Transcript Physics 122B - Institute for Nuclear Theory

Physics 122B
Electricity and Magnetism
Lecture 4
Computing Electric Fields
April 02, 2007
Martin Savage
Lecture 4 Announcements
 Labs begin this week. Buy your 122Z Lab Manual
at the University Bookstore before going to your lab.
.
 Lecture Homework #1 have been posted on the
Tycho system. It is due at 10 PM on Wednesday,
April 4.
Review of Field Lines
Field lines start on
positive charges.
Q
Field lines stop on
negative charges.
-Q
More charge 
more field lines.
2Q
Field line spacing
indicates field
strength
+
+
weak
strong
+
Field lines
never cross.
Closed loops?
Energy flow.
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Physics 122B - Lecture 4
3
Recall : Computing Charged Object
E-Fields Using Coulomb’s Law
1. Choose a coordinate system that will facilitate
integration.
2. Use any applicable symmetries to set E-field
components to zero or equal to each other.
3. Break up the object into point-like elements.
4. Write the Coulomb’s Law contribution to the E-field
from a representative point-like element.
5. Integrate over the entire object to get the E-field.
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Recall: The E-Field of a Charged Ring
A thin uniformly charged ring of radius R has a
total charge Q. Find the electric field on the axis
of the ring .. the z-axis …(perpendicular to the
page).
The linear charge density of the ring is l =
Q/(2pR). The system has cylindrical symmetry for
rotations about the axis, so along the z-axis Ex=Ey=0
and we need only to find Ez. Consider a small
segment of the circumference of the ring of width
dl = R df. The contribution to the electric field on
the z-axis is :
1 dQ
1 l Rdf
z
dEz 
cos


4p 0 ri 2
4p 0 R 2  z 2 R 2  z 2
Ez 

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1
l Rz
4p 0  R 2  z 2 3/ 2
1
2p

0
df 
1
l Rz
2 0  R 2  z 2 3/ 2
Qz
4p 0  R 2  z 2 3/ 2
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5
The on-axis E-Field of a Charged
Ring (2)
Ez 
1
Qz
4p 0  R  z
2

2 3/ 2
Near-field limit: linear
1 Qz
Ez ( z  R) 
4p 0 R3
Far-field limit: 1/r2 (same as point charge)
1
Q
Ez ( z  R) 
4p 0 z 2
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Example: A Charged Ring
A thin uniformly charged ring of radius R = 0.1 m
has a total charge Q = 10 nC.
Find the electric field on the axis of symmetry at
z = 0.2 m from the center the ring.
Ez 
1
Qz
4p 0  R 2  z 2 3/ 2
-8
(1.0

10
C)(0.2 m)
9
2
2
 (9.0 10 N m /C )
[(0.1 m) 2  (0.2 m) 2 ]3/ 2
 1.61103 N/C
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The E-Field of a Charged Disk
A thin uniformly charged disk of radius R has a
total charge Q. Find the electric field E at points
on the z axis perpendicular to the disk plane.
The charge area-density of the disk is h = Q/A
= Q/(pR2). It has cylindrical symmetry for
rotations about the z axis, so Ex=Ey=0, and we need
only find Ez. Consider the disk to be made of
successive rings, each having radius r, thickness dr,
and charge dQ=h(2prdr). Then the contribution of
each ring to Ez is:
1
zdQ
1 zh (2p rdr )
dEz 
Ez 
zh
2 0
zh

2 0
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R

0
r
rdr
2
 z2 
3/ 2

1
1


z 2  R2
z
zh
4 0
4p 0  r  z
2
z 2  R2

z2

2 3/ 2

4p 0  r 2  z 2 3/ 2
du
zh  2
2
2
(using
u

r

z
,
du

2
rdr
)=

u 3/ 2
4 0  u
 h 
1

1



1  ( R / z )2
 2 0 

Q

2
 2p 0 R
Physics 122B - Lecture 4
z 2  R2
z2


1
1 

2

1  ( R / z ) 
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The E-Field of a Charged Disk (2)
1

h 
1
Ez 
1 

2
2 0 
1  ( R / z ) 


1

1 

2
2
2p 0 R 
1  ( R / z ) 
Q
Near-field limit: z<<R
h
Q
Ez ( z  R) 

2 0 2p 0 R 2
Far-field limit: z>>R
(1   )  (1   ) if  <<1

Ez ( z  R) 
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
EZ
0.5
0
-0.5
-1
-4
-2
0
z R
2
4
(E field is constant, independent of z)
1
1  ( R / z)2

 [1  ( R / z ) ]
Q
2
1 Q
1
2


1

(1

(
R
/
z
)
)

2
 4p z 2
2p 0 R 2 
0
Physics 122B - Lecture 4
1
2
 1  12 ( R / z ) 2 if R<<z
(E field ~ 1/z2, same
as a point charge Q)
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Example: A Charged Disk
A thin uniformly charged disk of radius R = 0.1 m has a
total charge Q = 10 nC.
Find the electric field on the axis of symmetry at z = 0.2
m from the center the disk.
Edisk

1 2Q 
1

1 

2
2
4p 0 R 
1  ( R / z ) 


1
 (9.0 10 N m /C )2(1.0 10 C) 1 
 /(0.1 m) 2

1  (1/ 2) 2 
 1.90 103 N/C
9
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2
2
-8
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An Infinite Charged Plane
An infinite plane of charge is the limiting case
of a disk of charge when the disk radius is allow
to become very large, i.e., R. As we have seen,
in that case:
Ez ( R  ) 
h
 constant
2 0
The E field must change direction across the
plane, so:
Eplane
 h
 2

0

 h

 2 0
kˆ if z>0
(independent of distance)
kˆ if z<0
Note that the system has translational symmetry
along any direction in the plane and rotational
symmetry about any perpendicular. Therefore, E
cannot depend on (x,y) position and Ex=Ey=0.
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Example: A Charged Plane
An infinite plane has a charge per unit area of
h = 1.0 nC/m2.
What is the electric field at a distance of 1.0
km above the plane.
Eplane
h

2 0
 12 4p (9.0 109 N m2 /C2 )(1.0 10-9 C/m2 )
 56.5 N/C
Note that the electric field does not depend on the
distance from the plane (because the field is constant).
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Question 1
Consider an infinite, uniformly
charged plane How do the
magnitudes of the E-fields
compare at the points
shown?
(a) Eb>Ec>Ed>Ee>Ea;
(b) Eb=Ec>Ed=Ee>Ea;
(c) Eb=Ec=Ed=Ee=Ea;
(d) Ea=Eb=Ec>Ed=Ee;
(e) Cannot tell without more information.
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A Sphere as a Set of Rings*
Charge per unit area
dE 
1
dE 
Q
Qring zring

1
dQ( z  R cos  )
4p 0  z 2  R 2  2 4p 0 ( z  R cos  )2  ( R sin  )2  2
ring 
 ring


Q
dQ  hTring Lring 
( Rd )(2p R sin  )  12 Q sin  d
2
4p R
y
3
3
8p 0  z 2  R 2  2 zR cos   2 Ring Length
Ring thickness


E
x
( z  R cos  )sin  d
3
Q
p

 0 if z  R
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R sin 

R cos 
( z  R cos  ) sin  d
8p 0 0  z 2  R 2  2 zR cos  


1 Q

if z  R
2
4p 0 z
R
3
2
z
* This calculation is not in Knight.
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Integral Trick
Q
d
=
p
( z  R cos  ) sin  d
E 13

dz
8p 0 0  z 2  R 2  2 zR cos   2


1 Q

if z  R
2
4
p
z
0
d
|R-z| - |R+z|
dz
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zR
 0 if z  R
Physics 122B - Lecture 4
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E-Field of a Charged Sphere
Therefore, the electric field outside the
surface of a uniformly spherically charged
shell of charge Q with radius R is:
Esphere
1
Q

rˆ
2
4p 0 r
(independent of R)
In other words, it has the same E-field
as that produced by a point charge Q
located at the center of the sphere.
If h=Q/4pR2 is the surface charge
density, then the surface E field is
Esurf=h/0, i.e., twice as big as the E field
of an infinite plane of charge. (Why?)
The book won’t tell you what the E field is inside the spherical shell,
but I will. It is zero (because the forces from opposite sides cancel).
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Example: A Charged Sphere
A sphere with a radius R = 0.1 m has a charge
Q = 20.0 nC on its surface.
What is the electric field at a distance of 1.0
m from the sphere?
Esphere
1
Q

4p 0 r 2
 (9.0 109 N m2 /C2 )(2.0 10-8 C)/(1.0 m) 2
 1.8 102 N/C
Note that the electric field does not depend on the radius of the
sphere (because the E-field of a charged sphere is the same as for
the same charge at a point at the center of the sphere.) The external
E-field tells you the charge but not the radius of the sphere.
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A Parallel-Plate Capacitor
The parallel plate capacitor is an important component
of electronic circuits. It is constructed of two
oppositely charged electrodes separated by a small gap.
The net charge of the capacitor is zero.
We can model the parallel plate capacitor as two parallel
infinite charged planes placed a distance d apart. Let the x
axis go from + to -. Then we have only to superpose the
fields that we have previously calculated. We find that in
the two regions outside the gap the superposed fields
cancel to give 0, while in the gap they add. Therefore:
Eoutside  E  E  iˆh / 2 0  iˆh / 2 0  0
Egap  E  E  2iˆh / 2 0  iˆh /  0
h  Q / A , so E capacitor  iˆ
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Q
0 A
Physics 122B - Lecture 4
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Capacitor Edge Effects
Since the electrodes of a real parallel plate capacitor are not infinite,
there are “edge effects” at the ends of the electrodes.
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Example:
A Parallel-Plate Capacitor
The parallel plate capacitor consists of two circular electrodes
of radius R=0.1 m separated by a gap of d=1 mm, with opposite
charges of magnitude Q=20 nC on the two electrodes.
What is the electric field between the plates?
Ecapacitor 
Q
0 A
A  p R2
R
1

0
Ecapacitor
(2.0 10-8 C)
 4p (9.0 10 N m /C )
p (0.1 m) 2
9
2
2
 7.2 104 N/C
(Notice that the electric field does not depend on the gap d.)
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Motion of a Charged Particle
in an Electric Field
Fon q  qE
a
Fon q
m

q
E
m
qE
If E is uniform, then a 
 constant
m
In an electric field, does a charged particle follow the field lines?
NO! Only a massless charged particle would follow the field lines.
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Example: An electron moving
across a capacitor
Two 6.0 cm diameter electrodes are
spaced 5.0 mm apart. They are charged
by transferring 1.0 x 1011 electrons from
one electrode to the other.
A electron is released from rest at
the surface of the negative electrode.
How long does it take the electron to
cross to the positive electrode? Assume
the space between the electrodes is a
vacuum.
E
h
Q
Ne


 0  0 A  0p R 2
x  d  12 a(t )2
 4(9.0 109 Nm 2 /C2 )(1.0  1011 )(1.6 10 19 C) /(0.03 m) 2
 6.39  10 N/C
5
eE (1.6 1019 C)(6.39 105 N/C)
a

 1.12 1017 m/s 2
-31
m
(9.1110 kg)
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Physics 122B - Lecture 4
2d
2(5.0 103 m)
t 

a
(1.12 1017 m/s 2 )
 2.98 1010 s
22
Example:
Deflecting an electron beam
An “electron gun” creates a beam of
electrons moving horizontally with a speed
vx = 3.34 x 107 m/s. The electrons enter a
2.0 cm long gap between two parallel
electrodes producing a downward electric
field of E = 5.0x104 N/C.
In what direction and by what angle is
the electron beam deflected?
eE (1.6 1019 C)(5.0 104 N/C)
15
2
ay 


8.78

10
m/s
m
(9.1110-31 kg)
t  L / vx  (0.02 m) /(3.34 107 m/s)  5.99 10-10 s
vy  ay t  (8.781015 m/s2 )(5.99 10-10 s)  5.26 106 m/s
(5.26 106 m/s)
  Arctan  Arctan
 8.95
7
vx
(3.34 10 m/s)
vy
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Physics 122B - Lecture 4
The deflection
is upward because
the negatively
charged electron
experiences a
force in the
opposite direction
from E.
23
End of Lecture 4
 Before the next lecture, read
Knight, Chapters 26.6 through 27.1
 Lecture Homework #1 is
submitted on the Tycho system
and is due at 10 PM on Wednesday,
April 4.