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Chapter 23
Electric Charge and Electric Fields
• What is a field?
• Why have them?
• What causes fields?
Field Type
gravity
electric
magnetic
Caused By
mass
charge
moving charge
Electric Charge
• Types:
– Positive
• Glass rubbed with silk
• Missing electrons
– Negative
• Rubber/Plastic rubbed with fur
• Extra electrons
• Arbitrary choice
– convention attributed to ?
• Units: amount of charge is measured in
[Coulombs]
• Empirical Observations:
– Like charges repel
– Unlike charges attract
Charge in the Atom
•
•
•
•
Protons (+)
Electrons (-)
Ions
Polar Molecules
Charge Properties
• Conservation
– Charge is not created or destroyed, only transferred.
– The net amount of electric charge produced in any
process is zero.
• Quantization
– The smallest unit of charge is that on an electron or
proton. (e = 1.6 x 10-19 C)
• It is impossible to have less charge than this
• It is possible to have integer multiples of this charge
Q  Ne
Conductors and Insulators
• Conductor
transfers charge on contact
• Insulator
does not transfer charge on contact
• Semiconductor might transfer charge on contact
Charge Transfer Processes
• Conduction
• Polarization
• Induction
The Electroscope
Coulomb’s Law
• Empirical Observations
F  q1q 2
1
F 2
r
Direction of the force is along the line joining the two charges
• Formal Statement
kq1q 2
F12  2 rˆ21
r21
Active Figure 23.7
(SLIDESHOW MODE ONLY)
Coulomb’s Law Example
• What is the magnitude of the electric force
of attraction between an iron nucleus
(q=+26e) and its innermost electron if the
distance between them is 1.5 x 10-12 m
Hydrogen Atom Example
• The electrical force between the
electron and proton is found from
Coulomb’s law
– Fe = keq1q2 / r2 = 8.2 x 108 N
• This can be compared to the
gravitational force between the electron
and the proton
– Fg = Gmemp / r2 = 3.6 x 10-47 N
Subscript Convention
kq1q 2
F12  2 rˆ21
r21
F12
r̂21
+q2
+q1
r21
F12  force on charge q1 due to charge q2
r21  distance from charge q2 to charge q1
r̂21  unit vector oriented from charge q2 to charge q1
More Coulomb’s Law
kq1q 2
F12  2 rˆ21
r21
r̂21 
F12
r̂21
r21
r21
+q2
r21
+q1
r21  r21
Coulomb’s constant:
permittivity of free space:
Charge polarity:
2
2
N

m
N

m
1
9
k  8.988x109

9.0x10

2
C2
C
4o
2
1
C
o 
 8.85x1012
4k
N  m2
Same sign
Opposite sign
Force is right
Force is Left
Electrostatics --- Charges must be at rest!
Superposition of Forces
F0  F01  F02  F03  ....
+Q1
+Q2
+Q3
r10
r20
r30
F03
F02
+Q0
F01
kq 0 q1
kq 0q 2
kq 0q 3
F0  2 rˆ10  2 rˆ20  2 rˆ30  ....
r10
r20
r30
N
 q1

q3
q2
qi
F0  kq0  2 rˆ10  2 rˆ20  2 rˆ30  ....   kq 0  2 rˆi0
r20
r30
i 1 ri0
 r10

Coulomb’s Law Example
y
+
Q
F1
L
F3
+
Q F2
L
Q
Q
+
+
x
• Q = 6.0 mC
• L = 0.10 m
• What is the magnitude
and direction of the net
force on one of the
charges?
Zero Resultant Force,
Example
• Where is the resultant
force equal to zero?
– The magnitudes of the
individual forces will be
equal
– Directions will be
opposite
• Will result in a
quadratic
• Choose the root that
gives the forces in
opposite directions
Electrical Force with Other
Forces, Example
• The spheres are in
equilibrium
• Since they are separated,
they exert a repulsive
force on each other
– Charges are like charges
• Proceed as usual with
equilibrium problems,
noting one force is an
electrical force
Electrical Force with Other
Forces, Example cont.
• The free body
diagram includes the
components of the
tension, the electrical
force, and the weight
• Solve for |q|
• You cannot determine
the sign of q, only
that they both have
same sign
The Electric Field
• Charge particles create forces on each other without ever
coming into contact.
» “action at a distance”
• A charge creates in space the ability to exert a force on a
second very small charge. This ability exists even if the
second charge is not present.
• We call this ability to exert a force at a distance a “field”
• In general, a field is defined:
Force
Why in the limit?
Field  lim
test quantity 0 test quantity
• The Electric Field is then:
F
N
E  lim
 C 
q 0 q
Electric Field near a Point Charge
kq o q
F  2 rˆ
r
F
E  lim
qo 0 q
o
kqq o
r̂
2
kq
r
E  lim
 2 rˆ
q o 0
qo
r
Electric Field Vectors
-Q
+Q
Electric Field Lines
Active Figure 23.13
(SLIDESHOW MODE ONLY)
Rules for Drawing Field Lines
• The electric field, E , is tangent to the field lines.
• The number of lines leaving/entering a charge is
proportional to the charge.
• The number of lines passing through a unit area normal
to the lines is proportional to the strength of the field in
that region.
# of electric field lines
E
Area 
• Field lines must begin on positive charges (or from
infinity) and end on negative charges (or at infinity).
The test charge is positive by convention.
• No two field lines can cross.
Electric Field Lines, General
• The density of lines
through surface A is
greater than through
surface B
• The magnitude of the
electric field is greater on
surface A than B
• The lines at different
locations point in different
directions
– This indicates the field is
non-uniform
Example Field Lines
+
Dipole
+
+
+
Line Charge
For a continuous linear charge distribution,
Linear Charge Density:
+

Q
dq

dx
Active Figure 23.24
(SLIDESHOW MODE ONLY)
More Field Lines
Q dq

A dA
Surface Charge Density:

Volume Charge Density:
Q dq
 
V dV
Superposition of Fields
E0  E01  E02  E03  ....
+q1
+q2
+q3
r10
E03
E02
r20
r30
0
E 01
kq 3
kq1
kq 2
E 0  2 rˆ10  2 rˆ20  2 rˆ30  ....
r10
r20
r30
N
 q1

q3
q2
qi
E0  k  2 rˆ10  2 rˆ20  2 rˆ30  ....   k  2 rˆi0
r20
r30
i 1 ri0
 r10

Superposition Example
• Find the electric field due
to q1, E1
• Find the electric field due
to q2, E2
• E = E1 + E2
– Remember, the fields add
as vectors
– The direction of the
individual fields is the
direction of the force on a
positive test charge
Electric Field of a Dipole (ex. 23.6)
E  E  E
E
kq
E  E 

y  a 
2
2
E
y
E  E x  E x  2E cos 


-q
2a
p
p  2aQ
+q
kq
E2 2 2
y a
y  2a
a
y a
2
2
kp

E
 y  a 
2
kp
y3
2
3
2
P23.19
Three point charges are
arranged as shown in
Figure P23.19.
(a) Find the vector electric
field that the 6.00-nC
and –3.00-nC charges
together create at the
origin.
(b) (b) Find the vector force
on the 5.00-nC charge.
Figure P23.19
P23.52
Three point charges are
aligned along the x axis
as shown in Figure
P23.52. Find the electric
field at
(a) the position (2.00, 0) and
(b) the position (0, 2.00).
Figure P23.52
FIG. P23.52(a)
E1 
keq
2
r
 8.99  10
ˆ
r
9
N m
2

C 2 4.00  109 C
 2.50 m 
2
 iˆ
 5.75ˆ
iN C
E2 
keq
2
r
 8.99  10
ˆ
r
9
 8.99  10

9
E3
N m
2
N m
2

C 2 5.00  109 C
 2.00 m 
2

C 2 3.00  109 C
 1.20 m

2
ER  E1  E2  E3  24.2 N C
E1 
E2 
E3 
keq
 iˆ 11.2 N
 iˆ 18.7 N
ˆ
C i
in +x direction.


2
ˆ 0.970ˆ
ˆ  8.46 N C  0.243i
r
j
2
ˆ  11.2 N C   ˆ
r
j
r
keq
r
keq
r2
ˆ   5.81 N
r
 
ˆ+0.928ˆ
C   0.371i
j
Ex  E1x  E3x  4.21ˆ
iN C
ER  9.42 N C
ˆ
C i
Ey  E1y  E2y  E3y  8.43ˆ
jN C
  63.4 above  x axis
P23.19
(a)
E1 
E2 
ke q1
r12
ke q2
r22
 
 8.99  10  3.00  10 
 
8.99 10 6.00  10 
9
9
ˆ
j
 0.100
2
9
9
ˆ
i
 0.300

2
 ˆj    2.70  10

3
N C ˆ
j
 iˆ    5.99 10
ˆ
N C i
2
 

ˆ 2.70  103 N C ˆ
E  E 2  E1   5.99  102 N C i
j
(b)



ˆ 2 700ˆ
F  qE  5.00  109 C 599i
j N C


ˆ 13.5  106 ˆ
F  3.00  106 i
j N 
 3.00iˆ 13.5ˆj N

Continuous Charge Distributions
+Q1
+Q2
+Q3
kq
E 0  2 rˆ
r
kdq
dE 0  2 rˆ
r
Single charge
Single piece of a charge distribution
r10
E03
E02
r20
0
r30
E 01
+
+
+
+
dq
dE0
0
N
qi
E0  k  2 rˆi0
i 1 ri0
dq
E0  k 
rˆ
2
r
all charge
Discrete charges
Continuous charge distribution
Finding dq
kdq
dE  2 rˆ
r
Line charge

Q
dq

dx
Surface charge

Q dq

A dA
Volume charge
Q dq
 
V dV
Cartesian
Polar
dq  dx
dq  Rd
dq  dxdy
dq  rdrd
dq  dxdydz
dq  rdrddz
dq  r 2 sin drdd
Example – Infinitely Long Line of Charge
+
+
kdq
dE  2 rˆ
r
dq  dy
dy
r x y
2
2
y-components cancel by symmetry
2
y +

+
+
+
dEx
x
dE y
dE
+
dE x 
kdq
cos 
2
r
k  dy 
x
dE  2
x  y2 x 2  y2
+

E  kx 

dy
x
2
y
3
2 2

 2  2k
 kx  2  
x
x 
Example – Charged Ring (ex 23.8)
d
+ 
a
r  x a
2
+
2
y-components cancel by symmetry
2

+
+
kdq
dE  2 rˆ
r
dq  ds  ad
dEx
x
+
+
dE x 
k  ad 
x
dE  2
x  a2 x2  a2
dE
dE y
kdq
cos 
2
r
+
E
2
kxa
x
2
a
 d 
3
2 2 0

kxa
x
2
a
3
2 2

 2  
kQx
x
2
a
3
2 2

Check a Limiting Case
E
kQx
x
2
R
3
2 2

x  R
When:
The charged ring must look like a point source.
E
kQx
2
x
 R
3
2 2


kQx
0
 2  R 2 
 x 1  2  
x 
 
3
2

kQx kQ
 2
3
x
x
Uniformly Charged Disk (ex. 23.9)
E
R
kQx
x
2
r
3
2 2

r
x
dE 
dE
kxdq
x
2
r
3
2 2

dq  dA  rdrd  2rdr
dE 
kx2rdr
x
2
r
R
E
3
2 2

0
kx2rdr
x
2
r
3
2 2

R
 kx
0
x2 R2
2rdr
x
2
r
3
2 2

 kx

du
x
u
2
3
2
x2 R2
x2 R2
 kx

x2
3

2
u du  kx
u


1
2
1
2



1
1 
x
 2kx 

  k2 1 

2
2
2
2
2
x

R
x
x

R




x2
Binomial Expansion Theorem
1  x 
x  1
n
n  n  1 2
 1  nx 
x  ...
2!
Quadratic terms and higher are small
1  x 
n
 1  nx
Two Important Limiting Cases
R  x
Large Charged Plate:
R
r
x
dE


x
1

E  k2 1 

k

2



2



2
2
4o
2o
x

R


Very Far From the Charged Plate:




x
x

E  k2 1 

k

2

1


2
2

x

R
R2


 x 1 2
x

x  R

1


2 2 
  k2 1  1  R  
 

x2  




  1 R 2 
 1 R 2  kR 2 kQ
 k2 1  1 
 k2 

 2
2 
2 
2
x
x
2 x 
  2 x 
Parallel Plate Capacitor
E0
+Q

E
2 o
E

o
-Q
E0
Motion of Charged Particles in a Uniform
Electric Field
F
q 0 q
E  lim
F  qE  ma
qE
a
m
+Q
-Q
v2  v02  2a  x  x 0 
-e
x
v x  2a x x  2
eE
m
x
Example
• A proton accelerates from rest in
a uniform electric field of 500
N/C. At some time later, its
speed is 2.50 x 106 m/s.
+Q
-Q
– Find the acceleration of the
proton.
– How long does it take for the
proton to reach this speed?
– How far has it moved in this time?
– What is the kinetic energy?
x
e
Motion of Charged Particles in a Uniform
Electric Field
v
+Q
F
E  lim
q 0 q
F  qE  ma

vx0
-e
-Q
ay 
eE
m
e E t
2
2
2

v  vx  vy  vx0  
 m 


v x  v x0  a x t  v x0
v y  v y0  a y t 
eE
m
t
eE t

  tan 
 mv x0 


1
2
Active Figure 23.26
(SLIDESHOW MODE ONLY)
Motion of Charged Particles in a Uniform
Electric Field
+Q
+Q
-Q
Phosphor
Screen
-e
-e
x
This device is known as a
cathode ray tube (CRT)
-Q
Summary
Point Charges:
Coulomb’s Law
Electric Field
kq1q 2
F12  2 rˆ21
r21
F
qo 0 q
o
E  lim
These can be used to find the fields in the vicinity of continuous charge distributions:
+
+
+
+
Line of Charge:
Charged Plate:
R
r
x
dE
2k
E
x

E
2 o
Dipole:
p
kp
E
+
 r  a 
2
2
3
2