Transcript الشريحة 1 - جامعة فلسطين

General Physics II
1
General Physics II: Electricity & Magnetism
I.
Course Description
Coulomb's law, the electrostatic field, Gauss’s Law, the electrostatic
potential, capacitance and dielectrics, electric current, resistance and
electromotive force, direct current circuits, magnetic field and magnetic
forces, sources of magnetic fields, Ampere's Law, Faraday's Law, induction
and Maxwell's equations.
II. Course Objectives
1. To provide a foundation in physics necessary for further study in
science, engineering and technology.
2. To provide an appreciation of the nature of physics, its methods and
its goals.
3. To contribute to the development of the student's thinking process
through the understanding of the theory and application of this
knowledge to the solution of practical problems.
2
Textbook: the class notes beside the following textbooks:
1. Physics for Scientists and Engineers, Raymond A.
Serway, 6th Edition
2. University Physics, Sears, Zemansky and Young
Course Outline
 Charge and Matter: Charge and conservation of charges, Material and
charge, electric forces, Coulombs Law
The Electric Field: Electric field, The lines of forces, Electric dipole,
Continuous charge distribution, Effect of electric field on point charge, Milliken’s
Experiment.
 Electric Flux and Gauss’s Law: Flux of electric field, Gauss’s Law and its
application.
3
 Electric Potential: Definition of electric potential, Potential difference
between two points, Calculation of electric potential, Electric potential energy,
Electric field and potential.
 Capacitors: Capacitor and capacitance, Parallel plate capacitor, Cylindrical
capacitor, Spherical capacitor, Capacitors connections, Energy stored in
capacitor, Effect of insulator inside a capacitor.
 Electric current and Ohm’s Law: Electric current, Current density,
Resistivity, Ohm’s Law, Electric power, Electromotive force and electric circuits,
Kirchoff´s Law, RC circuit
 The Magnetic Field: Definition and introduction, Flux of magnetic field,
Magnetic force, Hall effect, Torque on a current loop, charge in a magnetic field,
Biot-Savart Law, Helmholtz coils, Ampere’s Law.
4
Faraday's Law : Faraday's Law of Induction, Motional emf, Lenz's Law,
Induced emf and Electric Fields, Generators and Motors/ Eddy Currents,
Maxwell's Equations
 Inductance : Self-Inductance, RL Circuits, Energy in a Magnetic Field,
Mutual Inductance, Oscillations in an LC Circuit, The RLC Circuit.
Alternating Current Circuits :AC Sources, Resistors in an AC Circuit,
Inductors in an AC Circuit, Capacitors in an AC Circuit, The RLC Series Circuit.
Power in an AC Circuit, Resonance in a Series RLC Circuit, The Transformer
and Power Transmission, Rectifiers and Filters.
5
GRADING POLICY
Your grade will be judged on your performance in Home work, Quizzes, tow
tests and the Lab. Points will be allocated to each of these in the following
manner:
GRADING SCALE:
Grade Component
Weight
HW/Quizzes
20
Midterm Exam
20
Final Exam
60
Total
100
6
Lecture I
Electrostatic
7
Introduction



Knowledge of electricity dates back to Greek
antiquity (700 BC).
Began with the realization that amber when
rubbed with wool, attracts small objects.
This phenomenon is not restricted to
amber/wool but may occur whenever two
non-conducting substances are rubbed
together.
8
Net Electrical Charge
Matters are made of atoms. An atom is basically
composed of three different components : electrons,
protons, and neutrons. An electron can be removed easily
from an atom
Normally, an atom is electrically neutral, which means that
there are equal numbers of protons and electrons. Positive
charge of protons is balanced by negative charge of
electrons. It has no net electrical charge.
When atoms gain or lose electrons, they are called "ions."
A positive ion is a cation that misses electrons.
A negative ion is an anion that gains extra electrons.
9
What is charge?
Objects that exert electric forces are said to have charge.
Charge is the source of electrical force. There are two kinds
of electrical charges, positive and negative. Same charges
(+ and +, or - and -) repel and opposite charges (+ and -)
attract each other.
10
The Law of Conservation of Charge
The Law of conservation of charge states that the
net charge of an isolated system remains
constant.
Charged Objects
When two objects are rubbed together, some electrons
from one object move to another object. For example,
when a plastic bar is rubbed with fur, electrons will move
from the fur to the plastic stick. Therefore, plastic bar will
be negatively charged and the fur will be positively
charged.
11
12
Quantization

Robert Millikan found, in 1909, that
charged objects may only have an integer
multiple of a fundamental unit of charge.
 Charge is quantized.
 An object may have a charge ±e, or ± 2e,
or ± 3e, etc but not say ± 1.5e.
"charge is
quantized" in terms
 Proton has a charge +1e.
of an equation, we
say:
q=ne
 Electron has a charge –1e.
 Some particles such a neutron have no
(zero) charge.
13
Unit of Electrical Charge:
The Coulomb " C "
The symbol for electric charge is written q, - q or Q.
The unit of electric charge is coulomb "C". The charge of
one electron is equal to the charge of one proton, which is
1.6 * 10-19 C. This number is given a symbol "e".
Example:
How many electrons are there in 1 C of charge?
14
Insulators and Conductors( Material classification)




Materials/substances may be classified
according to their capacity to carry or conduct electric
charge
Conductors are material in which electric charges move
freely.
Insulator are materials in which electrical charge do not
move freely.
 Glass, Rubber are good insulators.
 Copper, aluminum, and silver are good conductors.
Semiconductors are a third class of materials with
electrical properties somewhere between those of
insulators and conductors.
 Silicon and germanium are semiconductors used
widely in the fabrication of electronic devices.
15
16
Example:
Identify substances or materials that can be
classified as


Conductors ?
Insulators?
Why? is static electricity more apparent in winter?
17
Lecture 2
Coulomb’s law
18
Coulomb’s Law
Coulomb discovered in 1785 the fundamental law
of electrical force between two stationary charged particles.
An electric force has the following properties:
 Inversely proportional to the square of the separation, r,
between the particles, and is along a line joining them.
 Proportional to the product of the magnitudes of the
charges |q1| and |q2| on the two particles.
 Attractive if the charges are of opposite sign and
repulsive if the charges have the same sign.
19
Coulomb’s Law (Mathematical Formulation)
F  q1 q2
1
F 2
r




ke known as the Coulomb constant.
Value of ke depends on the choice of units.
SI units



Force: the Newton (N)
Distance: the meter (m).
Charge: the coulomb ( C).
20
1
1
9
2
2
Ke 


9

10
Nm
/
C
4e
4  8.85 1012
where eo is known as the Permittivity constant of free space.
eo = 8.85 x 10-12 C2/N.m2
Experimentally measurement: ke = 8.9875×109 Nm2/C2.
Reasonable approximate value: ke = 8.99×109 Nm2/C2.
21
Example: the Coulomb constant unit
Then the Coulomb constant unit is
22
The electrostatic force


The electrostatic force
is often called Coulomb
force.
It is a force (thus, a
vector):


a magnitude
a direction.
F12  F21
F12   F21
23
Example:
The electron and proton of a hydrogen atom are
separated (on the average) by a distance of about
5.3x10-11 m. Find the magnitude of the electric force
that each particle exerts on the other.
q1 =-1.60x10-19 C
q2 =1.60x10-19 C
r = 5.3x10-11 m
Fe  k e
e
2
r2
 9 10
9 Nm 2
C2
1.6 10
 5.3 10
19
11

m
C
2
2
 8.2 108 N
Attractive force with a magnitude of 8.2x10-8 N.
24
Gravitational vs. Electrical Force
q1
m1
1 q1 q2
Felec =
4e0 r 2
Fgrav
m1m2
=G
r2
For an electron:
|q| = 1.6 × 10-19 C
m = 9.1 × 10-31 kg
F
F
q2
m2
r


Felec
Fgrav
q1q2
=
m1 m2
1
4 e0
G
Felec
 4.17 × 10 + 42
Fgrav
25
Superposition of Forces
FT  F10 + F20 + F30 + ....
+Q1
+Q2
+Q3
r10
r20
r30
F30
F20
+Q0
F10
kq 0q1
kq 0q 2
kq 0q3
FT  2 rˆ10 + 2 rˆ20 + 2 rˆ30 + ....
r10
r20
r30
N
 q1

q3
q2
qi
FT  kq 0  2 rˆ10 + 2 rˆ20 + 2 rˆ30 + ....   kq 0  2 rˆi0
r20
r30
i 1 ri0
 r10

26


The net force on q3 is the vector sum of the
forces F32 and F31.
The magnitude of the forces F32 and F31 can
calculated using Coulomb’s law.
9
9
5

10
C
2

10
C
q3 q 2

9 Nm 2 
9
F32  k e

9

10

5.62

10
N
2
2
C2
r
 4m 
F31  k e
q 3 q1
r
2
 5 10 C  6 10 C   1.08 10
9
 9 109
Nm 2
C2
9
 5m 
2
8
N
Fx  F32  F31 cos 37.0o  3.01109 N
Fy  F31 sin 37.0o  6.50 109 N
F  Fx2 + Fy2  7.16  109 N
  65.2o
27
Example: Two fixed charge, 1µC and -3µC are
separated by 10 cm. Where may a third charge
be located so that no force act on it ?
F31  F32
q 3q1
q 3q 2
k
k
2
r31
r32 2
1106
3 106

2
2
d
(d + 10)
Solve the eq. to find d.
28
Examination of the geometry of Figure leads to
tan  
x/2
L2 + x / 2
2
If L is much larger than x (which is the case if Ө is very small), we may
neglect x/2 in the denominator and write tan Ө = x/2L. This is equivalent to
approximating tanӨ by sinӨ. The magnitude of the electrical force of one
ball on the other is
q2
Fe 
4e 0 x 2
by Eq. When these two expressions are used in the equation mg tan Ө =
Fe, we obtain
29
 q2L 
m gx
1 q

x

2
2L
4e o x
2
e
m
g
o


2
1/ 3
b) We solve x3 for the charge:
q
3
mgx

2kL
 0.010 kg   9.8 m s 2   0.050 m 
3
2  8.99 109 N  m 2 C 2  1.20 m 
  2.4  108 C.
Thus, the magnitude is
| q | 2.4 108 C.
30
Lecture 3
The Electric Field
31
Electric Field



Suggests the notion of electrical field (first
introduced by Michael Faraday (1791-1867).
An electric field is said to exist in a region of
space surrounding a charged object.
If another charged object enters a region
where an electrical field is present, it will be
subject to an electrical force.
32
Electric Field & Electric Force
Consider a small charge q0 near a larger charge Q.
We define the electric field E at the location of the small
test charge as a ratio of the electric force F acting on it
and the test charge q0
F
E  N /C
q0
This is the field produced by the charge Q, not
by the charge q0
33
Electric Field Direction

The direction of E at a point is the
direction of the electric force that would be
exerted on a small positive test charge
placed at that point.
E
E
- -
-
-
-
-
- -
+
+
-
+
-
-
+
+
+
+
+
+
+
+
+
+ +
34
Electric Field from a Point Charge
Suppose we have two charges,
q and q0, separated by a
distance r. The electric force
between the two charges is
We can consider q0 to be a test
charge, and determine the
electric field from charge q as
F  ke
q qo
E  ke
r
2
q
r
2
35
• If q is +ve, field at a given point is radially outward
from q.
• If q is -ve, field at a given point is radially inward from q.
36
Electric Field Lines
To visualize electric field patterns, one can draw lines
pointing in the direction of the electric field vector at
any point.
These lines are called electric field lines.
1. The electric field vector is tangent to the electric field
lines at each point.
2. The number of lines per unit area through a surface
perpendicular to the lines is proportional to the
strength of the electric field in a given region.
3. No two field lines can cross each other . Why?
37
The electric field lines for a point charge.
(a) For a positive point charge, the lines are directed radially outward.
(b) For a negative point charge, the lines are directed
radially inward.
Note that the figures show only those field lines that lie in the plane of
the page.
38
The electric field lines for two positive point
charges.
The electric field lines for two point charges of equal
magnitude and opposite sign (an electric dipole)
39
Question:
Two charges q1 and q2, fixed along the x-axis as shown,
produce an electric field E at the point (x,y)=(0,d), which
is the directed along the negative y-axis.
Which of the following is true?
1. Both charges are positive
2. Both charges are negative
3. The charges have opposite signs
40
Electric Field from an Electric Dipole
A system of two oppositely charged point particles is
called an electric dipole.
The vector sum of the electric field from the two charges
gives the electric field of the dipole (superposition
principle).
We have shown the electric field lines from a dipole
41
Example:

Two charges on the x-axis a distance
d apart
 Put -q at x = -d/2
 Put +q at x = +d/2
Calculate the electric field at a point P a distance x from
the origin
42
Principle of superposition:The electric field
at any point x is the sum of the electric
fields from +q and -q
1
q
1 q
E  E+ + E 

2
2
4e 0 r+ 4e 0 r
Replacing r+ and r- we get


1
1


E

2
2
1 d 
4e 0  x  1 d 

x
+


2
2
q
This equation gives the electric field everywhere
on the x-axis (except for x = d/2)
43
Example:
Electric Field Due to Two Point Charges
Charge q1=7.00 mC is at the origin, and charge q2=-10.00
mC is on the x axis, 0.300 m from the origin. Find the
electric field at point P, which has coordinates (0,0.400) m.
y
E1
0.400 m
P
q1
E
E2
0.300 m
q2
x
44
E1  k e
E2  ke
q1
r12
q2
2
2
r
 8.99 10
9 Nm 2
C2

7.00 106C
 0.400m 
2
  3.9310 N / C
5
10.00 10 C   3.60 10 N / C
6
 8.99 10
9 Nm 2
C2
 0.500m 
5
2
E x  53 E 2  2.16 105 N / C
E y  E 1  E 2 sin   E 1  54 E 2  1.05 105 N / C
E  E x2 + E y2  2.4 105 N / C
  tan 1 (E y / E x )  25.9o
45
Example
In Figure, determine the point
(other than infinity) at which the
total electric field is zero.
Solution: The sum of two vectors can be zero only if the two vectors have
the same magnitude and opposite directions.
46
Motion of charge particles in a uniform
electric field
An electron moving horizontally passes between two horizontal
planes, the upper plane charged negatively, and the lower positively.
A uniform, upward-directed electric field exists in this region. This
field exerts a force on the electron. Describe the motion of the
electron in this region.
-
-
- -
E0
-
+
+
+ +
-
- - - - - - - - - - - - - - - - -
ve
+
+ + + + + + + + + + + + + + + + +
47

Horizontally:





No electric field
No force
No acceleration
Constant horizontal
velocity
Vertically:




Constant electric field
Constant force
Constant acceleration
Vertical velocity increase
linearly with time.
Ex  0
Fx  0
ax  0
v x  vo
x  vo t
E y  Eo
Fy  eE o
ay  eE o / m e
v y  eE o t / m e
1
y  eE o t 2 / m e
2
48
-
-
- -
-
- - - - - - - - - - - - - - - - -
-
+
+
+ +
+
+ + + + + + + + + + + + + + + + +
Conclusions:
 The charge will follow a parabolic path downward.
 Motion similar to motion under gravitational field
only except the downward acceleration is now
larger.
49
+Q
+Q
-Q
-e
Phosphor
Screen
-e
x
-Q
This device is known as a
cathode ray tube (CRT)
50
Continuous Charge Distributions
kq
E 0  2 rˆ
r
kdq
dE 0  2 rˆ
r
Single charge
+Q1
+Q2
+Q3
r10
Single piece of a charge distribution
E 03
E 02
r20
0
r30
N
qi
E0  k  2 rˆi0
i 1 ri0
Discrete charges
E 01
+
+
+
+
dq
dE0
0
dq
E0  k 
rˆ
2
r
all charge
Continuous charge distribution
51
kdq
dE  2 rˆ
r
Cartesian
Polar
dq  dx
dq  Rd
Surface charge
Q dq
 
A dA
dq  dxdy
dq  rdrd
Volume charge
Q dq
 
V dV
dq  dxdydz
dq  rdrddz
Line charge

Q

dq
dx
dq  r 2 sin drdd
52
Example: Electric Field Due to a Charged Rod
A rod of length l has a uniform positive charge per unit
length λ and a total charge Q. Calculate the electric field
at a point P that is located along the long axis of the rod
and a distance a from one end.
dq  dx
dq
dx
dE  ke 2  ke 2
x
x
l +a
E

a
dx
ke  2  ke 
x
l +a

a
l +a
dx
 1
 ke   
2
x
 x a
keQ
Q1
1 
E  ke
 

l  a l + a  a(l + a)
53
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
dEy
dE
+
+

E  kx 

dy
2
x
 +y
3
2 2

dE x 
kdq
cos 
2
r
k  dy 
x
dE  2
x + y2 x 2 + y2
 2  2k
 kx  2  
x
x 
54
Example – Charged Ring
d
+ 
a
r  z +a
2
+
2
2
perpendicular-components cancel by symmetry

+
+
kdq
dE  2 rˆ
r
dq  ds  ad
dE z
z
+
+
E
2
kza
z
2
+a
 d 
3
2 2 0

kdq
cos 
2
r
k  ad
z
dE  2 2
z +a
z2 + a 2
dE
dE
+
dE z 
kza
z
2
+a
3
2 2

 2 
kQz
z
2
+a
3
2 2

55
When:
z  a
The charged ring must look like a point source.
E
kQz
2
z
 +a
3
2 2


kQz
0
 2  a 2 
 z 1 + 2  
  z 
3
2
kQz kQ
 3  2
z
z
56
Example – Uniformly Charged Disk
E
R
kQz
z
+r
2
3
2 2

r
z
kzdq
dE 
dE
z
2
+r
3
2 2

dq  dA  rdrd  2rdr
dE 
kz  2rdr 
z
z2 + R 2
 kz

z2
2
+r
3

2
R
E
3
2 2

u du  kz
0
z2 + R 2
u


1
2
1
2
kz2rdr
z
2
+r
3
2 2

R
 kz
0
z2 + R 2
2rdr
z
2
+r
3
2 2

 kz

du
z
u
2
3
2



1
1 
z
 2kz 

  k2 1  2

2
2
2
2
z
+
R
z
z
+
R




z2
57
Two Important Limiting Cases
R  z
Large Charged Plate:
R
r
z
dE


z
1

E  k2 1 

k

2



2



2
2
4eo
2eo
z
+
R


Very Far From the Charged Plate:




z
z

E  k2 1 

k

2

1



z2 + R 2 
R2

 z 1+ 2
z

z  R

1


2 2 
  k2 1  1 + R  
 

z2  




  1 R 2 
 1 R 2  kR 2 kQ
 k2 1  1 
 k2 

 2
2 
2 
2
z
z
2 z 
  2 z 
58
Lecture 4
Discussion
59
[1] In figure, two equal positive charges q=2x10-6C interact
with a third charge Q=4x10-6C. Find the magnitude and
direction of the resultant force on Q.
FQq 1  FQq 2  K e
Q q1
r2
6
6
(
4

10
)(
2

10
)
9
 9  10
 0.29N
2
(0.5)
4
Fx  2 F cos  0.29( )  2  0.23  0.46N
5
Fy  F sin   F sin   0
FT  0.46N
 0
60
[2] A charge Q is fixed at each of two opposite corners of
a square as shown in figure. A charge q is placed at each of
the other two corners. If the resultant electrical force on Q is
Zero, how are Q and q related.
Fx  0 
F12  F13 cos   0
kQq
kQQ
1

.
a2
2a 2
2
Q
q 
....................................(1)
2 2
Fy  0  F13 sin   F14  0
kQQ
1
kQq
.

2a 2
a2
2
Q
q 
....................................(1)
2 2
then
Q 2
2q
61
[3] Two fixed charges, 1µC and -3µC are separated by
10cm as shown in figure (a) where may a third charge be
located so that no force acts on it? (b) is the equilibrium
stable or unstable for the third charge?
F31  F32
Ke
q 3q1
 r31 
2
q 3q 2
 Ke
 r32 
2
1106
3 106

2
2
d
d + 10 
3d 2  d + 10   3d 2  d 2 + 20d + 100
2
62
2d 2  20d  100  0  d 2  10d  50  0
a  1.b  10, c  50
b  b 2  4ac 10  100  4(1)( 50)
d 

2a
2
10  10 3
10  300
2
d 

2
2
d  5 + 5 3  13.66
d  5  5 3  13.66
b )the equilibrium s unstable
63
[4] Find the electric field at point p in figure due to the
charges shown.
E  E1  E 2
x
Solution:
 36 104 N / C
E y  E3
 28 104 N / C
E p  (E x ) + (E y )
2
2
 46.1N / C
  141
64
[5] A charged cord ball of mass 1g is suspended on a
light string in the presence of a uniform electric
field as in figure. When E=(3i+5j) *105N/C, the ball
is in equilibrium at Θ=37o. Find (a) the charge on
the ball and (b) the tension in the string.
E x  3 105 N / C
E y  5 105 N / C
Fx  qE x T sin 37  0....(1)
Substitute T from equation (1) into equation (2)
Fy  qE y T cos 37  0....(2)
Substitute T from equation (1) into equation (2)
Substitute by q into equation (1) to find T=5.44*10-3N
65
[6] A 1.3µC charge is located on the x-axis at x=-0.5m, 3.2µC charge is
located on the x-axis at x=1.5m, and 2.5µC charge is located at the
origin. Find the net force on the 2.5µC charge
6
6
q1q2
1
.
3

10

2
.
5

10
9
3
F21  K e

9

10

177

10
r21 2
0.52
6
6
q2 q3
3
.
2

10

2
.
5

10
9
3
F23  K e

9

10

32

10
r23 2
1.52

F  F21 + F23  117103 + 32103  149103 N
t
66
[7] Two free point charges +q and +4q are a distance 1cm apart. A third
charge is so placed that the entire system is in equilibrium. Find the
location, magnitude and sign of the third charge. Is the equilibrium
stable?
E1  E 2
Ke
q
4q

K
e
d2
d  12
4d 2  d  1
2
4d 2 _ d 2 + 2d  1  0
3d 2 + 2d  1  0
d 
2
4  4  3  1
6
d  1m
d  1 / 3m
67
[8] Two protons in a molecule are separated by a distance of 3.8*10-10m.
Find the electrostatic force exerted by one proton on the other.
9
9
9
q1q2
1
.
6

10

1
.
6

10
14
.
4

1
.
6

10
9
F  K e 2  9 109


1
.
6

10
21
10 2
r
14
.
4

10
3.8 10


[9] The electric force on a point charge of 4.0mC at some point is 6.9*104N in the positive x direction. What is the value of the electric field at
that point?
F 6.9 104
2
E 

1
.
725

10
N C
6
q 4.0 10
68
[10] Two point charges are a distance d apart . Find E points to the left
P. Assume q1=+1.0*10-6C, q2=+3.0*10-6C, and d=10cm
E P1
6
q1
1
.
0

10
 K e 2  9  109
r1
x2
EP2
6
q2
3
.
0

10
 K e 2  9  109
r2
x  102
ET  E P1 + E P 2
6
6


1
.
0

10
3
.
0

10
9

 9  10 
+
2
2 
x
x  10 

69
[11] Calculate E (direction and magnitude) at point P in Figure.
E x  E1  E3  0
2q
E y  E2  k e 2
r
2q
1
2q
1 4q
ET  E y  k e

. 2 
. 2
2
4e 0 a
4e 0 a
 a 


2
 2
q
E
From middle of the triangleaway fromit
2
e 0 a
70
[12] A uniform electric field exists in a region between two oppositely
charged plates. An electron is released from rest at the surface of the
negatively charged plate and strikes the surface of the opposite plate,
2.0cm away, in a time 1.5*10-8s. (a) What is the speed of the electron
as it strikes the second plate? (b) What is the magnitude of the electric
field .
Horizental :
first find the acceleration from the relation
x  v 0t + 1/ 2at 2
then find the velocity
v  v 0 + at
v  2.7 106 m / s
then find the electric field
E  F / q  ma / q  1103 N / C
71
[13] Three charges are placed on corners of an equilateral triangle as
shown in Figure 1. An electron is placed at the center of the
triangle. What is the magnitude of the net force on the electron?
0.5
cos30 
 r  0.577m
r
Thus, the m agnitude of each of the three forces is
F
k e 1c
 4.3 109 N
r2
and the net force on the electron is
Fnet  Fx2 + Fy2  2 F  8.6 109 N
72
[14] A uniform electric field exists in the region between two
oppositely charged plane parallel plates. An electron is
released from rest at the surface of the negatively
charged plate and strikes the surface of the opposite
plate 2x10-8 s later. If the magnitude of the electric field
is 4x103 N/C, what is the separation between the
plates?
F eE
a    7 104 m / s 2
me me
Since the electron starts from rest , the distnce is given by
1 2 1
d  at  (7 1014 )(2 108 ) 2  0.14m
2
2
73
Lecture 5
Electric Flux and Gauss’s Law
74
Electric Flux

Electric flux quantifies the notion
“number of field lines crossing a
surface.”
 The electric flux  through a flat
surface in a uniform electric field
depends on the field strength E, the
surface area A, and the angle 
between the field and the normal to the
surface.
 Mathematically, the flux is given by
  EA cos  E  A.

Here A is a vector whose
magnitude is the surface area A and
whose orientation is normal to the
surface.
A

E
75

When  < 90˚, the flux is positive (out of the surface), and
when   90˚, the flux is negative.

Units: Nm2/C in SI units, the electric flux is a SCALAR
quantity

Find the electric flux through the area A = 2 m2, which is
perpendicular to an electric field E=22 N/C
Answer: F = 44 Nm2/C.
76
Example:
Calculate the flux of a constant E field (along x) through a
cube of side “L”.
y
1
2
Solution
E
1   EA1 cos 1   EL2
x
z
 2  EA2 cos  2  EL2
 net   EL + EL  0
2
2
77
Question:
The flux through side B of the cube in the figure is the
same as the flux through side C. What is a correct expression for the
flux through each of these sides?
s3 E
 s2 E
 s3E cos45
 s2 E cos45
78
When we have a complicated surface, we can
divide it up into tiny elemental areas:
 
d  E  dA  E dAcos
 
    E.dA
79
Example:

What’s the total flux on a
closed surface with a
charge inside?


The shape and size don’t
matter!
Just use a sphere
 E  dA
 E  dA cos(0)
E 
1
q

4e o r 2

 dA
1
q
q
2
4

r

4e o r 2
eo
80
What is Gauss’s Law?
 Gauss’s Law does not tell us anything new, it is NOT a
new law of physics, but another way of expressing
Coulomb’s Law
 Gauss’s law makes it possible to find the electric field
easily in highly symmetric situations.
81
Gauss’ Law

The precise relation between flux
and the enclosed charge is given by
Gauss’ Law
d E  E  dA
 E   E  dA 


Qencl
 E   E.dA
eo
e0 is the permittivity of free space in the
Coulomb’s law
The  symbol has a little circle to
indicate that the integral is over a closed
surface.
The flux through a closed surface is equal to the total
charge contained divided by permittivity of free space
82
A few important points on Gauss’ Law
The integral is over the value of E on a closed
surface of our choice in any given situation
The charge Qencl is the net charge enclosed by the
arbitrary close surface of our choice.
It does NOT matter where or how much charge is
distributed inside the surface
The charge outside the surface does not contribute.
Why?
83
Question

What’s the total flux
with the charge
outside? Why?
Solution:


Zero.
Because the surface
surrounds no charge
84
Gauss Coulomb



Calculate E of point like (+) charge Q
Consider sphere radius r centered at the
charge
Spherical symmetry: E is the same
everywhere on the sphere, perpendicular to
the sphere
 
2
   E.dA  E  dA  E (4r )
Qen

e0
Qen
Q
Q
E (4r ) 
 E
k 2
2
e0
4r e 0
r
2
85
Application of Gauss’s law
Gauss’s law can be used to calculate the electric field if the
symmetry of the charge distribution is high. Here we concentrate
in three different ways of charge distribution



A linear charge distribution 
A surface charge distribution 
A volume charge distribution 
86
A linear charge distribution

Let’s calculate the electric field from
a conducting wire with charge per
unit length  using Gauss’ Law
1
 E   E  dA   q
eo

We start by assuming a Gaussian
surface in the form of a right cylinder
with radius r and length L placed
around the wire such that the wire is
along the axis of the cylinder
87



From symmetry we can see that the
electric field will extend radially from the
wire.
How?
 If we rotate the wire along its axis, the
electric field must look the same
 Cylindrical symmetry
 If we imagine a very long wire, the
electric field cannot be different
anywhere along the length of the wire
 Translational symmetry
Thus our assumption of a right cylinder as
a Gaussian surface is perfectly suited for
the calculation of the electric field using
Gauss’ Law.
88


The electric flux through the ends of the cylinder
is zero because the electric field is always
parallel to the ends.
The electric field is always perpendicular to the wall of the
cylinder so
 
   E  dA
 
  EdA cos(0)  E 2 rL 
 E 2 rL   q / e 0  L / e 0

… and now solve for the electric field

2k 
E

2e 0 r
r
89
A surface charge distribution

Assume that we have a thin, infinite
non-conducting sheet of positive charge
The charge density in this
case is the charge per unit
area, 
From symmetry, we can see
that the electric field will be
perpendicular to the surface
of the sheet
90



To calculate the electric field using Gauss’ Law,
we assume a Gaussian surface in the form of a right
cylinder with cross sectional area A and height 2r, chosen
to cut through the plane perpendicularly.
Because the electric field is perpendicular to the plane
everywhere, the electric field will be parallel to the walls of the
cylinder and perpendicular to the ends of the cylinder.
Using Gauss’ Law we get
   E  dA  EA + EA
 q / e 0  A / e 0

… so the electric field from an infinite
non-conducting sheet with charge density 

E
2e 0
91
Assume that we have a thin, infinite conductor
(metal plate) with positive charge

 The “charge density” in this case is
also the charge per unit area, , on
either surface; there is equal surface
charge on both sides.
From symmetry, we can see that the
electric field will be perpendicular to
the surface of the sheet
To calculate the electric field using
Gauss’ Law, we assume a Gaussian
surface in the form of a right cylinder
with cross sectional area A and
height r, chosen to cut through one
side of the plane perpendicularly.

92




The field inside the conductor is zero so the end inside the
conductor does not contribute to the integral.
Because the electric field is perpendicular to the plane
everywhere, the electric field will be parallel to the walls of the cylinder
and perpendicular to the end of the cylinder outside the conductor.
Using Gauss’ Law we get
A
EA 
e0
… so the electric field from an infinite
conducting sheet with surface charge density  is

E
e0
93
A volume charge distribution
let’s calculate the electric field from
charge distributed uniformly throughout charged sphere.


Assume that we have insolating a solid sphere of
charge Q with radius r with constant charge density
per unit volume .
We will assume two different spherical
Gaussian surfaces
 r2 > r
(outside)
 r1 < r
(inside)
94



Let’s start with a Gaussian surface with
r1 < r.
From spherical symmetry we know that the electric
field will be radial and perpendicular to the Gaussian
surface.
Gauss’ Law gives us
 4 3
 E  dA  E4 r   e  3  r1 
2
1

Solving for E we find
 r1
E
3e 0
inside
95
In terms of the total charge Q …
Q
r1
E 4
3

r
3e 0
3
Qr1
kQr1
E
 3
3
4e 0 r
r
inside
96



Now consider a Gaussian surface with
radius r2 > r.
Again by spherical symmetry we know that the electric
field will be radial and perpendicular to the Gaussian
surface.
Gauss’ Law gives us
 4 3
 E  dA  E4 r   e  3  r 
2
2

Solving for E we find
kQ
E 2
r2
outside
same as a point charge!
97
Electric field vs. radius for a conducting
sphere
Er  r
Er
Er 
1
r2
98
Properties of Conductors
E is zero within conductor
If there is a field in the conductor, then the free electrons would feel a
force and be accelerated. They would then move and since there are
charges moving the conductor would not be in electrostatic
equilibrium. Thus E=0
net charge within the surface is zero
How ?
99
Lecture 6
Application
(Gauss’s Law)
100
[1] A solid conducting sphere of radius a has a net charge +2Q. A
conducting spherical shell of inner radius b and outer radius c is
concentric with the solid sphere and has a net charge –Q as shown in
figure. Using Gauss’s law find the electric field in the regions labeled 1,
2, 3, 4 and find the charge distribution on the spherical shell.
Region (1) r < a
To find the E inside the solid sphere
of radius a we construct a Gaussian
surface of radius r < a
E = 0 since no charge inside the
Gaussian surface
Region (3) b > r < c
E=0 How?
101
Region (2) a < r < b
we construct a spherical Gaussian surface of radius r
 E.dA 
qencl
e0
E (4r ) 
2
,
2Q
e0
where   0, why ?
1 2Q
E
, where a < r < b
2
4e 0 r
Region (4) r > c
we construct a spherical Gaussian surface of radius r > c, the total net
charge inside the Gaussian surface is q = 2Q + (-Q) = +Q Therefore
Gauss’s law gives
 E.dA 
qencl
e0
E (4r ) 
2
Q
e0
,
1
Q
E
, where r  c
2
4e 0 r
102
[2] A long straight wire is surrounded by a hollow cylinder whose axis
coincides with that wire as shown in figure. The solid wire has a charge
per unit length of +λ , and the hollow cylinder has a net charge per unit
length of +2λ . Use Gauss law to find (a) the charge per unit length on
the inner and outer surfaces of the hollow cylinder and (b) the electric

field outside the hollow
cylinder, a distance r from the axis.
(a) Use a cylindrical Gaussian surface S1 within
the conducting cylinder where E=0
 E.dA 
qencl
e0
0
Also inner  
thus inner + outer  2
 outer  3
103
(b) For a Gaussian surface S2 outside the conducting cylinder
 E.dA 
qencl
E ( 2rl ) 
e0
1
e0
3
E
2e 0 r
(   + 3 )l
104
[3] Consider a long cylindrical charge distribution of radius R with a
uniform charge density ρ. Find the electric field at distance r from the
axis where r < R.
If we choose a cylindrical Gaussian surface of length L and radius r, Its
volume is πr²L , and it encloses a charge ρπr²L . By applying Gauss’s
law we get,
Thus
radially outward from the cylinder axis
Notice that the electric field will increase as r increases, and also the
electric field is proportional to r for r<R. For the region outside the
cylinder (r>R), the electric field will decrease as r increases.
105
Two Parallel Conducting Plates



When we have the situation shown in the left two panels (a positively charged
plate and another negatively charged plate with the same magnitude of charge),
both in isolation, they each have equal amounts of charge (surface charge
density ) on both faces.
But when we bring them close together, the charges on the far sides move to
the near sides, so on that inner surface the charge density is now 2.
A Gaussian surface shows that the net charge is zero (no flux through sides —
dA perpendicular to E, or ends — E = 0). E = 0 outside, too, due to shielding, in
just the same way we saw for the sphere.
106
Two Parallel Nonconducting Sheets




The situation is different if you bring two nonconducting sheets of charge close
to each other.
In this case, the charges cannot move, so there is no shielding, but now we can
use the principle of superposition.
In this case, the electric field on the left due to the positively charged sheet is
canceled by the electric field on the left of the negatively charged sheet, so the
field there is zero.
Likewise, the electric field on the right due to the negatively charged sheet is
canceled by the electric field on the right of the positively charged sheet.

The result is much the same as
before, with the electric field in
between being twice what it was
previously.
107
[4] Two large non-conducting sheets
of +ve charge face each other as
shown in figure. What is E at points
(i) to the left of the sheets (ii) between
them and (iii) to the right of the
sheets?
We know previously that for each sheet,
the magnitude of the field at any point is

E
2e 0
a) At point to the left of the two parallel sheets
E   E1 + ( E2 )  2 E

E
e0
108
b) At point between the two sheets
E  E1 + (E2 )  0
(c) At point to the right of the two parallel
sheets
E  E1 + E2  2 E

E
e0
109
[5] A square plate of copper of sides 50cm is
placed in an extended electric field of
8*104N/C directed perpendicular to the
plate. Find (a) the charge density of each
face of the plate
E  8  104 N / C
A  0.25m 2
EA 
q
e0
q  0.17  106 C
  0.68 10
6
C
2
m
110
[6] An electric field of intensity 3.5*103N/C is applied the x
axis. Calculate the electric flux through a rectangular plane
0.35m wide and 0.70m long if (a) the plane is parallel to the yz
plane, (b) the plane is parallel to the xy plane, and (c) the plane
contains the y axis and its normal makes an angle of 40o with
the x axis.
(a) the plane is parallel to the yz plane
  EAcos0  857.5Nm2 / C
(b) the plane is parallel to the xy plane
The angel 90
c) the plane is parallel to the xy plane
The angel 40
111
[7] A long, straight metal rod has a radius of 5cm and a charge per unit
length of 30nC/m. Find the electric field at the following distances from
the axis of the rod: (a) 3cm, (b) 10cm, (c) 100cm.

use E 
to
4e 0 r
find
a ) zero
b) 5.4  103 N / C
c ) 540N / C
112
[8] The electric field everywhere on the surface
of a conducting hollow sphere of radius
0.75m is measured to be equal to
8.90*102N/C and points radially toward the
center of the sphere. What is the net charge
within the surface?
  EA  E 4r 2
  6.3  103 N .m 2 / C
Now to find the net ch arg e

q
e0
q  5.5  108 C
113
[9] A point charge of +5mC is located at the center of a sphere with
a radius of 12cm. What is the electric flux through the surface of
this sphere?
q
5
2
   5.5 10 Nm / C
e
[10] (a) Two charges of 8mC and -5mC are inside a cube of sides
0.45m. What is the total electric flux through the cube? (b) Repeat
(a) if the same two charges are inside a spherical shell of radius 0.
45 m.
6
6
q (8 10  5 10 )
5
2
 
 3.4 10 Nm / C
12
e
8.8510
114
[12] A solid copper sphere 15cm in radius has a total charge of
40nC. Find the electric field at the following distances measured from
the center of the sphere: (a) 12cm, (b) 17cm, (c) 75cm.
(a) At 12 cm the charge in side the Gaussian
surface is zero so the electric field E=0
(b)
qencl
 E.dA  e 0 ,   0, why ?
qencl
EA 
e0
q
E (4r ) 
e0
2
E
q
4

E

125

10
N / C radially outward
2
4e 0 r
115
[13] Two long, straight wires are separated by a distance d = 16 cm, as shown
below. The top wire carries linear charge density 3 nC/m while the bottom wire
carries -5 nC/m.
1- What is the electric field (including direction) due to the top wire at a point
exactly half-way between the two wires?
2- Find the electric field due to the bottom wire at the same point, exactly halfway between the two wires (including direction).
3- Work out the total electric field at that point.
116
using Gauss’ Law and cylindrical Gaussian surface as
Lec.5 page 15

E
2e 0 r
3 109
1  Et 
 674N / C Down
12
2 (8.85410 )(0.08)
5 109
2  Eb 
 1123N / C Down
12
2 (8.85410 )(0.08)
  
3  E  Et + Eb  674 ˆj  1123ˆj  1797ˆjN / C
117
Lecture 7
The Electric Potential
118
Electric Potential Energy

The electric force, like the
gravitational force, is a
conservative force.
(Conservative force: The work is pathindependent.)
E

As in mechanics, work is
W  Fd cos 

Work done on the positive
charge by moving it from A to
B
B
A
d
W  Fd cos  qEd
119

The work done by a conservative force equals
the negative of the change in potential energy,
DPE
DPE  W  qEd
This equation is valid only for the case of a uniform
electric field
If a charged particle moves perpendicular to electric field lines, no
work is done.
if d  E
120

The potential difference between points A and B, VB-VA,
is defined as the change in potential energy (final minus
initial value) of a charge, q, moved from A to B, divided
by the charge
DPE
DV  VB  VA 
q



Electric potential is a scalar quantity
Electric potential difference is a measure of electric
energy per unit charge
Potential is often referred to as “voltage”
If the work done by the electric field is zero, then the electric potential
must be constant
121

Electric potential difference is the work done to
move a charge from a point A to a point B
divided by the magnitude of the charge. Thus the
SI units of electric potential difference
1V  1 J C

In other words, 1 J of work is required to move a 1 C of
charge between two points that are at potential
difference of 1 V

Question: How can a bird stand on a high voltage line
without getting zapped?
122

Units of electric field (N/C) can be expressed in
terms of the units of potential (as volts per meter)
1 N C  1V m

Because the positive tends to move in the direction of the
electric field, work must be done on the charge to move it
in the direction, opposite the field. Thus,


A positive charge gains electric potential energy when it is moved in
a direction opposite the electric field
A negative charge looses electrical potential energy when it moves
in the direction opposite the electric field
123
Example : A uniform electric field of magnitude 250 V/m is directed in the
positive x direction. A +12µC charge moves from the origin to the point
(x,y) = (20cm, 50cm). (a) What was the change in the potential energy
of this charge? (b) Through what potential difference did the charge
move?
Begin by drawing a picture of the situation, including the direction of
the electric field, and the start and end point of the motion.
(a) The change potential energy is given by the charge times the field
times the distance moved parallel to the field. Although the charge
moves 50cm in the y direction, the y direction is perpendicular to the
field. Only the 20cm moved parallel to the field in the x direction
matters for determining the change of potential energy. ∆ PE = -qEd =
-(+12µC)(250 V/m)(0.20 m) = -6.0×10-4 .
(b) The potential difference is the difference of electric potential,.
∆ V = ∆ PE / q = -6.0×10-4 J / 12µC = -50 V.
124
Analogy between electric and gravitational
fields

The same kinetic-potential energy theorem works here
A
d
E
q
B


A
d
g
m
B
If a positive charge is released from A, it accelerates in the direction
of electric field, i.e. gains kinetic energy
If a negative charge is released from A, it accelerates in the direction
opposite the electric field
KEi + PEi  KE f + PE f
125
Example: motion of an electron
What is the speed of an electron accelerated from
rest across a potential difference of 100V?
KEi + PEi  KE f + PE f
Given:
DV=100 V
me = 9.11×10-31 kg
mp = 1.67×10-27 kg
|e| = 1.60×10-19 C
Find:
ve=?
vp=?
KE f  KEi  DPE  qDV
Vab
1 2
m vf  qDV
2
 2qDV
vf 
m
ve  5.9 106 m / s
v p  1.3 105 m / s
126
Problem:
A proton is placed between two parallel conducting plates in a vacuum as shown.
The potential difference between the two plates is 450 V. The proton is released
from rest close to the positive plate.
What is the kinetic energy of the proton when it reaches the negative plate?
Example : Through what potential difference would an electron need to
accelerate to achieve a speed of 60% of the speed of light, starting
from rest? (The speed of light is 3.00×108 m/s.)
The final speed of the electron is vf = 0.6(3.00×108 m/s) = 1.80×108
m/s. At this speed, the energy (non-relativistic) is the kinetic energy,
KE =KEf =½mvf² = 0.5(9.11×10-31 kg)(1.80×108 m/s)² = 1.48×10-14 J.
This energy must equal the change in potential energy from moving
through a potential difference, KEf = ∆PE = -q ∆V. Therefore:
∆V = KE/q = (1.48×10-14 J)/(-1.6×10-19 C) = -9.25×104 V.
127
Electric potential and potential energy due
to point charges



Electric circuits: point of zero potential is defined by
grounding some point in the circuit
Electric potential due to a point charge at a point in
space: point of zero potential is taken at an infinite
distance from the charge
With this choice, a potential can be found as
f 
kq
V f  Vi    E.dr  V 
r
i

Note: the potential depends only on charge of an object,
q, and a distance from this object to a point in space, r.
128
Superposition principle for potentials

If more than one point charge is present, their
electric potential can be found by applying superposition
principle
The total electric potential at some point P due to several
point charges is the algebraic sum of the electric
potentials due to the individual charges.

Remember that potentials are scalar quantities!
129
Potential energy of a system of point charges


Consider a system of two particles
If V1 is the electric potential due to charge q1 at a point P,
then work required to bring the charge q2 from infinity to P
without acceleration is q2V1. If a distance between P and
q1 is r, then by definition
q2
P

q1
r
A
q1q2
PE  q2V1  ke
r
Potential energy is positive if charges are of the same
sign.
130
Example: potential energy of an ion
Three ions, Na+, Na+, and Cl-, located such, that
they form corners of an equilateral triangle of side 2 nm
in water. What is the electric potential energy of one of
the Na+ ions?
Cl?
qNa qCl
qNa qNa
qNa
PE  ke
+ ke
 ke
 qCl + qNa 
r
r
r
but : qCl  qNa !
Na+
Na+
qNa
PE  ke
 qNa + qNa   0
r
131
Potentials and charged conductors

Recall that work is opposite of the change in
potential energy,
W   PE  q VB  VA 


No work is required to move a charge between two points
that are at the same potential. That is, W=0 if VB=VA
Recall:
1.
2.

all charge of the charged conductor is located on its surface
electric field, E, is always perpendicular to its surface, i.e. no
work is done if charges are moved along the surface
Thus: potential is constant everywhere on the surface of a
charged conductor in equilibrium
… but that’s not all!
132

Because the electric field is zero inside the conductor, no
work is required to move charges between any two
points, i.e.
W  q VB  VA   0


If work is zero, any two points inside the conductor have
the same potential, i.e. potential is constant everywhere
inside a conductor
Finally, since one of the points can be arbitrarily close to
the surface of the conductor, the electric potential is
constant everywhere inside a conductor and equal to its
value at the surface!
 Note that the potential inside a conductor is not
necessarily zero, even though the interior electric
field is always zero!
133
The electron volt

A unit of energy commonly used in atomic,
nuclear and particle physics is electron volt (eV)
The electron volt is defined as the energy that electron
(or proton) gains when accelerating through a potential
difference of 1 V

Relation to SI:
1 eV = 1.60×10-19 C·V = 1.60×10-19 J
Vab=1 V
134
Example : ionization energy of the electron
in a hydrogen atom
In the Bohr model of a hydrogen atom, the electron, if it is
in the ground state, orbits the proton at a distance of r =
5.29×10-11 m. Find the ionization energy of the atom, i.e.
the energy required to remove the electron from the atom.
Note that the Bohr model, the idea of electrons as tiny balls orbiting
the nucleus, is not a very good model of the atom. A better picture is
one in which the electron is spread out around the nucleus in a cloud
of varying density; however, the Bohr model does give the right
answer for the ionization energy
135
In the Bohr model of a hydrogen atom, the electron, if it is in the ground state,
orbits the proton at a distance of r = 5.29 x 10-11 m. Find the ionization energy,
i.e. the energy required to remove the electron from the atom.
Given:
r = 5.292 x 10-11 m
me = 9.11×10-31 kg
mp = 1.67×10-27 kg
|e| = 1.60×10-19 C
Find:
E=?
The ionization energy equals to the total energy of the
electron-proton system,
e2
v2
E  PE + KE with PE  ke , KE  m
r
2
The velocity of e can be found by analyzing the force
on the electron. This force is the Coulomb force;
because the electron travels in a circular orbit, the
acceleration will be the centripetal acceleration:
mac  Fc
or
v2
e2
m  ke 2 , or
r
r
e2
v  ke
,
mr
2
Thus, total energy is
e 2 m  ke e 2 
e2
18
E   ke + 


k


2.18

10
J  -13.6 eV

e
r 2  mr 
2r
136
Calculating the Potential from the Electric
Field

To calculate the electric potential from the
electric field we start with the definition of the work dW
done on a particle with charge q by a force F over a
 
displacement ds
dW  F  ds

In this case the force is provided by the electric field
 
F = qE
dW  qE  ds

Integrating the work done by the electric force on the
particle as it moves in the electric field from some initial
point i to some final point f we obtain
W
f
i

qE  ds
137

Remembering the relation between the change
in electric potential and the work done …
We
DV  
q

…we find

Taking the convention that the electric potential is zero at
infinity we can express the electric potential in terms of
the electric field as
f 

We
DV  Vf  Vi  
   E  ds
i
q
 


V ( x )   E  ds
i
138
Example - Charge moves in E field


Given the uniform electric field E,
find the potential difference Vf-Vi by
moving a test charge q0 along the
path icf.
Idea: Integrate E  ds along the
path connecting ic then cf.
(Imagine that we move a test
charge q0 from i to c and then from
c to f.)
139
Example - Charge moves in E field
c
 
f 

E  ds   E  ds
Vf  Vi   
i
c
c 

 i E  ds  0 (ds perpendicular t o E)
f 
f

 E  ds   E ds cos(45)  E  dist ance
c
c
1
2
Vf  Vi   Ed
140
Question

We just derived Vf-Vi for the path i → c → f. What
is Vf-Vi when going directly from i to f ?
A: 0
B: -Ed
C: +Ed
D: -1/2 Ed
Quick: DV is independent of path.
Explicit: DV = -  E . ds =  E ds = - Ed
141
Lecture 8
Application
(The Electric Potential)
142
[1] What potential difference is needed to stop an electron
with an initial speed of 4.2×105m/s?
W  DKE  KE f  KEi   KEi
1
 qDV   m v2
2
1 m v2 1 (9.11 1031 )(4.2 105 ) 2
DV  2
 2
q
1.6  1019
DV  0.57volt
[2] An ion accelerated through a potential difference of
115V experiences an increase in potential energy of
7.37×10-17J. Calculate the charge on the ion.
DPE
DV 
q
q  6.141019 C
143
[3] An infinite charged sheet has a surface charge density σ
of 1.0×10-7 C/m2. How far apart are the equipotential
surfaces whose potentials differ by 5.0 V?

E
 5650N / C
2e 0
V  Ed
V
d   8.8  10 4 m
E
[4] At what distance from a point charge of 8µC would the
potential equal 3.6×104V?
1 q
V
 r  2m
4e 0 r
144
[5] At a distance r away from a point charge q, the electrical
potential is V=400v and the magnitude of the electric field is
E=150N/C. Determine the value of q and r.
1
q
q
V
&r  k
4e 0 r
V
1
q
 E
4e 0 r 2
q
V2
 Ek

q 2 kq
(k )
V
 r  2.67m
145
[6] Calculate the value of the electric potential at point P due
to the charge configuration shown in Figure. Use the values
q1=5mC, q2=-10mC, a=0.4m, and b=0.5m.
V  V1 + V2 + V3 + V4
 q2 q1
 q2
V  k[
+
+ +
]
2
2
2
2
a
a
a +b
a +b
q1
By substitute find V
146
[7] Two large parallel conducting plates are 10cm apart and
carry equal and opposite charges on their facing
surfaces. An electron placed midway between the two
plates experiences a force of 1.6×1015N. What is the
potential difference between the plates?
V  Ed
F
E
q
F
1.6 1015
32
V  d 

0
.
5

5

10
volt
19
q
1.6 10
147
[8] Two point charges are located as shown in, where
ql=+4mC, q2=-2mC, a=0.30m, and b=0.90m. Calculate the
value of the electrical potential at points P1, and P2. Which
point is at the higher potential?
V p1  V1 + V2
1
q1 q2
V p1 
[ + ]
4e 0 r1 r2
5
6
4

10

2

10
V p1  9  109 [
+
]  70200volt
1 .2
0.3 2
148
[9] In figure prove that the work required to put four charges
together on the corner of a square of radius a is given by
 0.21q 2
W
e 0a
W  w12 + w13 + w14 + w23 + w24 + w34
 q2
q2
q2 q2
q2
q2
W
[
+


+

]
4e 0 a
a
2a a
2a a
1
 4q 2
q2
W
[
+
]
4e 0
a
2a
1
 2 4q 2 + 2q 2
 0.2q 2
W
[
]
4e 0
e 0a
2a
1
149
[10] Assume we have a system of three
point charges:
q1 = +1.50 mC
q2 = +2.50 mC
q3 = -3.50 mC.
q1 is located at (0,a)
q2 is located at (0,0)
q3 is located at (b,0)
a = 8.00 m and b = 6.00 m.
What is the electric potential at point P
located at (b,a)?
150
The electric potential at point P is
given by the sum of the
electric potential from the three
charges
r1
r2
r3
 q1
 q1 q2 q3 
kqi
q2
q3 
V 
 k + +   k +
+ 
2
2
a
 r1 r2 r3 
b
a +b
i 1 ri
3
 1.50 10 6 C
V  8.99 10 N/C 
+
 6.00 m

9

2.50 10 6 C
8.00 m  + 6.00 m 
2
2
3.50 10 6 C 

+
8.00 m 
V  562 V
151
[11] An electron initially has velocity 5 x 105 m/s. It is
accelerated through a potential of 2 V. What is its
final velocity?
This is aconservation of energy problem.
Since the initial velocity is 5  105 m / s the electron' s initial kinetic energy is
1 2
m v  1.141019 J .
2
Since the electron accelerated ,
ki 
its kinetic energy increases and its potential energy decreases
U i  U f  e (2V )  3.2 1019 J .
The final kinetic energy is
K f  K i + U i  U f  4.34 1019 J
 v f  10105 m / s
152
[12] A charge of -1x10-8C weighs 1 g. It is released at rest from
point P and moves to point Q. It's velocity at point Q is 1
cm/s. What is the potential difference, VP - VQ?
This is aconservation of energy problem.
1 2
K f  K i  m v  5 108  U i  U f  q(Vi  V f )
2
The initial electric potential, is V p ,
and the final electric potiential is VQ
8
5 10 J
V p  VQ  Vi  V f 
 5V
8
 110 C
153
[13] A proton with a speed of vi = 2.0 x 105 m/s enters a region of
space where source charges have created an electric potential.
What is the proton’s speed after it has moved through a potential
difference of ∆V = 100 V?
KEi + PEi  KE f + PE f
KE f  KEi   DPE   qDV
1 2 1 2
m vf  m vi  qDV
2
2
2qDV
2
5
v f  vi 
 1.4410 m / s
m
154
[14] Potential Due to a Charged Rod

A rod of length L located along the x axis has a uniform linear
charge density λ. Find the electric potential at a point P located on
the y axis a distance d from the origin.

Start with
dq  dx
1 dq
1
dx
dV 

2
2 1/ 2
4
e
r
4
e
(
x
+
d
)
then,
0
0


dx

2
2 1/ 2 L



ln
x
+
(
x
+
d
) 0
2
2 1/ 2
4e 0 ( x + d )
4e 0
0
L
V   dV  


lnL + ( L2 + d 2 )1/ 2   ln d 
4e 0
So
 L + ( L2 + d 2 )1/ 2 

V
ln 

4e 0 
d



155
[15] The Electric Potential of a Charged Ring
Find the potential of a
thin uniformly charged
ring of radius R and
charge Q at point P on
the z axis?
rP  R 2 + z 2 and dQ 
dV 
Q
Rd
2 R
1 dQ
1
Q

d
2
2
4e 0 rP
4e 0 2 R + z
VP   dV 
1
Q
4e 0 2 R 2 + z 2
2
 d 
0
1
Q
4e 0
R2 + z 2
156
[16] The Electric Potential of a Charged Disk
Find the potential V of a thin
uniformly charged disk of radius R
and charge density s at point P on
the z axis?
dV 
V 
R
0
kdq
z 2 + a2
2 ak da
z 2 + a2

k (2 ada)
z 2 + a2
R
2ada
0
z 2 + a2
  k 
u  z 2 + a2 ; du  2ada
z 2 + R2
V   k  2
z
du
  k  2 u

u
z 2 + R2
z2
 2 k

z 2 + R2  z 2

z 
2


 2 k z 1 + ( R / z )  1  kQ 2 1 + ( R / z ) 2  1



R 
157
Lecture 9
The Electric Field & Electric Potential
Due to Continuous Charge Distributions
158
Field & Potential Due to a Continuous
Charge Distribution
The electric field and electric potential due to a
continuous charge distribution is found by treating
charge elements as point charges and then summing
via integrating, the electric field vectors and the
electric potential produced by all the charge elements.
159
[1] Electric Field Due to a Charged Rod
A rod of length L has a uniform positive charge per unit
length λ and a total charge Q. Calculate the electric field
at a point P that is located along the long axis of the rod
and a distance a from one end.
dq  dx
dq
dx
dE  ke 2  ke 2
x
x
l +a
E

a
dx
ke  2  ke 
x
l +a

a
l +a
dx
 1
 ke   
2
x
 x a
keQ
Q1
1 
E  ke
 

l  a l + a  a(l + a)
160
[2] Infinitely Long Line of Charge
dq  dy
dq
dE  k e 2
r
+
+
dy
r x +y
2
2
y-components cancel by symmetry
2
y +

+
+
+
dEx
x
dEy
dE
dq
dE  dE x  ke 2 cos 
r
k  dy 
x
dE  2
x + y2 x 2 + y2
+
+

E  kx 

dy
2
x
 +y
3
2 2

 2  2k
 kx  2  
x
x 
161
[3] Electric Field on the Z-Axis of a Charged Ring
determine the field at point P on the axis of the ring.
dq  ds  ad
dq
dE  k e 2
r
d
+ 
a
r  z +a
2
+
2
perpendicular-components cancel by symmetry

+
+
2
dE z
z
+
+
E
2
kza
z
2
+a
 d 
3
2 2 0

kdq
cos 
2
r
k  ad
z
dE  2 2
z +a
z2 + a 2
dE
dE
+
dE z 
kza
z
2
+a
3
2 2

 2 
kQz
z
2
+a
3
2 2

162
When:
z  a
The charged ring must look like a point source.
E
kQz
2
z
 +a
3
2 2


kQz
0
 2  a 2 
 z 1 + 2  
  z 
3
2
kQz kQ
 3  2
z
z
Note that for z >> R (the radius of the ring), this reduces to a simple
Coulomb field.
163
[4] Electric Field on the Axis of an Uniformly
Charged Disk
E
Using the charged ring result,
R
kQz
z
+r
2
3
2 2

r
z
kzdq
dE 
dE
z
2
+r
3
2 2

dq  dA  rdrd  2rdr
dE 
R
E
0
kz2rdr
z
2
+r
3
2 2

R
 kz
0
kz  2rdr 
z
+r
2
2rdr
z
2
+r
3
2 2

3
2 2

 kz
z2 + R 2

du
z
u
2
3
2
164
E  kz
z2 +R2


3
2
u du
z2
z2 +R2
 1 
u 2 
 kz 
1
 
 2  z2

1
1 
 2kz 


2
2
2
z 
 z +R


1
 k 2 1 

2
2
z +R 

165
Two Important Limiting Cases
R  z
Large Charged Plate:
R
r
z
dE


z
1

E  k2 1 

k

2



2



2
2
4eo
2eo
z
+
R


Very Far From the Charged Plate:




z
z

E  k2 1 

k

2

1



z2 + R 2 
R2

 z 1+ 2
z

z  R

1


2 2 
  k2 1  1 + R  
 

z2  




  1 R 2 
 1 R 2  kR 2 kQ
 k2 1  1 
 k2 

 2
2 
2 
2
z
z
2 z 
  2 z 
166
[5] Potential Due to a Charged Rod

A rod of length L located along the x axis has a uniform
linear charge density λ. Find the electric potential at a point P
located on the y axis a distance d from the origin.

Start with
dq  dx
1 dq
1
dx
dV 

2
2 1/ 2
4
e
r
4
e
(
x
+
d
)
0
0
then,


dx

2
2 1/ 2 L



ln
x
+
(
x
+
d
) 0
2
2 1/ 2
4e 0 ( x + d )
4e 0
0
L
V   dV  




lnL + ( L2 + d 2 )1/ 2   ln d 
4e 0
So
 L + ( L2 + d 2 )1/ 2 

V
ln 

4e 0 
d

167
[6] The Electric Potential of a Charged Ring
Find an expression for the
electric potential at a point
P located on the
perpendicular central axis
of a uniformly
charged ring of radius a
and total charge Q
rP  R 2 + z 2 and dQ 
dV 
Q
Rd
2 R
1 dQ
1
Q

d
2
2
4e 0 rP
4e 0 2 R + z
VP   dV 
1
Q
4e 0 2 R 2 + z 2
2
 d 
0
1
Q
4e 0
R2 + z 2
168
[7] The Electric Potential of a Charged Disk
Find the potential V of a thin
uniformly charged disk of radius R
and charge density σ at point P on
the z axis?
dV 
V 
R
0
kdq
z 2 + a2
2 ak da
z 2 + a2

k (2 ada)
z 2 + a2
R
2ada
0
z 2 + a2
  k 
u  z 2 + a2 ; du  2ada
z 2 + R2
V   k  2
z
du
  k  2 u

u
z 2 + R2
z2
 2 k

z 2 + R2  z 2

z 
2


 2 k z 1 + ( R / z )  1  kQ 2 1 + ( R / z ) 2  1



R 
169
Lecture 10
Capacitance and capacitors
170
Capacitors



Capacitors are devices that store energy in an electric
field.
Capacitors are used in many every-day applications
 Heart defibrillators
 Camera flash units
Capacitors are an essential part of electronics.
 Capacitors can be micro-sized on computer chips or
super-sized for high power circuits such as FM radio
transmitters.
171
Definition of Capacitance

The definition of capacitance is
Q
C
V



The units of capacitance are coulombs per volt.
The unit of capacitance has been given the name
farad (abbreviated F) named after British physicist
Michael Faraday (1791 - 1867)
1C
1F 
1V
A farad is a very large capacitance
 Typically we deal with mF (10-6 F), nF (10-9 F),
or pF (10-12 F)
172
The parallel-plate capacitor

The capacitance of a device
depends on the area of the
plates and the distance
between the plates
A
C  e0
d

where A is the area of one of
the plates, d is the separation,
e0 is a constant (permittivity of
free space),
e0= 8.85×10-12 C2/N·m2
A
+Q
d
-Q
A
173
Example: A parallel plate capacitor has plates 2.00 m2 in area, separated by a
distance of 5.00 mm. A potential difference of 10,000 V is applied across the
capacitor. Determine
 the capacitance
 the charge on each plate
Solution:
Given:
DV=10,000 V
A = 2.00 m2
d = 5.00 mm
Find:
C=?
Q=?
Since we are dealing with the parallel-plate capacitor,
the capacitance can be found as
A
2.00 m 2
12
2
2
C  e 0  8.85 10 C N  m
d
5.00 103 m
 3.54 109 F  3.54 nF


Once the capacitance is known, the charge can be
found from the definition of a capacitance via charge
and potential difference:


Q  C DV  3.54 109 F 10000V   3.54 105 C
174
Cylindrical Capacitor




Consider a capacitor constructed of two collinear
conducting cylinders of length L.
The inner cylinder has radius r1 and
the outer cylinder has radius r2.
Both cylinders have charge per
unit length  with the inner cylinder
having positive charge and the outer
cylinder having negative charge.
We will assume an ideal cylindrical capacitor
 The electric field points radially from the inner
cylinder to the outer cylinder.
 The electric field is zero outside the collinear
cylinders.
175

We apply Gauss’ Law to get the electric field between
the two cylinder using a Gaussian surface with radius r
and length L as illustrated by the red lines
 
e 0  E  dA  q
e 0EA  L where A  2 rL

… which we can rewrite to get an
expression for the electric field
between the two cylinders

E
2e 0 r
176

As we did for the parallel plate capacitor, we define the
voltage difference across the two cylinders to be V=V1 – V2.
r2
V1  V2   r
1
 
r
E  ds   r
2
1

dr
2e 0r
 r2 


ln  
2e 0  r1 

The capacitance of a cylindrical capacitor is
C
q

V
L

r2
ln( )
2e 0
r1

2e 0 L
r2
ln( )
r1
177
Spherical Capacitor

Consider a spherical capacitor formed by two
concentric conducting spheres with radii r1 and r2

Let’s assume that the inner sphere has
charge +q and the outer sphere has
charge –q.
The electric field is perpendicular to the
surface of both spheres and points
radially outward

178

To calculate the electric field, we use a
Gaussian surface
consisting of a concentric sphere of
radius r such that r1 < r < r2

The electric field is always
perpendicular to the Gaussian surface
so

… which reduces to
E
q
4e 0 r
2
179

To get the electric potential we follow a method similar to the one
we used for the cylindrical capacitor and integrate from the
negatively charged sphere to the positively charged sphere
r1
V   Edr 
r2

r2
q  1 1
dr 
 
2

4e 0 r
4e 0  r1 r2 
q
Using the definition of capacitance we find
q
C 
V


r1
q
4e 0

q  1 1  1 1
    

4e 0  r1 r2   r1 r2 
The capacitance of a spherical capacitor is then
r1r2
C  4e 0
r2  r1
180
Combinations of capacitors


It is very often that more than one capacitor is
used in an electric circuit
We would have to learn how to compute the equivalent
capacitance of certain combinations of capacitors
C2
C1
C3
C2
C5
C3
C1
C4
181
a. Parallel combination
Connecting a battery to the parallel combination of capacitors is
equivalent to introducing the same potential difference for both
capacitors, V1  V2  V
A total charge transferred to the system from the battery is the sum of
charges of the two capacitors, Q1 + Q2  Q
By definition,
a
Q1  C1V1 & Q2  C2V2
C1
V=Vab
+Q1
C2
Q1
Q2
Thus, Ceq would be
Q  Q1 + Q2
+Q2
b
CeqV  C1V1 + C2V2
Ceq  C1 + C2 , sin ce V  V1  V2
182
Parallel combination:

Analogous formula is true for any number of
capacitors,
Ceq  C1 + C2 + C3 + ...

(parallel combination)
It follows that the equivalent capacitance of a parallel
combination of capacitors is greater than any of the
individual capacitors
183
Example: A 3 mF capacitor and a 6 mF capacitor are connected
in parallel across an 18 V battery. Determine the equivalent
capacitance and total charge deposited.
a
Given:
V = 18 V
C1= 3 mF
C2= 6 mF
Find:
Ceq=?
Q=?
C1
V=Vab
+Q1 C2
+Q2
Q1
Q2
b
First determine equivalent capacitance of C1 and C2:
C12  C1 + C2  9 m F
Next, determine the charge


Q  C DV  9 106 F 18V   1.6 104 C
184
Series combination
Connecting a battery to the serial combination of capacitors is equivalent
to introducing the same charge for both capacitors, Q1  Q2  Q
A voltage induced in the system from the battery is the sum of potential
differences across the individual capacitors,
By definition ,
V1 
V  V1 + V2
Q1
Q
& V2  2
C1
C2
Thus, Ceq would be
a
V  V1 + V2
Q
Q
Q
 1+ 2
Ceq
C1 C2
1
1
1

+
, sin ce Q  Q1  Q2
Ceq
C1 C2
+Q1
C1
V=Vab
Q1
c
C2
+Q2
Q2
b
185
Series combination:

Analogous formula is true for any number of
capacitors,
1
1 1
1
 +
+ + ...
Ceq C1 C2 C3

(series combination)
It follows that the equivalent capacitance of a series
combination of capacitors is always less than any of the
individual capacitance in the combination
186
Example: A 3 mF capacitor and a 6 mF capacitor are connected
in series across an 18 V battery. Determine the equivalent
capacitance and total charge deposited.
a
V = 18 V
C1= 3 mF
C2= 6 mF
Find:
Ceq=?
Q=?
+Q1
C1
Given:
V=Vab
Q1
c
C2
+Q2
Q2
b
First determine equivalent capacitance of C1 and C2:
C1C2
Ceq 
 2 mF
C1 + C2
Next, determine the charge


Q  C DV  2 106 F 18V   3.6 105 C
187
Example: Capacitors in Series and Parallel
Three capacitors are connected as
shown.
(a) Find the equivalent capacitance of
the 3-capacitor combination.
(b) The capacitors, initially uncharged,
are connected across a 6.0 V battery.
Find the charge and voltage drop for
each capacitor.
188
Ceq1  C1 + C2  6.0 mF
Ceq  1/ (1/ Ceq1 +1/ C3 )  1/[1/ (6.0 mF) +1/ (3.0 mF)]  2.0 mF
Q  CeqV  (2.0 mF)(6.0 V)  12.0 mC
V3  Q / C3  (12.0 mC) / (3.0 mF)  4.0 V
V24  V  V3  (6.0 V)  (4.0 V)  2.0 V
Q2  C2V24  (2.0 mF)(2.0 V)  4.0 mC
Q4  C4V24  (4.0 mF)(2.0 V)  8.0 mC
189
Energy stored in a charged capacitor



Consider a battery connected to a
capacitor
A battery must do work to move
electrons from one plate to the other.
The work done to move a small
charge Dq across a voltage V is
DW = V Dq.
As the charge increases, V increases
so the work to bring Dq increases.
Using calculus we find that the
energy (U) stored on a capacitor is
given by:
1
Q2 1
U  QV 
 CV 2
2
2C 2
V
V
q
Q
190
Example: electric field energy in parallel-plate capacitor
Find electric field energy density (energy per unit volume) in a
parallel-plate capacitor
+
+
+
+
+
+
+
+
+
+










=

Thus,
1
U  CV 2
2
e A
C 0
volume  Ad V  Ed
d
u  U / volume  energy density
+
Recall
1 e0 A
( Ed ) 2 /( Ad )
2 d
and so, the energy density is
1
u  e0E2
2
191
Example: In the circuit shown V = 48V, C1 = 9mF, C2 = 4mF and
C3 = 8mF.
(a) determine the equivalent capacitance of the circuit,
(b) determine the energy stored in the combination by
calculating the energy stored in the equivalent capacitance,
Ceq  5.14mF
C1
V
C2
C3
The energy stored in the capacitor C123 is then


1
1
2
2
6
U  CV  5.14 10 F  48V   5.9 103 J
2
2
192
Capacitors with dielectrics


A dielectrics is an insulating material (rubber, glass, etc.)
Consider an insolated, charged capacitor
Q
+Q
V0


+Q
Q
Insert a dielectric
V
Notice that the potential difference decreases (k = V0/V)
Since charge stayed the same (Q=Q0) → capacitance increases
Q0
Q0
 Q0
C


  C0
V V0  V0


dielectric constant: k = C/C0
Dielectric constant is a material property
193


Capacitance is multiplied by a factor k when the
dielectric fills the region between the plates completely
E.g., for a parallel-plate capacitor
A
C  e 0
d


The capacitance is limited from above by the electric
discharge that can occur through the dielectric material
separating the plates
In other words, there exists a maximum of the electric field,
sometimes called dielectric strength, that can be produced
in the dielectric before it breaks down
194
Dielectric constants and dielectric
strengths of various materials at room
temperature
Material
Dielectric
constant, k
Vacuum
1.00
--
1.00059
106 ×3
80
--
3.78
106 ×9
Air
Water
Fused quartz
Dielectric
strength (V/m)
195
Example: Take a parallel plate capacitor whose plates have an area of 2 m2 and
are separated by a distance of 1cm. The capacitor is charged to an initial voltage
of 3 kV and then disconnected from the charging source. An insulating material
is placed between the plates, completely filling the space, resulting in a decrease
in the capacitors voltage to 1 kV. Determine the original and new capacitance,
the charge on the capacitor, and the dielectric constant of the material.
Since we are dealing with the parallel-plate capacitor, the original capacitance
can be found as
2
C  e0


A
2.00 m
 8.85 1012 C 2 N  m2
 18 nF
3
d
1.00 10 m
The dielectric constant and the new capacitance are
DV1
C   C0 
C0  0.33 18nF  6nF
DV2
The charge on the capacitor can be found to be


Q  C DV  18 109 F  3000V   5.4 105 C
196
How does an insulating dielectric material reduce electric fields by
producing effective surface charge densities?
Reorientation of polar molecules
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Induced polarization of non-polar molecules
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
 +
Dielectric Breakdown: breaking of molecular bonds/ionization of molecules.
197
Solved Problems
(Electrostatic)
198
[1] Three charges are located in the XY plane as Shown in the figure.
Q1= - 3 nC, Q2 = 5 nC Q3 = 3 nC. The distance between Q1 and Q2 is
8 m and between Q2 and Q3 is 6 m.
Find:
a)The electric field due to the charges at the origin O.
b)The electric potential due to these charges at the origin O.
c) What is the direction of the net force on the point O.
199
a)The electric field due to the charges at the origin O.
3  109 27
E1  9  10 

N /C
25
25
5  109 9
9
E 2  9  10 
 N /C
25
5
3  109 27
9
E3  9  10 

N /C
25
25
 E x   E1 cos 2  E 2 cos 2  E3 cos 2
9
27 4 9 4 27 4
    
  3.2 N / C
25 5 5 5 25 5
E
x
 E 2 sin  2  E1 sin  2  E3 sin  2

9 3 27 3 27 3
 
 
  0.2 N / C
5 5 25 5 25 5
ET  E x2 + E y2  10.3 N / C
200
b) The electric potential due to these charges at the origin O.
 3 109
27
V1  9  10 
 V
5
5
5 109
9
V2  9 10 
 9V
5
3  109 27
9
V3  9 10 
 V
5
5
V  V1 + V2 + V3  9V
9
c) What is the direction of the net force on the point O.
T he direction of the net force is the direction of E
tan 
E
E
y
x

 0. 2
 0.0625
 3. 2
  180+ 0.06  180.06
201
[2] Two point charges are located on the x-axis as follows: charge Q1
= 8x10-9 C is at x = 0 and charge Q2 = -1x10-9 C is at x = 1cm. What is
the net force in the x direction on a third charge,Q3 = +2x10-9 C,
placed at x = 2 cm ?
using coulomb's law
F1 
F2 
k 2 109 8 109
(2cm) 2
 3.6 10 4 N in + ve x direction
k 2 109  1109
(2cm) 2
 1.8 10 4 N in  ve x direction
Fnet  1.8 10 4 N
202
[3] In a hydrogen atom an electron circles a proton. Since the mass of
the proton is much greater than the mass an electron, assume that
the proton stays fixed in space and the electron rotates around it
with radius 0.529x10-10 m. What is the velocity of the electron?
T he force on the electron is
m v2 ke2
F
 2
r
r
ke2
v
 2.2 106 m / s
mr
203
[4] A uniform electric field exists in the region between two oppositely
charged plane parallel plates. An electron is released from rest at
the surface of the negatively charged plate and strikes the surface
of the opposite plate 2x10-8 s later. If the magnitude of the electric
field is 4x103 N/C, what is the separation between the plates?
T his is problem involving constant acceleration
F eE
a 
 7 1014 m / s 2
m
m
ke2
v
 2.2 106 m / s
mr
Since the electron starts from rest
1 2
d  at  14cm
2
204
[5] Two charged particles are placed on the x-axis as follows: Q1 = 1x10-9
C at x = 0 m and Q2 = -2x10-9 C at x = 1.6 m. Where on the x-axis is
the electric field zero?
The electric field can only be zero where the field due to Q1 and Q2 point in
opposite direction. This occurs both for x<0m and for x>1.6 m, however since
Q2 is greater in magnitude, one must be further away from it in order to have
the magnitudes of the fields due to Q1 and Q2 be equal. This means the field is
zero in the region x<0.
Sitting the magnitude equal,
k Q1
( x)
2

k Q2
( x + 1.6) 2
( x + 1.6) 2  2 x 2
 x + 1.6   2 x
x  3.86m
or
x  0.66m
205
[6] Two stationary point charges are arranged on the x-axis as
follows: Q1 = 2.4x10-10 C at x = 0 m and Q2 = -1.2x10-10 C at x = 1
m. An electron is placed at x = 2 m and let go. What is the
velocity of the electron when it reaches x = 3 m.
This is a conservation of energy problem. The initial kinetic energy of the
electron is zero. The initial and final potential energy of the electron are:
kqe q1 kqe q2
PEi 
+
 0J
2
1
kqe q1 kqe q2
PE f 
+
 2.8810 20 J
3
2
Consequently, the final kinetic energy is
1
me v 2f  k f  ki + PE i  PE f  2.88 10 20 J
2
And the final velocity of the electron is
v  2.5 105 m / s
206
[7] Using the superposition principle determine the electric
field at point P in Fig.
Using the convention that a positive electric
field is up and a negative electric field is down,
the net electric field at p is
Ep  
3 10
2e 0
9
+
2 10
2e 0
9
+
110
2e 0
9
 0V / m
207
[8] A 1 Coulomb charge is located at the origin. Another
charge, Q, is located at x = 3 m. If the electric field is zero at
x = 1 m, what is Q?
The electric field due to the 1C charge is in positive x-direction at
x=1m. In order to cancel this field, the charge at x=3m must be
positive so that its electric field is in the negative x-direction at x=1m.
The two field will cancel if they have the same magnitude.
k (1) kQ

 Q  4C
2
2
(1)
(2)
208
[9]
C1
a) Find the equivalent capacitance of the
entire combination.
V
+
-
C3
C2
C1 and C2 are in series.
1
1
1
C1C 2
 +
 C12 
C12 C1 C 2
C1 + C 2
C12 
C1 = 10 mF
C2 = 5.0 mF
C3 = 4.0 mF
10  5 50

 3.3mF
10 + 5 15
C12 and C3 are in parallel.
Ceq  C12 + C 3  3.3 + 4.0  7.3mF
209
b) If V = 100 volts, what is the charge Q3 on C3?
C = Q/V
Q3  C3V  4.0 106 100
Q3  4.0 104 Coulombs
c) What is the total energy stored in the circuit?
1
1
U  CeqV 2  1.3 10 6 10 4  3.6 10  2 J
2
2
U  3.6 102 J
210
[10] A wire of uniform charge density and length L lies along the x
axis as shown in Figure. What is the electric potential at point A?
dq  dx
dq
dx
dV  ke
 ke
x
x
l +d
V

d
dx
ke 
 ke 
x
l +d

d
dx
l +d
 ke  ln xd
x
V  ke ln(l + d )  ln d   ke ln(1 + l / d )
211
[11] Determine the point nearest the charges where the total electric
field is zero. Take the -2.50 µC charge to be at the origin of the x axis.
1.00 m
0.39 m
-2.50 mC
-6.00 mC
[12] The plates of a parallel plate capacitor are separated by 1 x 10–4
m. If the material in between them is a jelly with a dielectric constant
of 2.26, what is the plate area needed to provide a capacitance of
1.50 pF?
7.50 x 10-6 m2
212
[13] Find the charge on the 2 mF capacitor in the following circuit.
18µC
2 mF
4 mF
2 mF
12 V
[14] Each of the protons in a particle beam has a kinetic energy of
3.25x10-15 J What are the magnitude and direction of the electric
field that will stop these protons in a distance of 1.25 m?
1.63 x 104 N/C and opposite to the motion
213
[15] Three point charges are fixed on the corners of an equilateral
triangle whose one side is b as shown in Figure.
1. What is the magnitude of the Coulomb force
acting on charge –q due to presence of other
charges?
q2
3k 2
b
2. What is the value of the electric potential at the center (point A) of
positive charges?
1
q
(4 
)k
3 b
3. What is the electric potential energy of system?
q2
k
b
214
[16] The stored energy of a capacitor is 3.0 J after having been charged
by a 1.5 V battery. What is the energy of the capacitor after it is
charged by 3.0 V battery?
12mJ
[17] Find the equivalent capacitance between
points a and b in the combination of capacitors
shown in Figure.
7µF and 5µF series
75
Ceq 
 2.9mF
7+5
4µF, 6µF and 2.9µF parallel
Ceq  4 + 6 + 2.9  12.9mF
215
[18] Two capacitors when connected in parallel give an equivalent
capacitance of 9.00 pF and give an equivalent capacitance of 2.00 pF
when connected in series. What is the capacitance of each capacitor?
in parallel
C1 + C2  9...................(1)
in series
C1C2
 2...................(2)
C1 + C2
Solve eq.1&2
C1  3 pF, C2  6 pF
or, C1  6 pF, C2  3 pF
216
Lecture 12
Current & Resistance
217
Electric Current
Definition: the current is the rate at
which charge flows through this
surface.
Given an amount of charge, DQ, passing through the area A
in a time interval Dt, the current is the ratio of the charge to
the time interval.
DQ
I 
Dt
The SI units of current is the ampere (A).
 1 A = 1 C/s
 1 A of current is equivalent to 1 C of charge passing
through the area in a time interval of 1 s.
218
Example: The amount of charge that passes through the
filament of a certain light bulb in 2.00 s is 1.67 c. Find the
current in the light bulb.
Find no. of electrons?
219
Current Density



When we care only about the total current I in a conductor, we do
not have to worry about its shape.
However, sometimes we want to look in more detail at the current
flow inside the conductor. Similar to what we did with Gauss’ Law
(electric flux through a surface), we can consider the flow of charge
through a surface. To do this, we consider (charge per unit time) per
unit area, i.e. current per unit area, or current density. The units are
amps/square meter (A/m2).
Current density is a vector (since it has a flow magnitude and

direction). We use the symbol J . The relationship between current
and current density is
 
I   J  dA,
or ,
 I
J
A
High current
density
Small current
density
220
Current and Drift Speed

Consider the current on a conductor of cross-sectional
area A.
221









Volume of an element of length Dx is : DV = A Dx.
Let n be the number of carriers per unit of volume.
The total number of carriers in DV is: n A Dx.
The charge in this volume is: DQ = (n A Dx)q.
Distance traveled at drift speed vd by carrier in time Dt:
Dx = vd Dt.
Hence: DQ = (n A vd Dt)q.
The current through the conductor:
I = DQ/ Dt = n A vd q.
The current density :
J = I/A = n vd q.
222
Example:
A copper wire of cross-sectional area 3.00x10-6 m2 carries a
current of 10 A. Assuming that each copper atom contributes
one free electron to the metal, find the drift speed of the
electron in this wire.
A = 3.00x10-6 m2 ; I = 10 A, q = 1.6 x 10-19 C.
n = 8.48 x 1022 electrons/ m3.
[Q] If the current density in a copper wire is equal to 5.8×106A/m2,
calculate the drift velocity of the free electrons in this wire.
223



Drift speeds are usually very small.
Drift speed much smaller than the average speed
between collisions.
 Electrons traveling at 2.46x10-6 m/s would take 68 min
to travel 1m.
So why does light turn on so quickly when one flips a
switch?
 The info (electric field) travels at roughly 108 m/s…
[Q] A silver wire 1 mm in diameter transfers a charge of 65 C in 1 hr,
15 min. Silver contains 5.80 x 1028 free electrons per cubic meter.
a) What is the current in the wire? b) What is the magnitude of the
drift velocity of the electrons in the wire?
Ans. a) 0.0144 A; b) 1.98 x 10-6 m/s
224
Resistance and Ohm’s Law
When a voltage (potential difference) is applied across the
ends of a metallic conductor, the current is found to be
proportional to the applied voltage.
In situations where the proportionality is exact,
one can write.
The proportionality constant R is called resistance of the conductor.
225
The resistance is defined as the ratio.
In SI, resistance is expressed in volts per ampere.
A special name is given: ohms
Example: if a potential difference of 10 V applied
across a conductor produces a 0.2 A current,
then one concludes the conductors has a resistance of
10 V/0.2 A = 50 W.
226
Ohm’s Law



Resistance in a conductor arises because of
collisions between electrons and fixed charges within
the material.
In many materials, including most metals, the
resistance is constant over a wide range of applied
voltages.
This is a statement of Ohm’s law.
Ohm’s Law
227
 The current–potential difference curve for an ohmic
material. The curve is linear, and the slope is equal to the
inverse of the resistance of the conductor.
 A nonlinear current–potential difference curve for a
junction diode. This device does not obey Ohm’s law.
228
Resistivity




Electrons moving inside a conductor subject to an
external potential constantly collide with atoms of the
conductor.
They lose energy and are repeated re-accelerated by the
electric field produced by the external potential.
The collision process is equivalent to an internal friction.
This is the origin of a material’s resistance.
229

The resistance of an ohmic conductor is proportional to
the its length, l, and inversely proportional to the cross
section area, A, of the conductor.
The constant of proportionality  is called the resistivity
of the material.



Every material has a characteristic resistivity that depends
on its electronic structure, and the temperature.
Good conductors have low resistivity.
Insulators have high resistivity.
230
Resistivity - Units
l
R
A





RA

l
Resistance expressed in Ohms,
Length in meter.
Area are m2,
Resistivity thus has units of W m.
231
Resistivity of various materials
Material
Silver
Copper
Gold
Aluminu
m
Pure
Silicon
Calcium
Sodium
Tungsten
Brass
Uranium
Mercury
Resistivity
(10-8 Wm)
1.61
1.70
2.20
2.65
Bismuth
Plutonium
Graphite
Germanium
Resistivity
(10-8 Wm)
106.8
141.4
1375
4.6x107
3.5
Diamond
2.7x109
3.91
Deionized
water
Iodine
Phosphorus
Quartz
Alumina
Sulfur
1.8x1013
4.75
5.3
7.0
30.0
98.4
Material
1.3x1015
1x1017
1x1021
1x1022
2x1023
232
Example
(a) Calculate the resistance per unit length of a nichrome wire
of radius 0.321 m.
Cross section:
Resistivity (Table): 1.5 x 106 Wm.
Resistance/unit length:
(b) If a potential difference of 10.0 V is maintained across a
1.0-m length of the nichrome wire, what is the current?
233
The reciprocal of the resistivity is called the
conductivity,
1


[Q] Stretching changes resistance: A wire of resistance R is stretched
uniformly until it is twice its original length. What happens to its
resistance? The resistance of the wire increases by a factor of four if the length increases twice
[Q] Speaker wires: Suppose you want to connect your stereo to remote
speakers. (a) If each wire must be 20m long, what diameter copper wire
should you use to keep the resistance less than 0.1Ω per wire? (b) If the
current on each speaker is 4.0A, what is the voltage drop across each
wire?
[Q] A 2.4m length of wire that is 0.031cm2 in cross section has a
measured resistance of 0.24Ω. Calculate the conductivity of the
material.
234
Temperature Variation of Resistance




The resistivity of a metal depends on many (environmental) factors.
The most important factor is the temperature.
For most metals, the resistivity increases with increasing temperature.
The increased resistivity arises because of larger friction caused by the
more violent motion of the atoms of the metal.
235
For most metals, resistivity increases approx. linearly
with temperature.
  is the resistivity at temperature T (measured in Celsius).
 o is the reference resistivity at the reference temperature To
(usually taken to be 20 oC).
 a is a parameter called temperature coefficient of resistivity.
For a conductor with fixed cross section.
236
Example:
A resistance thermometer, which measures temperature by
measuring the change in the resistance of a conductor, is made of
platinum and has a resistance of 50.0 W at 20oC. When the device
is immersed in a vessel containing melting indium, its resistance
increases to 76.8 W. Find the melting point of Indium.
Using a=3.92x10-3(oC)-1 from
table.
Ro=50.0 W.
To=20oC.
R=76.8 W.
237
[Q] A resistance thermometer using a platinum wire is used to
measure the temperature of a liquid. The resistance is 2.42
ohms at 0oC, and when immersed in the liquid it is 2.98 ohms.
The temperature coefficient of resistivity of platinum is
0.0038 . What is the temperature of the liquid?
238
Superconductivity


1911: H. K. Onnes, who had figured out how to make liquid helium,
used it to cool mercury to 4.2 K and looked at its resistance
At low temperatures the resistance of some metals0, measured
to be less than 10-16•ρconductor (i.e., ρ<10-24 Ωm)!
Resistance versus
temperature for a
sample of mercury
(Hg). The graph
follows that of a
normal metal above
the critical
temperature Tc. The
resistance drops to
zero at Tc, which is
4.2 K for mercury.
239
Electrical energy and power
In any circuit, battery is used to induce electrical current
 chemical energy of the battery is transformed into kinetic
energy of mobile charge carriers (electrical energy gain)
Any device that possesses resistance (resistor) present in
the circuit will transform electrical energy into heat
kinetic energy of charge carriers is transformed into heat
via collisions with atoms in a conductor (electrical energy
loss)
240
Electrical energy


Consider circuit on the right in detail
AB: charge gains electrical energy
form the battery
(battery looses chemical energy)
 CD: electrical energy lost (transferred into heat)
 Back to A: same potential energy (zero) as before
 Gained electrical energy = lost electrical energy on
the resistor
241
Power

Compute rate of energy loss (power dissipated on the
resistor)

Use Ohm’s law

Units of power : watt
delivered energy: kilowatt-hours


1 kWh  103W  3600 s   3.60 106 J
242
Example
A high-voltage transmission line with resistance of 0.31 W/km
carries 1000A , starting at 700 kV, for a distance of 160 km. What is
the power loss due to resistance in the wire?
Observations:
1. Given resistance/length, compute total resistance
2. Given resistance and current, compute power loss
Now compute power
243
Lecture 13
Direct Current Circuits
244
What is electromotive force (emf)?
A current is maintained in a closed circuit by a source of emf.
The term emf was originally an abbreviation for electromotive
force but emf is NOT really a force, so the long term is
discouraged.
A source of emf works as “charge pump” that forces electrons to
move in a direction opposite the electrostatic field inside the
source.
Examples of such sources are:
batteries
generators
thermocouples
photo-voltaic cells
245

246




Each real battery has some
internal resistance
AB: potential increases by on the
source of EMF, then decreases by
Ir (because of the internal
resistance)
Thus, terminal voltage on the
battery DV is
Note: EMF is the same as the
terminal voltage when the current
is zero (open circuit)
247



Now add a load resistance R
Since it is connected by a
conducting wire to the battery →
terminal voltage is the same as the
potential difference across the load
resistance
Thus, the current in the circuit is
[Q] Under what condition does the potential difference across the
terminals of a battery equal its emf?
248
Resistors in series
1. Because of the charge conservation, all charges going through the
resistor R2 will also go through resistor R1. Thus, currents in R1 and R2 are
the same,
2. Because of the energy conservation, total potential drop (between A and
C) equals to the sum of potential drops between A and B and B and C,
By definition,
Thus, Req would be
249

Analogous formula is true for any number of resistors,
(series combination)

It follows that the equivalent resistance of a series
combination of resistors is greater than any of the
individual resistors
[Q] How would you connect resistors so that the equivalent resistance
is larger than the individual resistance?
[Q] When resistors are connected in series, which of the following
would be the same for each resistor: potential difference, current,
power?
250
Example
In the electrical circuit below, find voltage across the resistor R1 in
terms of the resistances R1, R2 and potential difference between the
battery’s terminals V.
Energy conservation implies:
with
Then,
Thus,
This circuit is known as voltage divider.
251
Resistors in parallel
1. Since both R1 and R2 are connected to the same battery, potential
differences across R1 and R2 are the same,
2. Because of the charge conservation, current, entering the junction A,
must equal the current leaving this junction,
By definition,
Thus, Req would be
or
252

Analogous formula is true for any number of resistors,
(parallel combination)

It follows that the equivalent resistance of a parallel
combination of resistors is always less than any of the
individual resistors
[Q] How would you connect resistors so that the equivalent resistance is
smaller than the individual resistance?
[Q] When resistors are connected in parallel, which of the following
would be the same for each resistor: potential difference, current, power?
253
example
In the electrical circuit below, find current through the
resistor R1 in terms of the resistances R1, R2 and total
current I induced by the battery.
Charge conservation implies:
with
Then,
Thus,
This circuit is known as current divider.
254
Example
Find the currents in the circuit shown
255
Example
Find the currents I1 and I2 and the voltage Vx in the circuit
shown below.
I
7W
+
20 V
+
_
Vx
I2
4W
First find the equivalent resistance seen
by the 20 V source:
I1
12 W
Req  7W +
_
I 
4W(12W)
10 W
12W + 4W
20V 20V

 2A
Req
10W
We now find I1 and I2 directly from the current division rule:
2 A(4W)
I1 
 0.5 A, and I 2  I  I1  1.5 A
12W + 4W
Finally, voltage Vx is
Vx  I 2  4W  1.5 A  4W   6V
256
Kirchhoff’s Rules
1.
The sum of currents entering any junction must equal the
sum of the currents leaving that junction (current or
junction rule) .
I  0
2.
Charge conservation
 I1 + I 2 + I 3  0  I1  I 2 + I 3
The sum of the potential differences across all the
elements around any closed-circuit loop must be zero
(voltage or loop rule).
V  0
loop
Energy conservation
V  IR1  IR2  0  V  I R1 + R2 
257
V  0 for any loop
258

Rules for Kirchhoff’s loop rule
259
260

Solving problems using Kirchhoff’s rules
261

Example
262
263
264
Example

Find all three currents
 Need three equations for
three unknowns
 Note that current directions
are already picked for us
(sometimes have to pick for
yourself)
 Use the junction rule first
 Alternative two loops
V2
V1
R1
R3
265
266
RC Circuits




A direct current circuit may contain capacitors and resistors, the current
will vary with time
When the circuit is completed, the capacitor starts to charge
The capacitor continues to charge until it reaches its maximum charge
(Q = Cξ)
Once the capacitor is fully charged, the current in the circuit is zero
267
Charging Capacitor in an RC Circuit

The charge on the capacitor
varies with time


q = Q(1 – e-t/RC)
The time constant, =RC

The current I is

The time constant represents
the time required for the
charge to increase from zero to
63.2% of its maximum
dq  t RC
I
 e
dt R
268
Notes on Time Constant



In a circuit with a large time constant, the capacitor
charges very slowly
The capacitor charges very quickly if there is a small
time constant
After t = 10 , the capacitor is over 99.99% charged
269
Discharging Capacitor in an RC Circuit



When a charged capacitor is placed
in the circuit, it can be discharged
 q = Qe-t/RC
The charge decreases exponentially
The current I is
I

dq
Q t RC

e
 I 0 e t RC
dt
RC
At t =  = RC, the charge decreases
to 0.368 Qmax
 In other words, in one time
constant, the capacitor loses
63.2% of its initial charge
270
Example : charging the unknown capacitor
A series combination of a 12 kW resistor and an unknown capacitor is
connected to a 12 V battery. One second after the circuit is completed,
the voltage across the capacitor is 10 V. Determine the capacitance of
the capacitor.
I
C
R
271
Recall that the charge is building up according to
q  Q 1  et RC 
Thus the voltage across the capacitor changes as
q Q
V   1  et RC   E 1  et RC 
C C
This is also true for voltage at t = 1s after the switch is closed,
V
V
 1  e t RC  e t RC  1  
E
E
C
t
 V
 log 1  
RC
 E
t
1s

 46.5m F
 V
 10 V 
R log 1  
12, 000 W  log 1 

E


12
V


272
Lecture 14
Discussion
273
[1] What is the magnitude of the current
flowing in the circuit shown in Fig. ?
The net voltage drop due to the batteries
0V so no current flows. I=0A
[2] A copper wire has resistance 5 Ohms. Given that the resistivity
of silver is 85 percent of the resistivity of copper, what is the
resistance of a silver wire three times as long with twice the
diameter?
Given, 5W  Cu lCu / ACu ,
 Ag  0.85Cu ,
l Ag  3lCu , and d Ag  2dCu . 
 Agl Ag (0.85Cu )(3lCu ) (0.6375) Cu lCu
R


 (0.6375)(5W)  3.2W
2
2
2
rAg
 (2rCu )
rCu
274
[3] A resistor draws a current of 1A when connected across an
ideal 3V battery. Another resistor draws a current of 2A when
connected across an ideal 3V battery. What current do the two
resistors draw when they are connected in series across an ideal
3V battery?
The first resistor has a resistance of
V 3
R1    1.5W
I 2
The second resistor has a resistance of
R1 
V 3
  3W
I 1
The series combination of the two resistors is
R1 + R2  4.5W
Which when connected across a 3V battery will draw a current of
V
3
I 
 0.67 A
R 4.5
275
[4] consider an RC circuit in which the capacitor is being by a battery
connected in the circuit. In five time constant, what percentage of
final charge is on the capacitor?
q  Q(1  e
 t / RC
)
t  5 RC
q
t / RC
 1 e
Q
q
 1  e 5 RC / RC  1  e 5  99.3%
Q
276
[4] In fig. (a) find the time constant of the circuit
and the charge in the capacitor after the
switch is closed. (b) find the current in the
resistor R at time 10 sec after the switch is
closed. Assume R=1×106 Ω, emf =30 V and
C=5×10-6F
a)The tim e constan t  RC  (1106 )(5 106 )  5 sec
and the ch arg e on the capacitor Q  C  (5 106 )(30)  150mC
b)The current in ch arg ing of the capacitor is given by
I

R
e t / RC
10
30 ( (1106 )(510 6 ) )
6
I
e

4
.
06

10
A
6
110
277
[5] In the circuit shown in Fig. what is the current labeled I?
I  I1 + I 2
2  6 I1  2 I 2  0
2  2 I1  4 I 2  0

I 1  0 .2 A
I 2  0 .4 A
I  0 .6 A
278
[6] A certain wire has resistance R. What is the resistance of a
second wire, made of the same material, which is half as long and
has 1/3 the diameter?
The resistance is proportional to the length of the wire and inversely
proportional to the area. Since area is proportional to the diameter squared, the
resistance is
Rnew  R / 2  9  9R / 2
279
[7] The quantity of charge q (in coulombs) that has passed through
a surface of area 2.00 cm2 varies with time according to the
equation q =4t3 +5t +6, where t is in seconds. (a) What is the
instantaneous current through the surface at t = 1.00 s ? (b) What is
the value of the current density?
[a] I
t 1sec
dq

dt
t 1sec
 (12t 2 + 5)
t 1sec
 17 A
[b] A  2cm2  2 104 m 2
I
17 A
3
2
J 

85

10
A
/
m
A 2 10 4 m 2
280
[8] An electric current is given by the expression I(t) =100 sin(120πt),
where I is in amperes and t is in seconds. What is the total charge
carried by the current from t = 0 to t= (1/240) s?
t2
1/ 240
100
q   Idt   100sin(120t )dt  
cos(120t )
120
t1
0
1 / 240
0
100


[cos( )  cos(0)]  0.265C
120
2
281
[9] Suppose that you wish to fabricate a uniform wire out of 1.00
g of copper. If the wire is to have a resistance of R = 0.500 Ω,
and if all of the copper is to be used, what will be (a) the length
and (b) the diameter of this wire?
M
M
M
M
 A
 Al 
V 
[a]  d (m ass density) 
l d
d
d
V
 d l 2
l
l
R
Now, R    R 
M
M
A
l d
l
(0.5)(1103 )
 1.82m

3
8
(1.7 10 )(8.9210 )
 d
MR
[b] find r
282
[10] Compute the cost per day of operating a lamp that draws a
current of 1.70 A from a 110V line. Assume the cost of energy
from the power company is $ 0.060 0/kWh.
p  IDV  1.7 110  187W
Energy used in a 24H day  (0.187kW )(24h)  4.49kWh
cost  4.49 $0.06  $0.269
283
Lecture 15
Magnetic Fields & Force
284
Magnets ... two poles: N and S
Unlike poles attract
Like poles repel
285
PERMANENT MAGNETS
Figure (a) Magnetic field pattern surrounding a bar magnet as displayed
with iron filings. (b) Magnetic field pattern between unlike poles of two bar
magnets. (c) Magnetic field pattern between like poles of two bar magnets
286
Magnetic Field lines:
(defined in same way as electric field lines, direction and
density)
287
Broken Permanent Magnet


If we break a permanent magnet in half, we do not get a
separate north pole and south pole.
When we break a bar magnet in half, we always get two
new magnets, each with its own north and south pole.
288
Source of Magnetic Fields
What is the source of magnetic fields?
Answer: electric charge in motion
e.g., current in wire surrounding cylinder (solenoid) produces
very similar field to that of magnets.
Therefore, understanding source of field generated by bar
magnet lies in understanding currents at atomic level
within bulk matter.
Orbits of electrons about nuclei
Intrinsic “spin” of
electrons (more
important effect)
289
Magnetic Materials
•
Materials can be classified by how they respond to an applied magnetic field, Bapp.
• Paramagnetic (aluminum, tungsten, oxygen,…)
• Atomic magnetic dipoles (~atomic bar magnets) tend to line up with
the field, increasing it. But thermal motion randomizes their
directions, so only a small effect persists: Bind ~ Bapp •10-5
• Diamagnetic (gold, copper, water,…)
• The applied field induces an opposing field; again, this is usually
very weak; Bind ~ -Bapp •10-5 [Exception: Superconductors exhibit perfect
diamagnetism  they exclude all magnetic fields]
• Ferromagnetic (iron, cobalt, nickel,…)
• Somewhat like paramagnetic, the dipoles prefer to line up with the
applied field. But there is a complicated collective effect due to
strong interactions between neighboring dipoles  they tend to all
line up the same way.
•
Very strong enhancement. Bind ~ Bapp •10+5
290
Magnetic Field Direction
A vector quantity: magnitude and direction…
The letter B is used to represent magnetic fields.
291
Magnetic Field of the Earth



A small magnetic bar should be said to have north and
south seeking poles. The north of the bar points towards
the North of the Earth.
The geographic north corresponds to a south
magnetic pole and the geographic south
corresponds to a magnetic north.
The configuration of the Earth magnetic resemble that of
a (big) magnetic bar one would put in its center.
292
Magnetic Field of the Earth
293
Magnetic Fields in analogy with Electric Fields
Electric Field:


Distribution of charge creates an electric field E in
the surrounding space.
Field exerts a force F=q E on a charge q
Magnetic Field:


Moving charge or current creates a magnetic field
B in the surrounding space.
Field exerts a force F on a charge moving q
294
The magnetic force
Observations show that the force is proportional to
 The field
 The charge
 The velocity of the particle
 The sine of the angle between the field and the
direction of the particle’s motion.
295
Strength and direction of the Magnetic Force on a
charge in motion
[Q] The figure shows a charged particle with velocity v travels through
a uniform magnetic field B. What is the direction of the magnetic force
F on the particle?
296
Magnetic Field Magnitude
[Q] An electron (q = -1.6x10-19 C) is moving at 3 x 105 m/s in the positive
x direction. A magnetic field of 0.8T is in the positive z direction. The
magnetic force on the electron is:
297
Magnetic Field Units





[F] = Newton
[v] = m/s
[q] = C
[B] = tesla (T).
 Also called weber (Wb) per square meter.
 1 T = 1 Wb/m2.
 1 T = 1 N s m-1 C-1.
 1 T = 1 N A-1 m-1.
CGS unit is the Gauss (G)
 1 T = 104 G.
298
Right Hand Rule
299
Magnetic Force on Current- carrying conductor.



A magnetic force is exerted on a single charge in motion
through a magnetic field.
That implies a force should also be exerted on a
collection of charges in motion through a conductor I.e. a
current.
The force on a current is the sum of all elementary
forces exerted on all charge carriers in motion.
300
Magnetic Force on Current

If B is directed into the
page we use blue
crosses representing the
tail of arrows indicating
the direction of the field,

If B is directed out of the
page, we use dots.

If B is in the page, we use
lines with arrow heads.
301
Force on a wire carrying current in a magnetic field.
302
303
Force on a wire carrying current in a magnetic field.
304
Force on a wire carrying current in a magnetic
field.

General Case: field at angle 
relative to current.
B

B sin 
I
Note: If wire is not straight, compute force on differential
elements and integrate:

 
dF  i dL  B
305
Example: Wire in Earth’s B Field
A wire
carries a current of 22 A from east to west.
Assume that at this location the magnetic field of the
earth is horizontal and directed from south to north,
and has a magnitude of 0.50 x 10-4 T. Find the
magnetic force on a 36-m length of wire. What
happens if the direction of the current is reversed?
306
For the straight portion:
F1=IB2R out of the page
For the curved portion: If  is the angle between B and ds,
then the magnitude of dF2 is dF2= IdSXB= I B sin ds
Because s=R we have ds=Rd  dF2= I B R sin d
The resultant force F2 on the curved wire must be into
the page. Integrating our expression for dF2 over the limits
=0 to  (that is, the entire semicircle) gives
Ftotal  F1 + F2  2IBR  2IRB  0
The net force acting on a closed
current loop in a uniform
magnetic field is zero.
307
Example:
Wire with current i.
Magnetic field out of page.
What is net force on wire?
F1  F3  iLB
dF  iBdL  iBRd
By symmetry, F2 will only have a vertical component,


0
0
F2   sin( )dF iBR  sin( )d 2iBR
Ftotal  F1 + F2 + F3  iLB + 2iRB + iLB  2iB( L + R)
308
Lecture 16
Magnetic Fields& Force
309
Torque on a Current Loop
Imagine a current loop in a magnetic field as follows:
B
I
B
F
F
a/2
b
a
F1  F2  BIb
 max  BIba  BIA
F
F
In a motor, one has
“N” loops of current
  BIA sin 
310
Example:
A circular loop of radius 50.0 cm is oriented
at an angle of 30.0o to a magnetic field of
0.50 T. The current in the loop is 2.0 A. Find
the magnitude of the torque.
30.0o
B
311
Galvanometer
Device used in the construction of ammeters and
voltmeters.
312
Galvanometer used as Ammeter



Typical galvanometer have an internal resistance of
the order of 60 W - that could significantly disturb
(reduce) a current measurement.
Built to have full scale for small current ~ 1 mA or
less.
Must therefore be mounted in parallel with a small
resistor or shunt resistor.
Galvanometer
60 W
Rp
313
Galvanometer used as Voltmeter
• Finite internal resistance of a galvanometer must
also addressed if one wishes to use it as voltmeter.
• Must mounted a large resistor in series to limit the
current going though the voltmeter to 1 mA.
• Must also have a large resistance to avoid disturbing
circuit when measured in parallel.
Galvanometer
60 W
Rs
314
Motion of Charged Particle in magnetic field



Consider positively charge
particle moving in a uniform
magnetic field.
Suppose the initial velocity
of the particle is
perpendicular to the
direction of the field.
Then a magnetic force will
be exerted on the particle
and make follow a circular
path.
315
The magnetic force produces a centripetal acceleration.
The particle travels on a circular trajectory with a radius:
What is the period of revolution of the motion?
The frequency
316
Example : Proton moving in uniform magnetic field
A proton is moving in a circular orbit of radius 14 cm in a
uniform magnetic field of magnitude 0.35 T, directed
perpendicular to the velocity of the proton. Find the
orbital speed of the proton.
317
Example: If a proton
moves in a circle of
radius 21 cm
perpendicular to a B
field of 0.4 T, what is
the speed of the
proton and the
frequency of motion?
f 
qB
2m
1.6 1019 C  0.4T
f 
2 1.671027 kg
f 
1.6  0.4
10 8 Hz  6.110 6 Hz
6.28 1.67
f  6.1106 Hz
v
qBr
m
1.6 1019 C  0.4T  0.21m
v
1.671027 kg
x
v
x
r
x
x
v
1.6  0.4  0.21
10 8
1.67
v  8.1106
m
s
 8.110 6
m
s
m
s
318
Example: Mass Spectrometer


Suppose that B=80mT,
V=1000V. A charged ion
(1.6022 10-19C) enters
the chamber and strikes
the detector at a distance
x=1.6254m. What is the
mass of the ion?
Key Idea: The uniform
magnetic field forces the
ion on a circular path and
the ion’s mass can be
related to the radius of
the circular trajectory.
319
320
Cyclotrons


A cyclotron is a particle
accelerator
The D-shaped pieces
(descriptively called “dees”)
have alternating electric
potentials applied to them
such that a positively
charged particle always
sees a negatively charged
dee ahead when it emerges
from under the previous
dee, which is now positively
charged
321



The resulting electric field
accelerates the particle
Because the cyclotron sits
in a strong magnetic field,
the trajectory is curved
The radius of the trajectory
is proportional to the
momentum, so the
accelerated particle spirals
outward
322
Example: Deuteron in Cyclotron
Suppose a cyclotron is operated at frequency f=12
MHz and has a dee radius of R=53cm. What is the
magnitude of the magnetic field needed for deuterons to
be accelerated in the cyclotron (m=3.34 10-27kg)?
323
Example
Suppose a cyclotron is operated at frequency f=12 MHz and
has a dee radius of R=53cm. What is the kinetic energy of
the deuterons in this cyclotron when they travel on a
circular trajectory with radius R (m=3.34 10-27kg, B=1.57
T)?
A) 0.9 10-14 J
B) 8.47 10-13 J
C) 2.7 10-12 J
D) 3.74 10-13 J
mv
r
qB
implies
RqB
v
 3.99  107 m/s
m
K  12 mv 2  2.7  10 12 J
324
Lecture 17
Sources of the Magnetic Field
325
History


1819 Hans Christian Oersted discovered that a compass
needle was deflected by a current carrying wire
Then in 1920s Jean-Baptiste Biot and Felix Savart
performed experiments to determine the force exerted
on a compass by a current carrying wire
326
Biot & Savart’s Results





dB the magnetic field
produced by a small
section of wire
ds a vector the length of
the small section of wire in
the direction of the current
r the positional vector from
the section of wire to where
the magnetic field is
measured
I the current in the wire
 angle between ds & r
327





dB perpendicular to ds
|dB| inversely proportional to |r|2
|dB| proportional to current I
|dB| proportional to |ds|
|dB| proportional to sin 
328
Biot–Savart Law
All these results could be summarised by one “Law”
Putting in the constant
Where m0 is the permeablity of free space
329
Magnetic Field due to Currents
The passage of a steady current in a wire produces a magnetic
field around the wire.
 Field form concentric lines around the wire
 Direction of the field given by the right hand rule.
If the wire is grasped in the right hand with the thumb in
the direction of the current, the fingers will curl in the
direction of the field.

Magnitude of the field
330
Magnetic Field of a current loop
Magnetic field produced by a wire can be enhanced
by having the wire in a loop.
Dx
1
1 loop Current  I
B
I
N loops Current  NI
Dx2
331
Ampere’s Law
Consider a circular path surrounding a current, divided in
segments ds, Ampere showed that the sum of the products
of the field by the length of the segment is equal to mo times
the current.

 B.Ds  m0 I 

 B.ds  m0 Ienc
332
Example
By way of illustration, let us use Ampere's law
to find the magnetic field at a distance r
from a long straight wire, a problem we
have solved already using the Biot-Savart
law
 Bds cos   B  ds  B2r )
B(2r)  m o i
B
m oi
2r
for r  R
333
Now consider the interior of the wire, where r < R. Here the current i
passing through the plane of circle 2 is less than the total current I.
Because the current is uniform over the cross section of the wire, the
fraction of the current enclosed by circle 2 must equal the ratio of the
area πr2 enclosed by circle 2 to the cross-sectional area πR2 of the
wire
i ' r 2
 2
i R
2
r
i'  2 i
R
2
r
'
B
.
ds

B
(
2

r
)

m
i
0  m0 ( 2 i)

R
for r < R
334
Magnetic Force between two parallel
conductors
335
336
Definition of the SI unit Ampere
Used to define the SI unit of current called
Ampere.
If two long, parallel wires 1 m apart carry the same current,
and the magnetic force per unit length on each wire is
2x10-7 N/m, then the current is defined to be 1 A.
337
Example
Two wires, each having a weight per units length of 1.0x10-4
N/m, are strung parallel to one another above the surface of
the Earth, one directly above the other. The wires are
aligned north-south. When their distance of separation is
0.10 mm what must be the current in each in order for the
lower wire to levitate the upper wire. (Assume the two wires
carry the same current).
1
2
l
d
I1
I2
338
F1
1
I1
B2
mg/l
2
l
d
I2
339
Magnetic Field of a solenoid


Solenoid magnet consists of a wire coil with multiple
loops.
It is often called an electromagnet.
340
Solenoid Magnet





Field lines inside a solenoid magnet are parallel,
uniformly spaced and close together.
The field inside is uniform and strong.
The field outside is non uniform and much weaker.
One end of the solenoid acts as a north pole, the other
as a south pole.
For a long and tightly looped solenoid, the field inside
has a value:
341
Since the field lines are straight inside the solenoid, the
best choice for amperian loop is a rectangle: abcd.
Winding density: n=N/L where N = total number of
windings and L = total length. Integrate:
b
c
d
a
 B  ds   B  ds +  B  ds +  B  ds +  B  ds  m i
0 enc
a
b
c

d
b
 B  ds  Bh  m inh  B  m in
0
0
a
342
Solenoid Magnet
The field inside has a value:
n = N/L : number of (loop) turns per unit length.
I : current in the solenoid.
343
Example:
Consider a solenoid consisting of 100 turns of wire and
length of 10.0 cm. Find the magnetic field inside when it
carries a current of 0.500 A.
344
Lecture 18
Applications
345
[1] What is the net force on the
rectangular loop of wire in Fig. ?
The force on the top and bottom of the loop
are equal and opposite so they cancel.
Fright 
Fleft 
1A  2m  m  1A
 4  107 N to the left
2  1m
1A  2m  m  1A
 8  107 N to the right
2  0.5m
Fnet  4 107 right
346
[2] What is the net force per
unit length on the wire carrying
2A in Fig. ?
The 2A wire is attracted to the 1A wire (force up) and attracted to the 3A
wire ( force down). Thus, the force per unit length on the 2A wire is
m 0 (2 A)(3 A)
F m 0 (2 A)(1A)

up +
down
l
2 (1m)
2 (1m)
 8  107 N / m down
347
[3] What is the magnitude and direction
of the magnetic field at point P in Fig. ?
m 0 (1A)
B1 
 2 107 T
2 (1m)
m 0 (2 A)
B2 
 4  107 T
2 (1m)
m (3 A)
B3  0
 3  107 T
2 (2m)
m (4 A)
B4  0
 4  107 T
2 (2m)
int o
out
out
out
BT   B1 + B2 + B3 + B4  0.9  106 T out
348
[4] What is the magnitude and
direction of the magnetic field at
point Q in Fig.?
Bloop 
mI
int o
2R
mI
Bwire 
out
2R
1 mI
Bnet  (1  )
 8.6 107 T int o
 2R
349
[5] An electron moves is a circular orbit with diameter 10 cm
in a 5 Tesla magnetic field. What is the time it takes the
electron to complete one orbit?
T he timeto completean orbit is thedistance traveledin one orbit,2r,
divided by thespeed of theelectron,v.
2r
v
Since theelectronis movingin acircular orbit perpendicular
T
the magneticfield,
mv2
v eB
F
 qvB  
r
r m
Thus, the tim e to com plete an orbit is
2r 2m
T

 7 1012 s
v
eB
350
[6] A proton moves with a velocity of v = (2iˆ- 4jˆ +kˆ ) m/s in
a region in which the magnetic field is B " (iˆ + 2jˆ - 3kˆ ) T.
What is the magnitude of the magnetic force this charge
experiences?
F  qvBsin   q(v  B)
i
j
k
Now, v  B  + 2  4 + 1  10i + 7 j + 8k
+1 + 2  3
v  B  14.6T .m / s
F  1.6 1019 14.6  2.341018 N
351
[7] A conductor suspended by two flexible wires as shown in Figure
has a mass per unit length of 0.040 0 kg/m. What current must
exist in the conductor in order for the tension in the supporting
wires to be zero when the magnetic field is 3.60 T into the page?
What is the required direction for the current?
Equilibrium condition
 

2T + FB + mg  0


FB  mg
F m g IlB


l
l
l
m g 0.04 9.8
I

 0.109A
lB
3.6
The direction of I in the bar is to the right
Now,
352
[8] A positive charge q = 3.20 × 10-19 C moves with a velocity
v = (2iˆ + 3jˆ - kˆ ) m/s through a region where both a uniform
magnetic field and a uniform electric field exist.
(a) Calculate the total force on the moving charge (in unit-vector notation), taking
B = (2iˆ + 4jˆ + kˆ ) T and E =(4iˆ - jˆ - 2kˆ ) V/m. (b) What angle does the force
vector make with the positive x axis?
a ] Lo r e ntz for c e is given by
F  Felectric + Fmagnetic
F  qE + q (v  B )  q ( E + v  B )
find ( E + v  B) ??
F  (3.52i  1.6 j )  1018 N
b]   tan
1
Fy
Fx
 24.4
353
[9] A nonconducting sphere has mass 80.0 g and radius 20.0 cm. A flat compact
coil of wire with 5 turns is wrapped tightly around it, with each turn concentric
with the sphere. As shown in Figure, the sphere is placed on an inclined plane
that slopes downward to the left, making an angle θ with the horizontal, so that
the coil is parallel to the inclined plane. A uniform magnetic field of 0.350 T
vertically upward exists in the region of the sphere. What current in the coil will
enable the sphere to rest in equilibrium on the inclined plane? Show that the
result does not depend on the value of !.
354
[10] The segment of wire in Figure carries a current of I =5.00 A, where the
radius of the circular arc is R = 3.00 cm. Determine the magnitude and direction
of the magnetic field at the origin.
The quarter circle m akes one forth the field
a full loop
1 m0 I
B (
)  26.2mT int o
4 2R
355
[11]Figure shows a section of a long hallow cylindrical of radii
a=5cm and b=10 cm, carrying a uniform distributed current I,
the magnitude of the magnetic field on its outer surface at r=b
is measured to be B=0.2T, where r is the radial distance from
the cylindrical axis.
(a) Find the current in the wire?
Am per' s Law on the outer surface (r  b)
 B.dl  m i
0 enc
B(2b)  m0i
i
2Bb
m0
2  0.2  0.1
5


10
A
7
4 10
356
(b) Find the magnitude of the magnetic field at r=20 cm?
 B.dl  m i
0 enc
B(2r )  m0i
m0i 4 10 10
B

 0.1T
2r
2  0.2
7
5
357
(c) Find the magnitude of the magnetic field at r=8 cm?
 B.dl  m i
0 enc
r 2  a 2
B(2r )  m0 ( 2
)i
2
b  a
m 0i r 2  a 2
B
( 2
)
2
2 r (b  a )
4 107 105
0.082  0.052
B
(
)  013T
2
2
2
0.08(0.10  0.05 )
358
Lecture 19
Faraday’s Law
359
magnetic flux
Definition of Magnetic Flux

The flux, , is defined as the product of the
field magnitude by the area crossed by the
field lines.
  B A  BA cos 
 
 B   B  dA

where B is the component of B
perpendicular to the loop,  is the angle
between B and the normal to the loop.
Units: T·m2 or Webers (Wb)
360
(a) The flux through the plane is zero when the magnetic
field is parallel to the plane surface.
(b) The flux through the plane is a maximum when the
magnetic field is perpendicular to the plane.
361
Example: A square loop 2.00m on a side is placed in a
magnetic field of strength 0.300T. If the field makes an angle
of 50.0° with the normal to the plane of the loop, determine
the magnetic flux through the loop.
  BA cos   0.300T  2.00m  cos 50.0
2
 0.386 Tm2
362
Faraday’s Law: Experiments

A current appears only if there is relative
motion between the loop and the magnet;
the current disappears when the relative
motion between them ceases.

Faster motion produces a greater current.

If moving the magnet’s north pole toward
the loop causes, say, clockwise current,
then moving the north pole away causes
counterclockwise current. Moving the
south pole toward or away from the loop
also causes currents, but in the reversed
directions.

An emf is induced in the loop when the number of magnetic field lines
that pass through the loop is changing.
363
The needle deflects momentarily when the switch is
closed
364
Faraday’s Law of Induction



The magnitude of the emf induced in a conducting loop is equal to
the rate at which the magnetic flux through that loop changes with
time，
If a coil consists of N loops with the same area,
the total induced emf in the coil is given by
In uniform magnetic field, the induced emf can be
expressed as
365
To induce an emf we can change,
•
•
•
•
the magnitude of B
the area enclosed by the loop
the angle between B and the normal to the area
any combination of the above over time.
366
Example 1: EMF in a loop
A wire loop of radius 0.30m lies so that an external magnetic field
of strength +0.30T is perpendicular to the loop. The field changes
to -0.20T in 1.5s. (The plus and minus signs here refer to opposite
directions through the loop.) Find the magnitude of the average
induced emf in the loop during this time.
B
367
The loop is always perpendicular to the field, so the normal to the loop
is parallel to the field, so cos  = 1. The flux is then
  BA  B r 2
Initially the flux is
i   0.30T    0.30m  =0.085 T  m 2
2
and after the field changes the flux is
 f   0.20T    0.30m  =-0.057 T  m2
2
The magnitude of the average induced emf is:
D  f  i 0.085 T  m 2 -0.057 T  m 2
emf 


 0.095V
Dt
Dt
1.5s
368
Example: One way to Induce an emf in a coil
A coil consists of 200 turns of wire having a total resistance of 2.0
W. Each turn is a square of side 18 cm, and a uniform magnetic
field directed perpendicular to the plane of the coil is turned on. If
the field changes linearly form 0 to 0.50 T in 0.80s, what is the
magnitude of the induced emf in the coil while the field is
changing.
 The area of one turn of the coil is (0.18m)2 = 0.0324 m2.
 The magnetic flux through the coil at t=0 is zero because B=0 at that time.
 At t=0.80s, the magnetic flux through one turn is :
B = BA = (0.50T)(0.0324m2) = 0.0162T.m2
 Therefore, the magnitude of the induced emf is :
NDB 200(0.0162T.m 2  0T.m 2 )
e

 4.1T.m 2 / s  4.1V
Dt
0.80s
369
Example : An Exponentially Decaying B Field
A loop of wire enclosing an area A is placed in a region where the
magnetic field is perpendicular to the plane of the loop. The
magnitude of B varies in time according to the expression B =
Bmax e-at, where a is some constant. That is, at t=0 the field is Bmax,
and for t>0, the field decreases exponentially. Find the induced
emf in the loop as a function of time.
370
Solution
 Because B is perpendicular to the plane of the loop, the
magnetic flux thought the loop at time t>0 is : B =
BAcos0 = ABmaxe-at
 Because Abmax and a are constants, the induced emf is :
e
d B
d
 AB max e at  aAB maxe at
dt
dt
 This expression indicates that the induced emf decays
exponentially in time.
 Note that the maximum emf occurs at t=0, where emax =
aABmax.
371
Motional EMF
As the wire moves,
FB  qv  B
Which sets the charges in motion in the
direction of FB and leaves positive charges
behind.
As they accumulate on the bottom, an electric field is set up inside.
In equilibrium,
FB  FE or qvB  qE or E  vB
DV  El  Blv
372
Motional EMF in a Circuit
 B  BA  Blx
d B
d
dx
E 
  Blx    Bl
dt
dt
dt
E   Blv
E
Blv
I

R
R
If the bar is moved with constant velocity,
Fapp  FB  IlB
B 2l 2 v 2 E 2
P  Fapp v  IlBv 

R
R
373
Example: Motional emf Induced in a Rotating Bar
A conducting bar of length L rotates with a constant angular
speed ω about a pivot at one end. A uniform magnetic field B is
directed perpendicular to the plane of rotation, as shown in Figure.
Find the motional emf induced between the ends of the bar.
374
Solution
 Consider a segment of the bar of length dr having a velocity v.
 The magnitude of the emf induced in this segment is :
de  Bvdr
 Because every segment of the bar is moving perpendicular to B, an emf de of
the same form is generated across each.
 Summing the emfs induced across all segments, which are in series, gives the
total emf between the ends of the bar :
e   Bvdr
 To integrate this expression, we must note that the linear speed of an element is
related to the angular speed  through the relationship v = r.
 Therefore, because B and  are constants, we find that :
1
e  B vdr  B rdr  B 2
2

0
375
Example : Magnetic Force Acting on a Sliding Bar

The conducting bar illustrated in Figure, of mass m and length
,
moves on two frictionless parallel rails in the presence of a uniform
magnetic field directed into the page. The bar is given an initial
velocity vi to the right and is released at t=0. Find the velocity of the
bar as a function of time.
376
Solution
 The induced current is counterclockwise, and the magnetic force is
FB = -I  B, where the negative sign denotes that the force is to the left and
retards the motion.
 This is the only horizontal force acting on the bar, and hence Newton’s second
law applied to motion in the horizontal direction gives :
Fx  ma  m
dv
 IB
dt
 We know that I=B v/R, and so we can
 write this expression as :
dv
B2  2
m

v
dt
R
 B2  2 
dv
dt
 
v
 mR 
377
 Integrating this equation using the initial condition that v=vi at t=0, we find
that:
dv  B2  2 t


0 dt
v
mR
v
vi
v
 B2  2 
t
t  
ln   

 mR 
 vi 

where the constant  
mR
B2 2
 From this result, we see that the velocity can be expressed in the exponential
form : v  vi e t / 
 This expression indicates that the velocity of the bar decreases exponentially
with time under the action of the magnetic retarding force.
378
Lenz’s Law
The polarity of the induced emf is such that it tends to
produce a current that creates a magnetic flux to oppose the
change in magnetic flux through the area enclosed by the
current loop.
As the bar is slid to the right,
the flux through the loop
increases.
This induces an emf that will
result in an opposing flux.
Since the external field is into
the screen, the induced field has
to be out of the screen.
Which means a
counterclockwise current
379
Energy Considerations
Suppose, instead of flowing counterclockwise, the induced current flows
clockwise:
Then the force will be towards the right
which will accelerate the bar to the right
which will increase the magnetic flux
which will cause more induced current to flow
which will increase the force on the bar
… and so on
All this is inconsistent with the conservation of energy
380
Moving Magnet and Stationary Coil
 Right moving magnet
increases flux through
the loop.
 It induces a current that
creates it own magnetic
field to oppose the flux
increase.
 Left moving magnet
decreases flux through
the loop.
 It induces a current that
creates it own magnetic
field to oppose the flux
decrease.
381
Application of Lenz’s Law
 When the switch is closed, the flux goes
from zero to a finite value in the direction
shown.
 To counteract this flux, the induced current
in the ring has to create a field in the opposite
direction.
 After a few seconds, since there is no change in the flux, no current
flows.
 When the switch is opened again, this time flux decreases, so a current
in the opposite direction will be induced to counter act this decrease.
382
A Loop Moving Through a Magnetic Field
383
Induced EMF and Electric Fields
Changing Magnetic Flux
EMF
Electric Field Inside a
Conductor
This induced electric field is non-conservative and time-varying
E 
d B
dt
d B
 E.ds   dt
General Form of
Faraday’s Law
W  qE  FE 2r 
qE  qE2r 
E
2r
1 d B
1 d

E

r 2 B 
2r dt
2r dt
r dB
E
2 dt
E
384
Example : Electric Field Induced by a Changing Magnetic
Field in a Solenoid
A long solenoid of radius R has n turns of wire per unit length and
carries a time-varying current that varies sinusoidally as I = Imax cos
t, where Imax is the maximum current and  is the angular
frequency of the alternating current source Figure. (a) Determine
the magnitude of the induced electric field outside the solenoid, a
distance r > R from its long central axis. (b) What is the magnitude
of the induced elelctric field inside the solenoid, a distance r from
its axis?
385
Solution for (a)
Consider an external point and take the path for our line integral to be a circle
of radius r centered on the solenoid as illustrated in Figure (31.18).
The magnitude of E is constant on this path and that E is tangent to it.
The magnetic flux through the area enclosed by this path is BA = BR2; hence :
 
d
2
2 dB
E

d
s


(
B

R
)



R

dt
dt
 
2 dB
E

d
s

E
(
2

r
)



R

dt
The magnetic field inside a long solenoid is given by ,
(1)
B = monI.
Substitute I = Imax cos t into this equation and then substitute the result into
Eq. (1), we find that :
386
d
(cost )
dt
 R 2m o nImax sin t
E(2r )  R 2m o nImax
m o nImaxR 2
E
sin t
2r
(for r > R)
(2)
387
Solution for (b)
 For an interior point (r < R), the flux threading an integration loop is given
by Br2.
 Using the same procedure as in part (a), we find that :
dB
E(2r )  r
 r 2m o nI max sin t
dt
2
m o nI max
E
r sin t
2
(3)
(for r < R)
388
Generators and Motors
 B  BAcos  BAcost
d B
d
  NAB cost   NAB sin t
dt
dt
E max  NAB
E  N
389
Example: An AC generator consists of 8 turns of wire,
each of area A = 0.090 0 m2 , and the total resistance of
the wire is 12.0 (. The loop rotates in a 0.500-T magnetic
field at a constant frequency of 60.0 Hz.
(A) Find the maximum induced emf.
  2f  377s 1
e max  NAB  136V
(B) What is the maximum induced current when the output
terminals are connected to a low-resistance conductor?
I max 
e max
R
 11 .3 A
390
Maxwell’s Equations
Q
 E.dA  e
Gauss’ Law
0
 B.dA  0
d B
 E.ds   dt
Gauss’ Law for Magnetism
no magnetic monopoles
Faraday’s Law
d E
 B.ds  m0 I + m0e 0 dt
F  qE + qv  B
Ampère-Maxwell Law
Lorentz Force Law
391
Lecture 20
Discussion
392
[1] A rectangular coil of 150 loops forms a
closed circuit with a resistance of 5 and
measures 0.2 m wide by 0.1 m deep, as shown
below. The circuit is placed between the poles
of an electromagnet which is producing a
uniform magnetic field of 40 T. The magnet is
switched off, causing the magnetic field to
drop to zero in 2 s. (The loops are parallel with
the faces of the electromagnet.)
a) Compute the average induced potential in the circuit.
b) Determine the average current in the circuit.
c) Indicate on the drawing the direction in which the
induced current flows.
393
a] D  (0.1)(0.2)[0  40]  0.8T .m 2
d
 0.8
  N
 (150)[
]  60V
dt
2
V 60
b] I  
 12A
R 5
C] The magnetic field from the magnet is up, but the flux is
decreasing. So the magnetic field from the current induced in
the loop(s) will also be up. Thus the current flow is left-toright across the front of the loop.
394
[2] A conducting rod of length l moves on two (frictionless)
horizontal rails, as shown to the right. A constant force of magnitude

Fapp  1.0 N

moves the bar at a constant speed of v  2.0m / s

through a magnetic field B
directed into the page. The resistor has a value R  8.0W
a) What is the current through the resistor
R?
(b) What is the mechanical power delivered
by the
constant force?
(
395
When the conducting rod moves to the right, this serves to increase the flux as
time passes , so any induced current wants to stop this change and decrease the
magnetic flux. Therefore, the induced current will act in such a way to oppose
the external field (i.e., the field due to the induced current will be opposite to the
external field). This must be a counterclockwise current.
To find the current, we only need to find the motionally-induced voltage and
then apply Ohm’s law. The rod is just a bar of length l moving at velocity v in a
magnetic field B. This gives us an voltage ∆V , and Ohm’s law gives us I:
ΔV  Blv  IR
Blv
 I
R
396
Keep in mind that v is velocity, while ∆V is voltage. Of course, the problem is
now that we don’t know B. We do know that the external and magnetic forces
must balance for the rod to have a constant velocity however. Constant
velocity implies zero acceleration, which implies no net force. If this is to be
true, the applied force must exactly balance the magnetic force on the rod.
For a conductor of length l carrying a current I in a field B, we know how to
calculate the magnetic force
Fm  BIl  Fapp
 B
Fapp
Il
Plug that into the first equation:
DV Blv Fapp lv Fapp v
I



R
R
IlR
IR
 I 
2
Fapp v
R
You should get I=0.5 A.
397
What about the power? Conservation of energy tells us that the mechanical
power delivered must be the same as the power dissipated in the resistor, or
PF  Fv cos
You should get 2 W.
Alternatively, you note that power delivered by a force is
P F  PR  I 2 R
where θ is the angle between the force and velocity. In this case, θ = 0, so
PF  Fv  2W
398
[4] Figure shows a right view of a bar that can slide
without friction. The resistor is 6.00 Ω and a 2.50T
magnetic field is directed perpendicularly downward,
into the paper. Let L= 1.20 m.
(a) Calculate the applied force required to move the
bar to the right at a constant speed of 2.00 m/s.
(b) At what rate is energy delivered to the resistor?
399
[6] Consider the mass spectrometer. The magnitude of the electric
field between the plates of the velocity selector is 2 500 V/m, and
the magnetic field in both the velocity selector and the deflection
chamber has a magnitude of 0.035 0 T. Calculate the radius of the
path for a singly charged ion having a mass m =2.18 * 10-26 kg.
400
[7] An electron moves in a circular path perpendicular to a
constant magnetic field of magnitude 1.00 mT. The angular
momentum of the electron about the center of the circle is 4.00
* 10-25 Js. Determine (a) the radius of the circular path and (b)
the speed of the electron.
401
[8] A proton moves with a velocity of v = (2iˆ - 4jˆ +kˆ ) m/s in a
region in which the magnetic field is B = (iˆ +2jˆ - 3kˆ ) T. What is
the magnitude of the magnetic force this charge experiences?
Give it in unit vector notation,
402