby conducting object
Some fish generate charge
separation - electric dipole.
Dipole is induced in nearby
(conducting) fish
Small changes detected by
fish.
18
Summary of conductors

E  0 everywhere inside a conductor
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Charge in conductor is only on the surface
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E  surface of conductor
---
Oct. 1, 2009
++
+
+
++
19
Electric forces, work, and energy

Consider positive particle charge q, mass m at rest
in uniform electric field E
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Force on particle from field
Opposite force on particle from hand
Let particle go - it moves a distance d
How much work was done on particle? W  Fd  qEd
1 2
K.E.

mv  W  qEd
How fast is particle moving?
2
v=0

Oct. 1, 2009
+

v>0
+
20
Work and kinetic energy

Work-energy theorem:
 Change in kinetic energy of isolated
particle = work done
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dW  F  ds  Fdscos
Total work
end

K 
end
 dW   F  ds
start
start
In our case, F  qE

Oct. 1, 2009
21
Electric forces, work, and energy

Same particle, but don’t let go
F  qE
Move particle distance d, keep speed ~0
How much work is done by hand on particle? W  Fd  qEd
What is change in K.E. 
of particle? K.E .  0
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How much force does hand apply?
Conservation of energy? W stored in field as potential energy


+
Oct. 1, 2009
+
22
Work, KE, and potential energy
If particle is not isolated,

Wexternal  K  U
Work done
on system
Change in
kinetic energy
Change in
electric potential energy
Works for constant electric field if U  qE  r



Only electric potential energy difference

Sometimes a reference
point is chosen
E.g. U r   0 at r  (0,0,0)
 Then U r   qE  r for uniform electric field

Oct. 1, 2009
23
Electric potential V


Electric potential difference V is the electric
potential energy / unit charge = U/q
For uniform electric field,
U r  qE  r
V r  

 E  r
q
q
This is only valid for a uniform electric field

Oct. 1, 2009
24
Quick Quiz
Two points in space A and B have electric potential
VA=20 volts and VB=100 volts. How much work
does it take to move a +100µC charge from A to
B?
A. +2 mJ
B. -20 mJ
C. +8 mJ
D. +100 mJ
E. -100 mJ
Oct. 1, 2009
25
Check for uniform E-field
Push particle against E-field, or across E-field
Which requires work?
+
Increasing electric
potential in this direction
Oct. 1, 2009
Constant electric potential
in this direction
+
Decreasing electric
potential in this direction
26
Potential from electric field
dV  E  d


Potential changes
largest in direction of V  V d
o
E-field.
Smallest (zero)
E
perpendicular to


E-field

d
V  Vo  E d

d
V=Vo
 V  Vo  E d

Oct. 1, 2009
27
Electric potential: general
U 
F
Coulomb
 ds 
 qE  ds  q  E  ds
Electric potential energy difference U
proportional to charge q that work is done on
U /q  V  Electric potential difference 
 E  ds
Depends only on charges that create E-fields

Electric field usually created 
by some charge
distribution.
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V(r) is electric potential of that charge distribution
V has units of Joules / Coulomb = Volts
Oct. 1, 2009
28
Electric potential of point charge

kQ
Electric field from point charge Q is E  2 rˆ
r

What is the electric potential difference?
V 
end

r final
E  ds 
start
 k
Define V r    0

Oct. 1, 2009

Q
k
dx
 
2
rinitial r
r final
Q
Q
Q
k
k
r rinitial
rinital
rfinal
Then
Q
V r  k
for point charge
r
29
Electric Field and equipotential lines for +
and - point charges

The E lines are directed away
from the source charge
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Oct. 1, 2009
A positive test charge would be
repelled away from the positive
source charge
The E lines are directed toward
the source charge
A positive test charge would
be attracted toward the
negative source charge
Blue dashed lines are equipotential
30