Physics 122B
Electricity and Magnetism
Lecture 5
E-fields and Charged Particles in E-Fields
April 04, 2007
Martin Savage
Lecture 5 Announcements
 Lecture Homework #1 have been posted on the
Tycho system. It is due at 10 PM tonight.
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
4 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/4R2 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 electric field inside the spherical-shell is zero (because the sum of
the forces cancel).
7/19/2016
Physics 122B - Lecture 5
3
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

4 0 r 2
 (9.0  109 N m 2 /C 2 )(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 that
of the same charge at a point at the center of the sphere.) Measuring
the external E-field tells you the charge but not the radius of the
sphere.
7/19/2016
Physics 122B - Lecture 5
4
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
7/19/2016
Q
0 A
Physics 122B - Lecture 5
5
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.
7/19/2016
Physics 122B - Lecture 5
6
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   R2
R
1

0
Ecapacitor
(2.0 10-8 C)
 4 (9.0 10 N m /C )
 (0.1 m) 2
9
2
2
 7.2 104 N/C
(Notice that the electric field does not depend on the gap d.)
7/19/2016
Physics 122B - Lecture 5
7
Motion of a Charged Particle
in an Electric Field
Fon q  qE
a
Fon q

q
E
m
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.
7/19/2016
Physics 122B - Lecture 5
8
Example: An electron moving
across a capacitor
Two 6.0 cm diameter circular 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  0 R 2
x  d  12 a(t )2
 4(9.0  109 Nm 2 /C 2 )(1.0  1011 )(1.6 1019 C) /(0.03 m) 2
 6.39  105 N/C
eE (1.6 1019 C)(6.39 105 N/C)
a

 1.12 1017 m/s 2
-31
m
(9.1110 kg)
7/19/2016
Physics 122B - Lecture 5
2d
2(5.0 103 m)
t 

a
(1.12 1017 m/s 2 )
 2.98 1010 s
V/c = a  t/c ~ 0.1
9
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? (Neglect fringing)
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
7/19/2016
Physics 122B - Lecture 5
The deflection
is upward because
the negatively
charged electron
experiences a
force in the
opposite direction
from E.
10
Motion in a Nonuniform Field
The motion of a charged particle in a non-uniform
field can be quite complicated. However, one case is
easy to analyze: the circular orbit of a charged
particle around another point charge, a charged
sphere, or a long straight charged wire (so that the
field at some distance r is constant). In such cases,
mv 2
Fnet 
qE
r
Example – An electron in a hydrogen atom orbits at a radius
of r=0.053 nm. What is the orbital velocity of the electron?
qE 
2
1 e
mv

4 0 r 2
r
7/19/2016
2
ve
1
1
4 0 mr
 (1.6 10
-19
(9.0 109 N m 2 /C 2 )
6
C)

2.18

10
m/s
-31
-11
(9.1110 kg)(5.3 10 m)
Physics 122B - Lecture 5
11
Question 1
Which field is responsible for the trajectory
of the proton as shown?
7/19/2016
Physics 122B - Lecture 5
12
Dipole in a Uniform Field
Consider a permanent electric dipole of dipole
moment p=qs in a uniform external electric field E.
F  qE
F  qE
Fnet  F  F  0
p
p
If the axis of the dipole is not aligned with
the field (a), the dipole will experience a
torque. If the dipole is aligned with the field
(b) it will not experience a torque, and it is in
stable equilibrium. In either case, there is no
net force on the dipole.
When dipoles, e.g., water molecules, are
placed in an external field, they will tend to
align with the field as shown, so that p||E.
7/19/2016
Physics 122B - Lecture 5
13
Torque on a Dipole
in a Uniform Field
If the axis of the dipole is not aligned
with the field, the dipole will experience a
torque . The dipole moment is:
p  qs where s goes from - q to  q
  F  ( s sin  )(qE )  pE sin 
  p E
7/19/2016
Physics 122B - Lecture 5
14
Example: Angular acceleration
of a dipole dumbbell
Two 1.0 g spheres are connected by a 2.0 cm
long insulating rod of negligible mass. The
spheres have equal and opposite charges of
10 nC. The rod is held in a uniform field of
E = 1.0x104 N/C at an angle of 300 to the field,
as shown, then released.
What is the angular acceleration a of the
system?
p  qs  (1.0 10-8 C)(0.02 m)  2.0 10-10 C m.
  pE sin   (2.0 10-10 C m)(1.0 104 N/C )sin 30  1.0 10-6 N m
I  2m(s / 2)2  2(0.001 kg)(0.01 m)2  2.0 10-7 kg m2
a   / I  (1.0 10-6 N m) /(2.0 10-7 kg m2 )  5.0 rad/s 2
7/19/2016
Physics 122B - Lecture 5
15
Dipole in a Nonuniform Field
F  qE
F  qE
Fnet  F  F  0 if E  E
In the non-uniform field of a point charge,
the dipole will first align with the field (p||E)
and then be attracted by it.
The net force on the dipole will be
attractive, independent of whether the field is
from a negative or a positive point charge. In
general, a dipole will experience a net force
toward any charged object.
7/19/2016
Physics 122B - Lecture 5
16
Example:
The force on a water molecule
The water molecule H2O has a
permanent dipole moment of
p = 6.2x10-30 C m. A water molecule
is located 10 nm from a Na+ ion in a
salt water solution.
What force does the ion exert on
the water molecule?
Edipole 
1
2p
4 0 r 3
Trick: Calculate force on
Na ion from dipole field.
Fdipole on ion  eEdipole 
1 2ep
4 0 r 3
2(1.6 10-19 C)(6.2 10-30 C m)
 (9.0 10 N m /C )
(1.0 10-8 m)3
9
2
2
 1.79 1014 N
 Fion on dipole  Fdipole on ion  1.79 1014 N
7/19/2016
Physics 122B - Lecture 5
17
End of Lecture 5
 Before the next lecture on Wednesday, read
Knight, Chapters 27.2 through 27.4
 Lecture Homework #1 should be submitted on
the Tycho system by tonight at 10 PM .