Electricity and Magnetism
Electric Fields
Lana Sheridan
De Anza College
Jan 12, 2018


Last time
• Forces at a fundamental level
• Electric field
• net electric field
• electric field lines


Warm Up Questions
Which of the following could be the charge on the particle hidden by the question mark?
far more work cian and physifields of a hypothetical ssian surface, calculations of ibution. For exsphere with a we discuss in
dE:
ace to the
on a Gaussian ited example, tward from the tely tells us that er a particle or
Fig. 23-1 A spherical Gaussian surface. If the electric field vectors are of uniform magnitude and point radially outward at all surface points, you can conclude that a net positive
Spherical Gaussian surface
?
E
(A) 0 C (B) −1 C
(C) −1.6 × 10−19 C (D) +1 μC
1Figure from Halliday, Resnick, Walker


Warm Up Questions
Which of the following could be the charge on the particle hidden by the question mark?
far more work cian and physifields of a hypothetical ssian surface, calculations of ibution. For exsphere with a we discuss in
dE:
ace to the
on a Gaussian ited example, tward from the tely tells us that er a particle or
Fig. 23-1 A spherical Gaussian surface. If the electric field vectors are of uniform magnitude and point radially outward at all surface points, you can conclude that a net positive
Spherical Gaussian surface
?
E
(A) 0 C (B) −1 C
(C) −1.6 × 10−19 C (D) +1 μC ←
1Figure from Halliday, Resnick, Walker


Overview
• Electric field lines
• Net electric field
• the effect of fields on charges
• the electric dipole


Field Lines
The electrostatic field caused by an electric dipole system looks something like:
are intersections of these surfaces with the page) and elecendicular to the electric field lines at every point.
metric electric a point charge
An electric field produced by an electric dipole
c
Notice that the lines point outward from a positive charge and inward toward a negative charge.
1Figure from Serway & Jewett


Field Lines
es. Figure 23.22 The electric field
!2q "q
one that terminates on "q.
!"
1Figure from Serway & Jewett


Field Lines
Compare the electrostatic fields for two like charges and two opposite charges:
FIELDS
tive r.
e ing age. ec
ic t it int.
positive test charge at any point near the of Fig. 22-3a, the net electrostatic force acti the test charge would be perpendicular t sheet, because forces acting in all other tions would cancel one another as a res the symmetry. Moreover, the net force o test charge would point away from the sh shown. Thus, the electric field vector at any in the space on either side of the sheet i perpendicular to the sheet and directed from it (Figs. 22-3b and c). Because the cha uniformly distributed along the sheet, a field vectors have the same magnitude. Such an electric field, with the same nitude and direction at every point, is a uniform electric field. Of course, no real nonconducting sheet (such as a flat expanse of plastic)
E
+
+
tric field it represents are said to have rotational symmetry about that axis.The electric field vector at one point is shown; note that it is tangent to the field line through that point.
Fig. 22-5 Field lines for a positive point charge and a nearby negative point charge that are equal in magnitude.The charges at
fi n
n n
s s fi
2
T a F
T
+

E


Field Lines
Compare the fields for gravity in an Earth-Sun system and electrostatic repulsion of two charges:
CTRIC FIELDS
ual positive
ach other.
gative
-dimen
lly rotate
xis passing
of the page.
d the elec
avreota
e electric
note that it
that point.
positive test charg
of Fig. 22-3a, the ne
the test charge wo
sheet, because for
tions would cance
the symmetry. Mo
test charge would
shown. Thus, the el
in the space on ei
perpendicular to
from it (Figs. 22-3b
uniformly distribu
field vectors have the same magnitude. Such an ele
nitude and direction at every point, is a uniform elec
E
+
+
14:16 Page 582
1Gravity figure from http://www.launc.tased.edu.au ; Charge from Halliday, Resnick, Walker


Field Lines: Uniform Field
Imagine an infinite sheet of charge. The lines point outward from the positively charged sheet.
F
E
++++
++++
++++
++++
Positive test charge
(a) (b)
+++
++++
+++
+
++
+
+++++++++
(c)
++
+
-2 were of uniform positive charge, the electric field the sphere would be directed radially away from field lines would also extend radially away from the llowing rule:
way from positive charge (where they originate) and re they terminate).
t of an infinitely large, nonconducting sheet (or plane) of positive charge on one side. If we were to place a
tic force a very unirge on ector e, and ce tend d
E:
1Figure from Halliday, Resnick, Walker.


Electric field due to an Infinite Sheet of Charge
Suppose the sheet is in air (or vacuum) and the charge density on the sheet is σ (charge per unit area):
E= σ
20
It is uniform! It does not matter how far a point P is from the sheet, the field is the same.
E
+
++
++
++
+++++++++
+
)


Field Lines: Uniform Field
The field from two infinite charged plates is the sum of each field.
s
P0
(24.8)
!
!
!
!
! ! ! ! ! ! ! !
.14 (Example 24.5) field lines due to an
!
!
!
!
! ! ! ! ! ! ! !
"
"
"
"
" " " " " " " "
Figure 24.15 (Example 24.5) The electric field lines between
The field in the center of a parallel plate capacitor is nearly uniform.


Free charges in an E-field
The force on a charged particle is given by F = qE.
If the charge is free to move, it will accelerate in the direction of the force.
Example: Ink-jet printing
Input signals Deflecting plate
G C Deflecting plate
E
e. When a charged oil drop drifted into chamber C through the hole in plate P1, its motion could be controlled by closing and opening switch S and thereby setting up or eliminating an electric field in chamber C. The microscope was used to view the drop, to permit timing of its motion.
Fig. 22-15 Ink-jet printer. Drops shot
ing plate P2 chamber C. charged dro ular, our ne By timi and thus de values of q w
in which e charge, 1.60 quantized, a Modern me experiment
Ink-Jet Pri
The need alternative


Motion of a Charged Particle in an E-field
n he

e cd d is a
(0, 0)
!
(x, y)
vi iˆ !
! vS
x
y
The electron undergoes a downward acceleration (opposite E), and its motion is parabolic while it is between the plates.
S
ES
""""""""""""
!!!!!!!!!!!!
Figure 23.24 (Example 23.11) An electron is projected horizontally into a uniform electric field produced by two charged plates.
Trajectory is a parabola: similar to projectile motion.


Motion of a Charged Particle in an E-field
(a) What is the acceleration of an electron in the field of strength E?
as shown N/C. The
he elec
ing one perpen he veloc tric field a curved lectron is tally in a
ecause the electric field is uniform, a constant electric force is electron, we can model it as a particle under a net force.
(0, 0)
!
(x, y)
vi iˆ !
! vS
x
y
The electron undergoes a downward acceleration (opposite E), and its motion is parabolic while it is between the plates.
S
ES
""""""""""""
!!!!!!!!!!!!
Figure 23.24 (Example 23.11) An electron is projected horizontally into a uniform electric field produced by two charged plates.
(b) The charge leaves the field at the point (`, yf ). What is yf in terms of `, vi , E , e, and me?


Motion of a Charged Particle in an E-field
(a) What is the acceleration of an electron in the field of strength E?
as shown N/C. The
he elec
ing one perpen he veloc tric field a curved lectron is tally in a
ecause the electric field is uniform, a constant electric force is electron, we can model it as a particle under a net force.
(0, 0)
!
(x, y)
vi iˆ !
! vS
x
y
The electron undergoes a downward acceleration (opposite E), and its motion is parabolic while it is between the plates.
S
ES
""""""""""""
!!!!!!!!!!!!
Figure 23.24 (Example 23.11) An electron is projected horizontally into a uniform electric field produced by two charged plates.
(b) The charge leaves the field at the point (`, yf ). What is yf in terms of `, vi , E , e, and me?
yf = − eE `2
2me v 2
i


Trick for working out Net field
Look for symmetry in the problem.
To find the E-field, usually several components (Ex , Ey , Ex ) must be found independently.
If the effects of charges will cancel out, you can neglect those charges.
If the effects of charges cancel out in one component, just worry about the other component(s).


Question about net field
charg magn
Fig. 22-20 Question 1.
2 Figure 22-21 shows two square arrays of charged particles. The squares, which are centered on point P, are misaligned. The particles are separated by either d or d/2 along the perimeters of the squares. What are the magnitude and direction of the net electric field at P?
+6q
–2q
+3q –2q
+3q
–q
+6q
–2q
–3q
–q
+2q –3q
+2q
–q
P
Fig. 22-21 Question 2.
5F axis. than a po field their
1Figure from Halliday, Resnick, Walker, page 597, problem 2.


Electric Dipole
electric dipole
A pair of charges of equal magnitude q but opposite sign, separated by a distance, d.
dipole moment:
p = qd rˆ
where rˆ is a unit vector pointing from the negative charge to the positive charge.
z
r(–)
r(+)
E(+)
d
z
–q
+q
P
++
–
p
E(–)
Dipole
center
Up here the +q field dominates.
Down here the –q
We are usually inte that are large compared that z # d. At such larg proximation, we can neg
The product qd, wh dipole, is the magnitude of the dipole. (The uni
The direction of is t dipole, as indicated in
p:
p:
E% q
4
&
'0z


Electric Dipole (Example 23.6, B)
Evaluate the electric field from the dipole at point P, which is at
position (0, y ).
uq1u 5 uq2u and a 5 b.
P
y
r
a
q a –q
x
u
u
uu
!"
ES
E2
S
E1
S
ure 23.13 (Example 23.6) en the charges in Figure 2 are of equal magnitude equidistant from the origin, situation becomes symmets shown here.
y 5 E 1y 1 E 2y 5 ke
0 q1 0
a 2 1 y 2 sin f 2 ke
0 q2 0
b 2 1 y 2 sin u


Electric Dipole (Example 23.7)
P
y
r
a
q a –q
x
u
u
uu
!"
ES
E2
S
E1
S
)
n,
q1 0
1 y 2 sin f 2 ke
0 q2 0
b 2 1 y 2 sin u
The y -components of the electric field cancel out, Ey = 0.
x -components:
Ex = E1,x + E2,x
Also E1,x = E2,x


Electric Dipole (Example 23.7)
P
y
r
a
q a –q
x
u
u
uu
!"
ES
E2
S
E1
S
)
n,
q1 0
1 y 2 sin f 2 ke
0 q2 0
b 2 1 y 2 sin u
The y -components of the electric field cancel out, Ey = 0.
x -components:
Ex = E1,x + E2,x
Also E1,x = E2,x
Ex = 2
( ke q
r 2 cos θ
)
= 2ke q
(a2 + y 2)
(
a
√a2 + y 2
)
= 2ke a q
(a2 + y 2)3/2


Electric Dipole (Example 23.7)
P
y
r
a
q a –q
x
u
u
uu
!"
ES
E2
S
E1
S
0
y 2 sin f 2 ke
0 q2 0
b 2 1 y 2 sin u
What happens as we move infinitely far from the dipole? (y >> a)


Electric Dipole (Example 23.7)
P
y
r
a
q a –q
x
u
u
uu
!"
ES
E2
S
E1
S
0
y 2 sin f 2 ke
0 q2 0
b 2 1 y 2 sin u
What happens as we move infinitely far from the dipole? (y >> a)
The constant a in the denominator has less and less affect on the function. We can see that the field function approaches Efar = 2ke a q
y3
yli→m∞
[E
Efar
]
= yli→m∞


2ke a q (a2+y 2)3/2
2ke a q y3


= yli→m∞

  
2ke a q
y3
(
(a
y )2+1
)3/2
2ke a q y3

  
=1


Electric Dipole (Example 23.7)
P
y
r
a
q a –q
x
u
u
uu
!"
ES
E2
S
E1
S
0
y 2 sin f 2 ke
0 q2 0
b 2 1 y 2 sin u
What happens as we move infinitely far from the dipole? (y >> a)
The constant a in the denominator has less and less affect on the function. We can see that the field function approaches Efar = 2ke a q
y3
yli→m∞
[E
Efar
]
= yli→m∞


2ke a q (a2+y 2)3/2
2ke a q y3


= yli→m∞

  

2ke a q
y3(
(a
y )2+1
)3/2


2ke a q y3

  
=1


Big-O Notation (Example 23.7)
P
y
r
a
q a –q
x
u
u
uu
!"
ES
E2
S
E1
S
0
y 2 sin f 2 ke
0 q2 0
b 2 1 y 2 sin u
y >> a
Recall that f (x) = O(g (x)) if
∣ ∣ ∣
f (x) g (x)
∣ ∣
∣6C
∀ x > k.


Big-O Notation (Example 23.7)
P
y
r
a
q a –q
x
u
u
uu
!"
ES
E2
S
E1
S
0
y 2 sin f 2 ke
0 q2 0
b 2 1 y 2 sin u
y >> a
Recall that f (x) = O(g (x)) if
∣ ∣ ∣
f (x) g (x)
∣ ∣
∣6C
∀ x > k.
∣ ∣ ∣ ∣
E
Efar
∣ ∣ ∣ ∣
=
∣ ∣ ∣ ∣ ∣ ∣
2ke a q (a2+y 2)3/2
2ke a q y3
∣ ∣ ∣ ∣ ∣ ∣
=
∣ ∣ ∣ ∣ ∣ ∣
(
(a
y
)2
+1
)−3/2∣
∣ ∣ ∣ ∣ ∣
Choosing k = a we can see:
∣ ∣ ∣ ∣
E
Efar
∣ ∣ ∣ ∣
61
2√2 ∀ y > a
Therefore, E = O
( 2ke a q y3
)
or simply O(y −3).


Electric Dipole (Example 23.7)
As we move away from the dipole (red line, r −3) the E-field falls off faster than it does for a point charge (blue line, r −2).
The negative charge partially shields the effect of the positive charge and vice versa.


Summary
• electric field lines
• the effect of fields on charges
• the electric dipole
Homework
• Collected homework 1, posted online, due on Monday, Jan 22.
Serway & Jewett:
• NEW: Ch 23, onward from page 716. Probs: 36, 51, 61, 79
• Understand examples 23.8 and 23.9.