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General Physics II
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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.
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.
II. Course Objectives
3
Course Outline
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
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.
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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.
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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.
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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 ComponentWeight
HW/Quizzes20
Midterm Exam20
Final Exam60
Total100
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Lecture I
Electrostatic
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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.
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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.
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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.
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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.
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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. Proton has a charge +1e. Electron has a charge –1e. Some particles such a neutron have no (zero)
charge.
"charge is quantized" in terms of an equation, we
say: q = n e
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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".
Unit of Electrical Charge: The Coulomb " C "
Example: How many electrons are there in 1 C of charge?
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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.
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Example:
Identify substances or materials that can be classified as Conductors ? Insulators?
Why? is static electricity more apparent in winter?
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Lecture 2Lecture 2
Coulomb’s law
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Coulomb’s LawCoulomb 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.
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Coulomb’s Law (Mathematical Formulation)
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).
1 2F q q
2
1F
r
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where is known as the Permittivity constant of free space.
= 8.85 x 10-12 C2/N.m2
9 2 212
1 19 10 /
4 4 8.85 10eK Nm C
Experimentally measurement: ke = 8.9875×109 Nm2/C2.Reasonable approximate value: ke = 8.99×109 Nm2/C2.
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Example: Example: the Coulomb constant unitthe Coulomb constant unit
Then the Coulomb constant unit isThen the Coulomb constant unit is
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The electrostatic force
The electrostatic force is often called Coulomb force.
It is a force (thus, a vector): a magnitude a direction. +
+r
q1
q2F21
F12
+-
r
q1
q2F21F12
12 21
12 21
F F
F F
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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.
qq11 =-1.60x10 =-1.60x10-19-19 C C
qq22 =1.60x10 =1.60x10-19-19 C C
r = 5.3x10r = 5.3x10-11-11 m m
2
2
22 19
9 822 11
1.6 109 10 8.2 10
5.3 10
Nme e C
CeF k N
r m
Attractive force with a magnitude of 8.2x10-8 NAttractive force with a magnitude of 8.2x10-8 N..
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Gravitational vs. Electrical Force
Felec
Fgrav =
q1q2
m1m2
140
G
r
F Fq1
m1
q2
m2
Felec = 1
40
q1q2
r2
Fgrav = G m 1 m 2
r 2
For an electron:
|q| = 1.6 10-19 Cm = 9.1 10-31 kg F
Felec
grav
417 10 42.
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Superposition of Forces
T 10 20 30F F F F ....
0 1 0 2 0 3T 10 20 302 2 2
10 20 30
kq q kq q kq qˆ ˆ ˆF r r r ....
r r r
N31 2 i
T 0 10 20 30 0 i02 2 2 2i 110 20 30 i0
qq q qˆ ˆ ˆ ˆF kq r r r .... kq r
r r r r
+Q3
+Q2
+Q1
+Q0
10F
30F
20F
10r
20r
30r
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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.
2
2
2
2
9 9
3 2 9 932 22
9 9
3 1 9 831 22
932 31
931
2 2 9
5 10 2 109 10 5.62 10
4
5 10 6 109 10 1.08 10
5
cos37.0 3.01 10
sin 37.0 6.50 10
7.16 10
65.2
Nme C
Nme C
ox
oy
x y
o
C Cq qF k N
r m
C Cq qF k N
r m
F F F N
F F N
F F F N
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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 ?
31 32
3 1 3 22 2
31 32
6 6
2 2
1 10 3 10
( 10)
F F
q q q qk k
r r
d d
Solve the eq. to find d.
29
Examination of the geometry of Figure leads to
22 2/
2/tan
xL
x
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
Fq
xe 2
024
by Eq. When these two expressions are used in the equation mg tan Ө = Fe, we obtain
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3/12
2
2
24
1
2
mg
Lqx
x
q
L
mgx
oo
b) We solve x3 for the charge:
3238
9 2 2
0.010 kg 9.8m s 0.050m2.4 10 C.
2 2 8.99 10 N m C 1.20 m
mgxq
kL
Thus, the magnitude is
8| | 2.4 10 C.q
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Lecture 3
The Electric Field
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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.
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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
This is the field produced by the charge Q, not by the charge q0
0
/F
E N Cq
Electric Field & Electric Force
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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
35
2o
e
q qF k
r
2e
qE k
r
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
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+
r
qqo
E
-
r
qqoE
• 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.
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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?
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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.
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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)
40
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
41
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
Electric Field from an Electric DipoleElectric Field from an Electric Dipole
42
ExampleExample::ExampleExample::
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
43
Principle of superposition:The electric field at any point x is the sum of the electric fields from +q and -q
Replacing r+ and r- we get
2 20 0
1 1
4 4
q qE E E
r r
2
212
210
11
4 dxdx
qE
This equation gives the electric field everywhere
on the x-axis (except for x = d/2)
44
Example: Electric Field Due to Two Point Charges
Charge q1=7.00 C is at the origin, and charge q2=-10.00 C 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.
x
y
0.300 mq1 q2
0.40
0 m
P
E1
E2
E
45
2
2
2
2
6
1 9 51 22
1
6
2 9 52 22
2
5325
541 2 1 25
2 2 5
1
7.00 108.99 10 3.93 10 /
0.400
10.00 108.99 10 3.60 10 /
0.500
2.16 10 /
sin 1.05 10 /
2.4 10 /
tan ( / ) 25.9
Nme C
Nme C
x
y
x y
oy x
CqE k N C
r m
CqE k N C
r m
E E N C
E E E E E N C
E E E N C
E E
46
ExampleIn Figure, determine the point (other than infinity) at which thetotal 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.
47
Motion of charge particles in a uniformelectric 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.
-ve
- - - - - - - - - - - - - - - - - - - - - -
+ + + + + + + + + + + + + + + + + + + + + +
0E
48
Horizontally: No electric field No force No acceleration Constant horizontal
velocity
0
0
0
x
x
x
x o
o
E
F
a
v v
x v t
Vertically: Constant electric field Constant force Constant acceleration Vertical velocity increase
linearly with time.2
/
/
1/
2
y o
y o
y o e
y o e
o e
E E
F eE
a eE m
v eE t m
y eE t m
49
-
- - - - - - - - - - - - - - - - - - - - - -
+ + + + + + + + + + + + + + + + + + + + + +
Conclusions: The charge will follow a parabolic path downward. Motion similar to motion under gravitational field only
except the downward acceleration is now larger.
50
-e-e
-Q +Q
+Q
-Qx
Phosphor Screen
This device is known as a cathode ray tube (CRT)
51
Continuous Charge Distributions
Ni
0 i02i 1 i0
qˆE k r
r
0 2all charge
dqˆE k r
r
Discrete chargesContinuous charge distribution
0 2
kqˆE r
r
Single charge
0 2
kdqˆdE r
r
Single piece of a charge distribution
+Q3
+Q2
+Q1
01E
03E
02E
10r
20r
30r 0 0
++
++
0dE
dq
52
Q dq
dx
Line charge dq dx dq Rd
Cartesian Polar
Surface chargeQ dq
A dA dq dxdy dq rdrd
Volume chargeQ dq
V dV dq dxdydz dq rdrd dz
2dq r sin drd d
2
kdqˆdE r
r
53
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.
Example: Electric Field Due to a Charged Rod
22 x
dxk
x
dqkdE
dxdq
ee
al
ae
al
a
e
al
a
e xk
x
dxk
x
dxkE
122
)(
11
ala
Qk
alal
QkE e
e
54
Example – Infinitely Long Line of Charge
+
+
+
+
dE
+
+
+
dy
x
dq dy
y
2 2 2r x y
2
kdqˆdE r
r
xdE
ydE
+
y-components cancel by symmetry
x 2
kdqdE cos
r
2 2 2 2
k dy xdE
x y x y
3 2
2 2 2
dy 2 2kE k x k x
x xx y
55
Example – Charged Ring
dE
z
dq ds ad
2 2 2r z a
zdE
dE
a
d
+
+
+
+ +
+
+
2
kdqˆdE r
r
perpendicular-components cancel by symmetry
z 2
kdqdE cos
r
2 2 2 2
k ad zdE
z a z a
2
3 3 32 2 2 2 2 202 2 2
k za k za kQzE d 2
z a z a z a
56
When: z a
The charged ring must look like a point source.
3 3 3 2
2 2 22 22
2
kQz kQz kQz kQE
z zz a a
z 1z
0
57
Example – Uniformly Charged Disk
dE
z
3
2 2 2
kQzE
z r
r
3
2 2 2
kzdqdE
z r
dq dA rdrd 2 rdr
3
2 2 2
kz 2 rdrdE
z r
R
2 2
2
R R z R
3 3 32 2 2 20 02 2 2z
kz 2 rdr 2rdr duE kz kz
z r z r u
2 2
2 2
2
2
z R1
z R 3 22
2 2 2 2 2z
z
u 1 1 zkz u du kz 2kz k 2 1
1 z R z z R2
58
Two Important Limiting Cases
2 2o o
z 1E k 2 1 k 2 2
4 2z R
Large Charged Plate: R z
dE
z
rR
Very Far From the Charged Plate: z R
12 2
22 2 2
2
z z RE k 2 1 k 2 1 k 2 1 1
zz R Rz 1
z
2 2 2
2 2 2 2
1 R 1 R k R kQk 2 1 1 k 2
2 z 2 z z z
59
Lecture 4
Discussion
60
[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.
0
46.0
0sinsin
46.023.02)5
4(29.0cos2
29.0)5.0(
)102)(104(109
2
669
21
1 2
NF
FFF
NFF
N
r
qQKFF
T
y
x
eQqQq
61
[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.
12 13
2 2
13 14
2 2
0 cos 0
1.
2 2
....................................(1)2 20 sin 0
1.
2 2
....................................(1)2 2
2 2
x
y
F F F
kQq kQQ
a a
F F F
kQQ kQq
a a
then
Q q
62
[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?
31 32
3 1 3 22 2
31 32
6 6
22
22 2 2
1 10 3 10
10
3 10 3 20 100
e e
F F
q q q qK K
r r
d d
d d d d d
63
2 2
2
2 20 100 0 10 50 0
1. 10, 50
10 100 4(1)( 50)4
2 2
10 10 310 300 2
2 2
5 5 3 13.66
5 5 3 13.66
)the equilibrium unstable
d d d d
a b c
b b acd
a
d
d
d
b s
64
[4 ]Find the electric field at point p in figure due to the charges shown.
Solution: 1 2
4
3
4
2 2
36 10 /
28 10 /
( ) ( )
46.1 /
141
x
y
p x y
E E E
N C
E E
N C
E E E
N C
65
[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.
Substitute T from equation (1) into equation (2)
5
5
3 10 /
5 10 /
sin 37 0....(1)
cos37 0....(2)
x
y
x x
y y
E N C
E N C
F qE T
F qE T
Substitute T from equation (1) into equation (2)
Substitute by q into equation (1) to find T=5.44*10-3N
66
[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
NFFF
r
qqKF
r
qqKF
t
e
e
3332321
32
669
223
3223
32
669
221
2121
10149103210117
10325.1
105.2102.3109
101775.0
105.2103.1109
67
[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?
md
md
d
dd
ddd
dd
d
qK
d
qK
EE
ee
3/1
16
13442
0123
012_4
14
1
4
2
22
22
22
21
68
[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] The electric force on a point charge of 4.0mC at some point is 6.9*10-
4N in the positive x direction. What is the value of the electric field at that point?
9
21
9
210
999
221 106.1
104.14
106.14.14
108.3
106.1106.1109
r
qqKF e
CNq
FE 2
6
4
10725.1100.4
109.6
69
[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
2
6
2
69
21
2
69
22
22
2
69
21
11
10
100.3100.1109
10
100.3109
100.1109
xxEEE
xr
qKE
xr
qKE
PPT
eP
eP
70
]11[ Calculate E (direction and magnitude) at point P in Figure.
22
31
2
0
r
qkEE
EEE
ey
x
it fromaway triangle theof middle From
4.
4
1
2
2.
4
1
2
2
20
20
20
2
a
qE
a
q
a
q
a
qkEE eyT
71
[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.
20
0
6
3
:
1/ 2
2.7 10 /
/ / 1 10 /
Horizental
first find the acceleration from the relation
x v t at
then find the velocity
v v at
v m s
then find the electric field E F q ma q N C
72
[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?
NFFFF
iselectrontheonforcenettheand
Nr
cekF
isforcesthreetheofeachofmagnitudetheThus
mrr
yxnet922
92
106.82
103.41
,
577.05.0
30cos
73
[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 ?
matd
bygivenisdistncetherestfromstartselectrontheSince
smm
eE
m
Fa
ee
14.0)102)(107(2
1
2
1
,
/107
28142
24
74
Lecture 5 Electric Flux and Gauss’s Law
75
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
Here A is a vector whose magnitude is the surface area A and whose orientation is normal to the surface.
cos .EA E A
E
A
76
When ˚, the flux is positive (out of the surface), and when ˚, 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 NmAnswer: F = 44 Nm22/C./C.
77
Example:
Calculate the flux of a constant E field (along x) through a cube of side “L.”
x
y
z
E
12
SolutionSolution
2 2 0net EL EL
21 1 1cosEA EL
22 2 2cosEA EL
78
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?
3s E
2s E
3 cos45s E
2 cos45s E
79
When we have a complicated surface, we can divide it up into tiny elemental areas:
AdEdAEAdEd
.cos
80
What’s the total flux on a closed surface with a charge inside? The shape and size don’t
matter! Just use a sphere
oo
o
E
qr
r
q
dAr
q
dE
d
22
2
44
1
4
1
)0cos(A
AE
Example:
81
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.
82
Gauss’ Law The precise relation between flux
and the enclosed charge is given by Gauss’ Law
0 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.
dAE
dd
E
E
.
AE
o
enclE
Qd
AE
The flux through a closed surface is equal to the total charge contained divided by permittivity of free space
83
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?
84
Question
What’s the total flux with the charge outside? Why?
Solution: Zero. Because the surface
surrounds no charge
85
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
20
20
2
0
2
4)4(
)4(.
r
Qk
r
QE
QrE
Q
rEdAEAdE
en
en
86
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
87
Let’s calculate the electric field from a conducting wire with charge per unit length using Gauss’ Law
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
A linear charge distribution
qdo
E 1
AE
88
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.
89
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
… and now solve for the electric field
E
20r
2kr
//2
2)0cos(
00
LqrLE
rLEAdE
AdE
90
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
A surface charge distribution
91
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
… so the electric field from an infinitenon-conducting sheet with charge density
E
20
E
20
// 00 Aq
EAEAdAE
92
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.
93
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
… so the electric field from an infiniteconducting sheet with surface charge density is
E 0
E 0
0A
EA
94
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)
A volume charge distribution
95
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
Solving for E we find
313
4214 rrEdAE
0
1
3 r
E inside
96
0
13
34 3
r
r
QE
inside
In terms of the total charge Q …
31
30
1
4 r
kQr
r
QrE
97
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
Solving for E we find
3342
24 rrEdAE
22rkQ
E outside
same as a point charge!
98
rE r
2r r1E rE
Electric field vs. radius for a conducting sphere
99
E is zero within conductorIf 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?
Properties of Conductors
100
Lecture 6
Application(Gauss’s Law)
101
[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 < aE = 0 since no charge inside the Gaussian surface
Region (3) b > r < c E=0 How?
102
Region (2) a < r < bwe construct a spherical Gaussian surface of radius r
brawherer
QE
QrE
whywhereq
dAE encl
,2
4
12)4(
?,0,.
200
2
0
Region (4) r > cwe 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
crwherer
QE
QrE
qdAE encl
,4
1)4(
,.
200
2
0
103
]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
32
0.0
outerouterinner
inner
encl
thus
Also
qdAE
104
(b) For a Gaussian surface S2 outside the conducting cylinder
rE
lrlE
qdAE encl
0
0
0
2
3
)3(1
)2(
.
105
]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
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.
radially outward from the cylinder axis
106
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.
107
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.
108
]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
a) At point to the left of the two parallel sheets02
E
0
21 2)(
E
EEEE
109
(c) At point to the right of the two parallel sheets
b) At point between the two sheets
0
21 2
E
EEEE
0)( 21 EEE
110
[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
26
6
0
2
4
1068.0
1017.0
25.0
/108
m
C
Cq
qEA
mA
CNE
111
[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.
CNmEA /5.8570cos 2
)a (the plane is parallel to the yz plane
)b (the plane is parallel to the xy plane
The angel 90
c) the plane is parallel to the xy plane
The angel 40
112
]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.
CNc
CNb
zeroa
findtor
Euse
/540)
/104.5)
)
4
3
0
113
[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?
Cq
q
echnetthefindtoNow
CmN
rEEA
8
0
23
2
105.5
arg
/.103.6
4
114
[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?
CNmq
/105.5 25
[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.
CNmq
/104.31085.8
)105108( 2512
66
115
[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
outwardradiallyCNEr
qE
qrE
qEA
whyq
dAE
encl
encl
/101254
)4(
?,0,.
42
0
0
2
0
0
(b)
116
[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 half-way between the two wires (including direction).
3 -Work out the total electric field at that point.
117
using Gauss’ Law and cylindrical Gaussian surface as Lec.5 page 15
CNjjjEEE
DownCNE
DownCNE
rE
bt
b
t
/ˆ1797ˆ1123ˆ6743
/1123)08.0)(10854.8(2
1052
/674)08.0)(10854.8(2
1031
2
12
9
12
9
0
118
Lecture 7The Electric Potential
119
Electric Potential Energy The electric force, like the
gravitational force, is a conservative force. (Conservative force: The work is path-independent.)
As in mechanics, work is
Work done on the positive charge by moving it from A to B
A B
E99999999999999
d
cosFdW
qEdFdW cos
120
The work done by a conservative force equals
the negative of the change in potential energy,
PE
This equation is valid only for the case of a uniform electric field
PE W qEd
If a charged particle moves perpendicular to electric field lines, no work is done.
if d E
121
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
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”
B A
PEV V V
q
If the work done by the electric field is zero, then the electric potential must be constant
122
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
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?
1 1V J C
123
Units of electric field (N/C) can be expressed in
terms of the units of potential (as volts per meter)
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
1 1N C V m
124
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.
125
Analogy between electric and gravitational fields
The same kinetic-potential energy theorem works here
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
A
B
qd
A
B
mdE
99999999999999g99999999999999
i i f fKE PE KE PE
126
Example: motion of an electron
Vab
What is the speed of an electron accelerated from
rest across a potential difference of 100V?
Given:
V=100 Vme = 9.1110-31 kg
mp = 1.6710-27 kg
|e| = 1.6010-19 C
Find:ve=?vp=? smv
smv
m
Vqv
Vqmv
VqPEKEKE
PEKEPEKE
p
e
f
f
if
ffii
/103.1
/109.5
2
2
1
5
6
2
127
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.
128
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
Note: the potential depends only on charge of an object, q, and a distance from this object to a point in space, r.
r
kqVdrEVV
f
i
if
.
129
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!
130
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
Potential energy is positive if charges are of the same sign.
P A
q1q2
r1 2
2 1 e
q qPE q V k
r
131
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-
Na+ Na+
? Na Cl Na Na Na
e e e Cl Na
q q q q qPE k k k q q
r r r
but : !Cl Naq q
0Nae Na Na
qPE k q q
r
132
Potentials and charged conductors
Recall that work is opposite of the change in
potential energy,
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. all charge of the charged conductor is located on its surface
2. 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
B AW PE q V V
… but that’s not all!
133
Because the electric field is zero inside the conductor, no work is required to move charges between any two points, i.e.
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!
0B AW q V V
134
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 = 1.6010-19 C·V = 1.60×10-19 J Vab=1 V
135
Example : ionization energy of the electron in a hydrogen atomIn 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
136
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 mme = 9.1110-31 kg
mp = 1.6710-27 kg
|e| = 1.6010-19 C
Find:
E=?
The ionization energy equals to the total energy of the electron-proton system,
E PE KE
22 2182.18 10 J -13.6 eV
2 2e
e e
k ee m eE k k
r mr r
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:
c cma F
2 2
,2e
e vPE k KE m
r with
or2 2
2,e
v em k
r r or
22 ,e
ev k
mr
Thus, total energy is
137
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
In this case the force is provided by the electric fieldF = qE
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
sdFdW
sdEqdW
f
isdEqW
138
Remembering the relation between the change
in electric potential and the work done …
…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
qW
V e
f
i
eif sdE
qW
VVV
i
sdExV
)(
139
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.)
140
Example - Charge moves in E field
EdVV
EdsEsdE
sdE
sdEsdEVV
if
f
c
f
c
c
i
f
c
c
iif
21distance)45cos(
E)lar toperpendicu (ds 0
141
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: V is independent of path.Explicit: V = - E . ds = E ds = - Ed
142
Lecture 8
Application(The Electric Potential)
143
]1[ What potential difference is needed to stop an electron with an initial speed of 4.2×105m/s?
]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.
voltV
q
mvV
mvVq
KEKEKEKEW iif
57.0
106.1
)102.4)(1011.9(21
21
2
1
19
25312
2
Cq
q
PEV
191014.6
144
]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?
]4[ At what distance from a point charge of 8µC would the potential equal 3.6×104V?
mrr
qV 2
4
1
0
mE
Vd
EdV
CNE
4
0
108.8
/56502
145
]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.
mr
kq
V
Vq
k
qkE
r
qE
V
qkr
r
qV
67.2
)(
4
1
&4
1
2
2
20
0
146
]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.
][22
212
22
1
4321
ba
q
a
q
a
q
ba
qkV
VVVVV
By substitute find V
147
]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?
voltdq
FV
q
FE
EdV
3219
15
1055.0106.1
106.1
148
]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?
voltV
r
q
r
qV
VVV
p
p
p
70200]2.1
102
23.0
104[109
][4
1
659
1
2
2
1
1
01
211
149
]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
a
qW
0
221.0
a
q
a
qqW
a
q
a
qW
a
q
a
q
a
q
a
q
a
q
a
qW
wwwwwwW
0
222
0
22
0
222222
0
342423141312
2.0]
2
242[
4
1
]2
4[
4
1
]22
[4
1
150
[10 ]Assume we have a system of three point charges:q1 = +1.50 Cq2 = +2.50 Cq3 = -3.50 C.
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)?
151
The electric potential at point P is given by the sum of the electric potential from the three charges
V kqi
rii1
3
kq1
r1
q2
r2
q3
r3
kq1
b
q2
a2 b2
q3
a
V 8.99 109 N/C 1.50 10 6 C
6.00 m
2.50 10 6 C
8.00 m 2 6.00 m 2
3.50 10 6 C
8.00 m
V 562 V
r1
r2
r3
152
smv
JUUKK
isenergykineticfinalThe
JVeUU
decreasesenergypotentialitsandincreasesenergykineticits
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,
.1014.12
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'/105
.
5
19
19
192
5
[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?
153
[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?
VC
JVVVV
Vispotientialelectricfinaltheand
VispotentialelectricinitialThe
VVqUUmvKK
problemenergyofionaconservatisThis
fiQp
Q
p
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8
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82
154
]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?
smm
Vqvv
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VqPEKEKE
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if
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ffii
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155
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
then,
So
[14 ]Potential Due to a Charged Rod
2/12200 )(4
1
4
1
dx
dx
r
dqdV
dxdq
ddLL
dxxdx
dxdVV
LL
ln)(ln4
)(ln4)(4
2/122
0
02/122
002/122
0
d
dLLV
2/122
0
)(ln
4
156
[15 ]The Electric Potential of a Charged RingFind the potential of a thin uniformly charged ring of radius R and charge Q at point P on the z axis?
2 2 and 2P
Qr R z dQ Rd
R
2 20 0
1 1
4 4 2P
dQ QdV d
r R z
2
2 2 2 20 00
1 1
4 42P
Q QV dV d
R z R z
157
[16 ]The Electric Potential of a Charged DiskFind the potential V of a thin uniformly charged disk of radius R and charge density s at point P on the z axis?
2 2 2 2
(2 )kdq k adadV
z a z a
2 2 2 20 0
2 2R Rak da adaV k
z a z a
2 2; 2u z a du ada
2 22 2
2 2
2 2 2
2 22
2 2
2 1 ( / ) 1 1 ( / ) 1
z Rz R
z z
duV k k u k z R z
u
zk z R z kQ R z
R
158
Lecture 9
The Electric Field & Electric Potential Due to Continuous Charge Distributions
159
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.
160
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.
[1 ]Electric Field Due to a Charged Rod
22 x
dxk
x
dqkdE
dxdq
ee
al
ae
al
a
e
al
a
e xk
x
dxk
x
dxkE
122
)(
11
ala
Qk
alal
QkE e
e
161
[2 ]Infinitely Long Line of Charge
2r
dqkdE
dydq
e
+
+
+
+
dE
+
+
+
dy
x
y
2 2 2r x y
xdE
ydE
+
y-components cancel by symmetry
2 2 2 2
k dy xdE
x y x y
3 2
2 2 2
dy 2 2kE k x k x
x xx y
cos2r
dqkdEdE ex
162
dE
z
2 2 2r z a
zdE
dE
a
d
+
+
+
+ +
+
+
perpendicular-components cancel by symmetry
z 2
kdqdE cos
r
2 2 2 2
k ad zdE
z a z a
2
3 3 32 2 2 2 2 202 2 2
k za k za kQzE d 2
z a z a z a
2r
dqkdE
addsdq
e
determine the field at point P on the axis of the ring.
[3 ]Electric Field on the Z-Axis of a Charged Ring
163
When: z a
The charged ring must look like a point source.
3 3 3 2
2 2 22 22
2
kQz kQz kQz kQE
z zz a a
z 1z
0
Note that for z >> R (the radius of the ring), this reduces to a simple Coulomb field.
164
dE
z
3
2 2 2
kQzE
z r
r
3
2 2 2
kzdqdE
z r
dq dA rdrd 2 rdr
3
2 2 2
kz 2 rdrdE
z r
R
2 2
2
R R z R
3 3 32 2 2 20 02 2 2z
kz 2 rdr 2rdr duE kz kz
z r z r u
[4 ]Electric Field on the Axis of an Uniformly Charged Disk
Using the charged ring result,
165
22
222
2
1
2
3
112
112
21
22
2
22
2
Rzk
zRzkz
ukz
duukzE
Rz
z
Rz
z
166
Two Important Limiting Cases
2 2o o
z 1E k 2 1 k 2 2
4 2z R
Large Charged Plate: R z
dE
z
rR
Very Far From the Charged Plate: z R
12 2
22 2 2
2
z z RE k 2 1 k 2 1 k 2 1 1
zz R Rz 1
z
2 2 2
2 2 2 2
1 R 1 R k R kQk 2 1 1 k 2
2 z 2 z z z
167
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
then,
So
[5 ]Potential Due to a Charged Rod
2/12200 )(4
1
4
1
dx
dx
r
dqdV
dxdq
ddLL
dxxdx
dxdVV
LL
ln)(ln4
)(ln4)(4
2/122
0
02/122
002/122
0
d
dLLV
2/122
0
)(ln
4
168
[6 ]The Electric Potential of a Charged Ring
Find an expression for the electric potential at a pointP located on the perpendicular central axis of a uniformlycharged ring of radius a and total charge Q
2 2 and 2P
Qr R z dQ Rd
R
2 20 0
1 1
4 4 2P
dQ QdV d
r R z
2
2 2 2 20 00
1 1
4 42P
Q QV dV d
R z R z
169
[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?
2 2 2 2
(2 )kdq k adadV
z a z a
2 2 2 20 0
2 2R Rak da adaV k
z a z a
2 2; 2u z a du ada
2 22 2
2 2
2 2 2
2 22
2 2
2 1 ( / ) 1 1 ( / ) 1
z Rz R
z z
duV k k u k z R z
u
zk z R z kQ R z
R
170
Lecture 10 Capacitance and capacitors
171
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.
172
Definition of Capacitance The definition of capacitance is
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)
A farad is a very large capacitance Typically we deal with F (10-6 F), nF (10-9 F),
or pF (10-12 F)
V 1
C 1 F 1
V
Q C
173
A
A
+Q
-Q
d
The parallel-plate capacitor
The capacitance of a device depends on the area of the plates and the distance between the plates
where A is the area of one of the plates, d is the separation, 0 is a constant (permittivity of free space),
0= 8.8510-12 C2/N·m2
0
AC
d
174
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
Given:
V=10,000 VA = 2.00 m2
d = 5.00 mm
Find:
C=?Q=?
Solution:Since we are dealing with the parallel-plate capacitor, the capacitance can be found as
2
12 2 20 3
9
2.008.85 10
5.00 10
3.54 10 3.54
A mC C N m
d m
F nF
9 53.54 10 10000 3.54 10Q C V F V C
Once the capacitance is known, the charge can be found from the definition of a capacitance via charge and potential difference:
175
Consider a capacitor constructed of two collinear conducting cylinders of length L.
The inner cylinder has radius r1 andthe outer cylinder has radius r2.
Both cylinders have charge perunit length with the inner cylinderhaving positive charge and the outercylinder 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.
Cylindrical Capacitor
176
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
… which we can rewrite to get anexpression for the electric fieldbetween the two cylinders
E
20r
rLALEA
qAdE
2 where0
0
177
As we did for the parallel plate capacitor, we define the
voltage difference across the two cylinders to be V=V1 – V2.
The capacitance of a cylindrical capacitor is
1
2
0
021
ln2
22
1
2
1
rr
drr
sdEVVr
r
r
r
)ln(
2
)ln(2 1
2
0
1
2
0 rr
L
rr
L
V
qC
178
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
179
To calculate the electric field, we use a Gaussian surfaceconsisting 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
40r2
180
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
Using the definition of capacitance we find
The capacitance of a spherical capacitor is then
V Edrr2
r1
q
40r2
drr2
r1
q
40
1
r1
1
r2
C q
V
q
q
40
1
r1
1
r2
40
1
r1
1
r2
12
2104
rr
rrC
181
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
C1
C2
C3
C5C1
C2
C3
C4
182
+Q1
Q1
C1
V=Vab
a
b
+Q2
Q2
C2
a. Parallel combination
1 2V V V
Connecting a battery to the parallel combination of capacitors is equivalent to introducing the same potential difference for both capacitors,
1 2Q Q Q A total charge transferred to the system from the battery is the sum of charges of the two capacitors,
2121
2211
21
222111
sin,
,
&
,
VVVceCCC
VCVCVC
QQQ
bewouldCThus
VCQVCQ
definitionBy
eq
eq
eq
183
Parallel combination:
Analogous formula is true for any number of
capacitors,
It follows that the equivalent capacitance of a parallel combination of capacitors is greater than any of the individual capacitors
1 2 3 ...eqC C C C (parallel combination)
184
Example: A 3 F capacitor and a 6 F capacitor are connected in parallel across an 18 V battery. Determine the equivalent capacitance and total charge deposited.
+Q1
Q1
C1
V=Vab
a
b
+Q2
Q2
C2Given:
V = 18 VC1= 3 FC2= 6 F
Find:
Ceq=?Q=?
First determine equivalent capacitance of C1 and C2:
12 1 2 9C C C F
Next, determine the charge
6 49 10 18 1.6 10Q C V F V C
185
Series combination
2121
2
2
1
1
21
2
22
1
11
sin,111
,
&
,
QQQceCCC
C
Q
C
Q
C
Q
VVV
bewouldCThus
C
QV
C
QV
definitionBy
eq
eq
eq
Connecting a battery to the serial combination of capacitors is equivalent to introducing the same charge for both capacitors,
1 2Q Q Q
A voltage induced in the system from the battery is the sum of potential differences across the individual capacitors,
+Q1
Q1
C1
+Q2
Q2
C2
V=Vab
a
c
b
21 VVV
186
Series combination:
Analogous formula is true for any number of
capacitors,
It follows that the equivalent capacitance of a series combination of capacitors is always less than any of the individual capacitance in the combination
1 2 3
1 1 1 1...
eqC C C C (series combination)
187
Example: A 3 F capacitor and a 6 F capacitor are connected in series across an 18 V battery. Determine the equivalent capacitance and total charge deposited.
+Q1
Q1
C1
+Q2
Q2
C2
V=Vab
a
c
b
Given:
V = 18 VC1= 3 FC2= 6 F
Find:
Ceq=?Q=?
First determine equivalent capacitance of C1 and C2:
1 2
1 2
2eq
C CC F
C C
Next, determine the charge
6 52 10 18 3.6 10Q C V F V C
188
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.
189
1 1 2 6.0 FeqC C C
1 31/ (1/ 1/ ) 1/ [1/ (6.0 F) 1/ (3.0 F)] 2.0 Feq eqC C C
(2.0 F)(6.0 V) 12.0 CeqQ C V
3 3/ (12.0 C) / (3.0 F) 4.0 VV Q C
24 3 (6.0 V) (4.0 V) 2.0 VV V V
2 2 24 (2.0 F)(2.0 V) 4.0 CQ C V
4 4 24 (4.0 F)(2.0 V) 8.0 CQ C V
190
1
2U QV
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 q across a voltage V is
W = V q. As the charge increases, V increases so the
work to bring q increases. Using calculus we find that the energy (U) stored on a capacitor is given by:
V
V
q
221
2 2
QCV
C
Q
191
Find electric field energy density (energy per unit volume) in a parallel-plate capacitor
2
0
20
20
1
2
/ energy density
1= ( ) /( )
21
2
volume A
U CV
AC V Ed
du U volume
AEd d
E
d
Ad
u
Example: electric field energy in parallel-plate capacitor
Recall
Thus,
and so, the energy density is
192
C1
C2
V
C3
Example: In the circuit shown V = 48V, C1 = 9F, C2 = 4F and C3 = 8F.
(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,
22 6 31 15.14 10 48 5.9 10
2 2U CV F V J
The energy stored in the capacitor C123 is then
FCeq 14.5
193
Q QQ Q
V0 V
Capacitors with dielectrics A dielectrics is an insulating material (rubber, glass, etc.) Consider an insolated, charged capacitor
Notice that the potential difference decreases (k = V0/V) Since charge stayed the same (Q=Q0) → capacitance increases
dielectric constant: k = C/C0
Dielectric constant is a material property
0 0 00
0 0
Q Q QC C
V V V
Insert a dielectric
194
Capacitance is multiplied by a factor k when the dielectric fills the region between the plates completely E.g., for a parallel-plate capacitor
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
0
AC
d
195
Dielectric constants and dielectric strengths of various materials at room temperature
MaterialDielectric constant, k
Dielectric strength (V/m)
Vacuum1.00 --
Air1.000593 106
Water80--
Fused quartz3.789 ×106
196
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
12 2 20 3
2.008.85 10 18
1.00 10
A mC C N m nF
d m
9 518 10 3000 5.4 10Q C V F V C
The charge on the capacitor can be found to be
The dielectric constant and the new capacitance are
10 0
2
0.33 18 6V
C C C nF nFV
197
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.
198
Solved Problems
(Electrostatic)
199
]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.
200
a)The electric field due to the charges at the origin O.
CNEEE
CN
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CN
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CNE
CNE
CNE
yxT
x
x
/3.10
/2.05
3
25
27
5
3
25
27
5
3
5
9
sinsinsin
/2.35
4
25
27
5
4
5
9
5
4
25
27
coscoscos
/25
27
25
103109
/5
9
25
105109
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27
25
103109
22
232122
232221
99
3
99
2
99
1
201
VVVVV
VV
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95
27
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103109
95
105109
5
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103109
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99
3
99
2
99
1
06.18006.0180
0625.02.3
2.0tan
EofdirectiontheisforcenettheofdirectionThe
x
y
E
E
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.
202
]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 ?
NF
directionxveinNcm
kF
directionxveinNcm
kF
net4
42
99
2
42
99
1
108.1
108.1)2(
101102
106.3)2(
108102
lawscoulomb'using
203
[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?
smmr
kev
r
ke
r
mvF
/102.2
iselectrontheonforceThe
62
2
22
204
[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?
cmatd
smmr
kev
smm
eE
m
Fa
142
1
restfromstartselectrontheSince
/102.2
/107
onacceleraticonstantinvolvingproblemisThis
2
62
214
205
[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,
mxormx
xx
xx
x
Qk
x
Qk
66.086.3
26.1
2)6.1(
)6.1()(22
22
21
206
[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.
smv /105.2 5
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:
Consequently, the final kinetic energy is
And the final velocity of the electron is
Jqkqqkq
PE
Jqkqqkq
PE
eef
eei
2021
21
1088.223
012
JPEPEkkvm fiiffe202 1088.2
2
1
207
[7 ]Using the superposition principle determine the electric field at point P in Fig.
mVEp /02
101
2
102
2
103
0
9
0
9
0
9
Using the convention that a positive electric field is up and a negative electric field is down, the net electric field at p is
208
]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.
CQkQk
4)2()1(
)1(22
209
V
C1
C2
+-
C3
C1 = 10 F
C2 = 5.0 F
C3 = 4.0 F
]9[
a) Find the equivalent capacitance of the entire combination.
C1 and C2 are in series.
21
2112
2112
111
CC
CCC
CCC
FC 3.315
50
510
51012
C12 and C3 are in parallel.
FCCCeq 3.70.43.3312
210
b) If V = 100 volts, what is the charge Q3 on C3?
C = Q/V
100100.4 633 VCQ
c) What is the total energy stored in the circuit?
CoulombsQ 43 100.4
JVCU eq2462 106.310103.1
2
1
2
1
JU 2106.3
211
[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?
x
dxk
x
dqkdV
dxdq
ee
dlde
dl
d
e
dl
d
e xkx
dxk
x
dxkV
ln
)/1ln(ln)ln( dlkddlkV ee
212
]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.
-6.00 C-2.50 C
1.00 m
0.39 m
]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
213
]13[ Find the charge on the 2 F capacitor in the following circuit.
]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
2 F4 F
2 F
12 V
18µC
214
[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?
2 .What is the value of the electric potential at the center (point A) of positive charges?
3 .What is the electric potential energy of system?
2
2
3b
qk
b
qk)
3
14(
b
qk
2
215
[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 ?
J12
FCeq 9.257
57
[17 ]Find the equivalent capacitance between points a and b in the combination of capacitors shown in Figure.
7µF and 5µF series
4µF, 6µF and 2.9µF parallel
FCeq 9.129.264
216
[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?
)1.........(..........921 CC
)2.........(..........221
21 CC
CC
in parallel
in series
Solve eq.1&2
pFCpFCor
pFCpFC
3,6,
6,3
21
21
217
Lecture 12Lecture 12Current & ResistanceCurrent & Resistance
218
Electric CurrentDefinition: the current is the rate at which charge flows through this surface.
Given an amount of charge, Q, passing through the area A in a time interval t, the current is the ratio of the charge to the time interval. Q
It
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.
219
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?Find no. of electrons?
220
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 . The relationship between current and current density is
J
A
IJorAdJI
,,
Small current density
High current density
221
Current and Drift Speed Consider the current on a conductor of cross-sectional
area A.
222
Volume of an element of length x is : V = A x. Let n be the number of carriers per unit of volume. The total number of carriers in V is: n A x. The charge in this volume is: Q = (n A x)q. Distance traveled at drift speed vd by carrier in time t:
x = vd t. Hence: Q = (n A vd t)q. The current through the conductor: I = Q/ t = n A vd q. The current density :
J = = n vd q.
223
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.
224
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
225
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.
226
The resistance is defined as the ratio.
In SI, resistance is expressed in volts per ampere.In SI, resistance is expressed in volts per ampere.A special name is given: ohmsA 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 .
227
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
228
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.
229
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.
230
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.
231
Resistivity - Units
Resistance expressed in Ohms, Length in meter. Area are m2, Resistivity thus has units of m.
lR
A
RA
l
232
Material Resistivity (10-8 m)
Material Resistivity (10-8 m)
Silver 1.61 Bismuth 106.8Copper 1.70 Plutonium 141.4Gold 2.20 Graphite 1375
Aluminum
2.65 Germanium 4.6x107
Pure Silicon
3.5 Diamond 2.7x109
Calcium 3.91 Deionized water
1.8x1013
Sodium 4.75 Iodine 1.3x1015
Tungsten 5.3 Phosphorus 1x1017
Brass 7.0 Quartz 1x1021
Uranium 30.0 Alumina 1x1022
Mercury 98.4 Sulfur 2x1023
Resistivity of various materials
233
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 m.
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?
234
]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.
The reciprocal of the resistivity is called the conductivity, 1
]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[ 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
235
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.
236
For most metals, resistivity increases approx. linearly with temperature.
is the resistivity at temperature T (measured in Celsius). is the reference resistivity at the reference temperature T
(usually taken to be 20 oC). is a parameter called temperature coefficient of resistivity.
For a conductor with fixed cross section.
237
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 =3.92x10-3(oC)-1 from table.
Ro=50.0 .
To=20oC.
R=76.8 .
238
[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?
239
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 metals0, measured to be less than 10-16•ρconductor (i.e., ρ<10-24 Ωm)!
Resistance versus temperature for asample of mercury (Hg). The graph follows that of a normal metal above the critical temperature Tc. Theresistance drops to zero at Tc, which is 4.2 K for mercury.
240
Electrical energy and power
In any circuit, battery is used to induce electrical currentIn any circuit, battery is used to induce electrical current chemical energy of the battery is transformed into kinetic chemical energy of the battery is transformed into kinetic energy of mobile charge carriers (electrical energy gain)energy of mobile charge carriers (electrical energy gain)Any device that possesses resistance (resistor) present in Any device that possesses resistance (resistor) present in the circuit will transform electrical energy into heat the circuit will transform electrical energy into heat kinetic energy of charge carriers is transformed into heat kinetic energy of charge carriers is transformed into heat via collisions with atoms in a conductor (electrical energy via collisions with atoms in a conductor (electrical energy loss)loss)
241
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
242
Power
Compute rate of energy loss (power dissipated on the resistor)
Use Ohm’s law
Units of power : watt
delivered energy: kilowatt-hours
3 61 kWh 10 3600 3.60 10W s J
243
Example
A high-voltage transmission line with resistance of 0.31 /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 resistance2. Given resistance and current, compute power loss
Now compute power
244
Lecture 13Direct Current Circuits
245
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:Examples of such sources are:batteriesbatteriesgeneratorsgeneratorsthermocouplesthermocouplesphoto-voltaic cellsphoto-voltaic cells
246
247
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 V is
Note: EMF is the same as the terminal voltage when the current is zero (open circuit)
248
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?
249
Resistors in series1. 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
250
Analogous formula is true for any number of resistors,
It follows that the equivalent resistance of a series combination of resistors is greater than any of the individual resistors
(series combination)
]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?
251
ExampleIn the electrical circuit below, find voltage across the resistor RIn the electrical circuit below, find voltage across the resistor R11 in in
terms of the resistances Rterms of the resistances R11, R, R22 and potential difference between the and potential difference between the
battery’s terminals V. battery’s terminals V.
Energy conservation implies:
with
Then,
Thus,
This circuit is known as voltage divider.
252
Resistors in parallel1. 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
253
Analogous formula is true for any number of resistors,
It follows that the equivalent resistance of a parallel combination of resistors is always less than any of the individual resistors
(parallel combination)
]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?
254
exampleIn the electrical circuit below, find current through the In the electrical circuit below, find current through the resistor Rresistor R11 in terms of the resistances R in terms of the resistances R11, R, R22 and total and total
current I induced by the battery. current I induced by the battery.
Charge conservation implies:
with
Then,
Thus,
This circuit is known as current divider.
255
Find the currents in the circuit shownFind the currents in the circuit shown
Example
256
ExampleFind the currents IFind the currents I11 and I and I22 and the voltage V and the voltage Vxx in the circuit in the circuit
shown below.shown below.
I1I2
4 1 2 2 0 V
7
+_
+
_
V x
I First find the equivalent resistance seen by the 20 V source:
4 (12 )7 10
12 4eqR
20 202
10eq
V VI A
R
We now find I1 and I2 directly from the current division rule:
1 2 1
2 (4 )0.5 , and 1.5
12 4
AI A I I I A
Finally, voltage Vx is 2 4 1.5 4 6xV I A V
257
Kirchhoff’s Rules1. The sum of currents entering any junction must equal the
sum of the currents leaving that junction (current or junction rule) .
2. The sum of the potential differences across all the elements around any closed-circuit loop must be zero (voltage or loop rule).
0 I321321 0 IIIIII
0loop
V
2121 0 RRIVIRIRV
Charge conservation
Energy conservation
258
loopany for 0V
259
Rules for Kirchhoff’s loop rule
260
261
Solving problems using Kirchhoff’s rules
262
Example
263
264
265
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
3R
1R1V
2V
266
267
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
268
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
RCteRdt
dqI
269
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
270
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
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
RCtRCt eIeRC
Q
dt
dqI 0
271
Example : charging the unknown capacitorA series combination of a 12 kA series combination of a 12 k resistor and an unknown capacitor is resistor and an unknown capacitor is connected to a 12 V battery. One second after the circuit is connected to a 12 V battery. One second after the circuit is completed, the voltage across the capacitor is 10 V. Determine the completed, the voltage across the capacitor is 10 V. Determine the capacitance of the capacitor.capacitance of the capacitor.
I
RC
272
Recall that the charge is building up according to
Thus the voltage across the capacitor changes as
This is also true for voltage at t = 1s after the switch is closed,
1 t RCq Q e
1 1t RC t RCq QV e e
C C E
1 t RCVe
E
146.5
10log 1 12,000 log 112
t sC F
V VRV
E
log 1t V
RC
E
1t RC Ve
E
273
Lecture 14
Discussion
274
[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?
.2,3,85.0,/5, CuAgCuAgCuAgCuCuCu ddandllAlGiven
2.3)5)(6375.0()6375.0(
)2(
)3)(85.0(222
Cu
CuCu
Cu
CuCu
Ag
AgAg
r
l
r
l
r
lR
275
The first resistor has a resistance of
The second resistor has a resistance of
The series combination of the two resistors is
Which when connected across a 3V battery will draw a current of
AR
VI 67.0
5.4
3
31
31 I
VR
]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?
5.12
31 I
VR
5.421 RR
276
[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?
%3.9911
1
5
)1(
5/5
/
/
eeQ
q
eQ
q
RCt
eQq
RCRC
RCt
RCt
277
[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
AeI
eR
I
bygiveniscapacitortheofingchincurrentTheb
CCQcapacitortheonechtheand
RCtconstimeThea
RCt
6)
)105)(101(
10(
6
/
6
66
1006.4101
30
arg)
150)30)(105(arg
sec5)105)(101(tan)
66
278
]5[ In the circuit shown in Fig. what is the current labeled I?
AI
AI
AI
II
II
III
6.0
4.0
2.0
0422
0262
2
1
21
21
21
279
[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
2/992/ RRRnew
280
[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?
2324
242
sec1
2
sec1sec1
/1085102
17
1022][
17)512(][
mAm
A
A
IJ
mcmAb
Atdt
dqIa
ttt
281
[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?
C
tdttIdtqt
t
265.0)]0cos()2
[cos(120
100
)120cos(120
100)120sin(100
240/1
0
240/1
0
2
1
282
[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?
rfindb
mMR
l
M
lR
lM
lR
A
lRNow
l
MA
MAl
MV
V
Mdensitymassa
d
d
d
dddd
][
82.1)1092.8)(107.1(
)101)(5.0(
,
)(][
38
3
2
283
[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.
269.0$06.0$49.4cos
49.4)24)(187.0(24
1871107.1
t
kWhhkWdayHainusedEnergy
WVIp
284
Lecture 15Magnetic Fields & Force
285
MagnetsMagnets ... two poles: N and S ... two poles: N and S
Like poles repelLike poles repelUnlike poles attractUnlike poles attract
286
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
287
Magnetic Field lines :
)defined in same way as electric field lines, direction and
density(
288
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.
289
Source of Magnetic FieldsWhat 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)
290
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
291
Magnetic Field Direction
A vector quantity: magnitude and direction…A vector quantity: magnitude and direction…The letter The letter BB is used to represent magnetic fields. is used to represent magnetic fields.
292
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.
293
Magnetic Field of the Earth
294
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
295
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.
296
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?
297
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:
298
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.
299
Right Hand Rule
300
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.
301
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.
302
Force on a wire carrying current in a magnetic field.
303
304
Force on a wire carrying current in a magnetic field.
305
Force on a wire carrying current in a magnetic field.
General Case: field at angle relative to current.
I
B
B sin
Note: If wire is not straight, compute force on differential elements and integrate:
BLdiFd
306
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?
307
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 dsBecause s=R we have ds=RddF2= I B R sin dThe 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
The net force acting on a closed current loop in a uniform magnetic field is zero.
02221total IRBIBRFFF
308
Example:Wire with current i.Magnetic field out of page.What is net force on wire?
iLBFF 31
iBRdiBRdFF 2)sin()sin(00
2
iBRdiBdLdF
By symmetry, F2 will only have a vertical component,
)(22321total RLiBiLBiRBiLBFFFF
309
Lecture 16Magnetic Fields& Force
310
Torque on a Current LoopImagine a current loop in a magnetic field as follows:
1 2F F BIb
BI
b
a
Ba/2
F
F
F
F
max BIba BIA
sinBIA
In a motor, one has “N” loops of current
311
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.
B
30.0o
312
GalvanometerDevice used in the construction of ammeters and voltmeters.
313
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
Rp
314
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
Rs
315
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.
316
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
317
Example : Proton moving in uniform magnetic fieldA 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.
318
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?
v
r
x x
x x
m
qBf
2
kg
TCf
27
19
1067.12
4.0106.1
HzHzf 68 101.61067.128.6
4.06.1
Hzf 6101.6
m
qBrv
kg
mTCv
27
19
1067.1
21.04.0106.1
s
m
s
mv 68 101.81067.1
21.04.06.1
s
mv 6101.8
319
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.
320
321
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
322
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
323
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)?
324
ExampleSuppose 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 JJ 107.2
m/s 1099.3 implies
12221
7
mvK
mRqB
vqBmv
r
325
Lecture 17Sources of the Magnetic Field
326
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
327
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
328
dB perpendicular to ds |dB| inversely proportional to |r|2 |dB| proportional to current I |dB| proportional to |ds| |dB| proportional to sin
329
Biot–Savart Law
All these results could be summarised by one “Law”
Putting in the constant
Where 0 is the permeablity of free space
330
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
331
Magnetic Field of a current loop
Magnetic field produced by a wire can be enhanced by having the wire in a loop. x1
I
x2
B
N loops Current NI
1 loop Current I
332
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 o times the current.
encIdsBIsB 00 ..
333
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
)2cos rBdsBBds
i)r2(B o
r2
iB o
Rrfor
334
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
Rrfor
)()2(.2
2
0'
0
2
2'
2
2'
iR
rirBdsB
iR
ri
R
r
i
i
335
Magnetic Force between two parallel conductors
336
337
Definition of the SI unit 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.
Used to define the SI unit of current called Ampere.
338
ExampleTwo 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).
l
d
1
2I1
I2
339
l
d
1
2
F1
B2I1
I2
mg/l
340
Magnetic Field of a solenoid Solenoid magnet consists of a wire coil with multiple
loops. It is often called an electromagnet.
341
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:
342
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:
0
0 0
b c d a
enc
a b c d
b
a
B ds B ds B ds B ds B ds i
B ds Bh inh B in
343
Solenoid Magnet
n = N/L : number of (loop) turns per unit length.
I : current in the solenoid.
The field inside has a value:
344
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.
345
Lecture 18
Applications
346
[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 .
rightthetoNm
AmAFleft
71085.02
121
leftthetoNm
AmAFright
710412
121
rightFnet7104
347
[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
downmN
downm
AAup
m
AA
l
F
/108
)1(2
)3)(2(
)1(2
)1)(2(
7
00
348
[3 ]What is the magnitude and direction of the magnetic field at point P in
Fig ? .
outTBBBBB
outTm
AB
outTm
AB
outTm
AB
oTm
AB
T6
4321
704
703
702
701
109.0
104)2(2
)4(
103)2(2
)3(
104)1(2
)2(
int102)1(2
)1(
349
[4 ]What is the magnitude and direction of the magnetic field at point Q
in Fig ?.
oTR
IB
outR
IB
oR
IB
net
wire
loop
int106.82
)1
1(
2
int2
7
350
[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 ?
seB
m
v
rT
isorbitancompletetotimetheThusm
eB
r
vqvB
12
2
10722
,r
mvF
field, magnetic the
lar perpendicuorbit acircular in moving iselectron theSincev
r2T
v.electron, theof speed by the divided
r,2 orbit, onein traveleddistance theisorbit an complete to timeThe
351
[6 ]A proton moves with a velocity of v = (2iˆ- 4jˆ +kˆ ) m/s ina region in which the magnetic field is B " (iˆ + 2jˆ - 3kˆ ) T.What is the magnitude of the magnetic force this chargeexperiences?
NF
smTBv
kji
kji
BvNow
BvqqvBF
1819 1034.26.14106.1
/.6.14
8710
321
142,
)(sin
352
[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?
rightthetoisbartheinIofdirectionThe
AlB
mgI
l
IlB
l
mg
l
FNow
gmF
gmFT
conditionmEquilibriu
B
B
109.06.3
8.904.0
,
02
353
[8 ]A positive charge q = 3.20 × 10-19 C moves with a velocityv = (2iˆ + 3jˆ - kˆ ) m/s through a region where both a uniformmagnetic 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?
4.24tan]
10)6.152.3(
??)(
)()(
]
1
18
x
y
magneticelectric
F
Fb
NjiF
BvEfind
BvEqBvqqEF
FFF
bygivenisa
forceLorentz
354
[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.!
355
]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.
oTR
IB
loopfulla
fieldtheforthonemakescirclequarterThe
int2.26)2
(4
1 0
356
[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?
ABb
i
ibB
idlB
brsurfaceoutertheonLawsAmper
enc
57
0
0
0
10104
1.02.022
)2(
.
)('
357
)b (Find the magnitude of the magnetic field at r=20 cm?
Tr
iB
irB
idlB enc
1.02.02
10104
2
)2(
.
570
0
0
358
)c (Find the magnitude of the magnetic field at r=8 cm?
TB
abr
ariB
iab
arrB
idlB enc
013))05.010.0(08.0
05.008.0(
2
10104
))(
(2
)()2(
.
22
2257
22
220
22
22
0
0
359
Lecture 19Faraday’s Law
360
magnetic flux
The flux, , is defined as the product of the field magnitude by the area crossed by the field lines.
where 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)
cosB A BA
Definition of Magnetic Flux
B
AdBB
361
(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.
362
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.
cosBA 2
2
0.300 2.00 cos50.0
0.386
T m
Tm
363
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.
364
The needle deflects momentarily when the switch is closed
365
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
366
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.
367
Example 1: EMF in a loop
A wire loop of radius 0.30m lies so that an external magnetic field 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 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 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 directions through the loop.) Find the magnitude of the average induced emf in the loop during this time. induced emf in the loop during this time.
B99999999999999
368
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
2BA B r
Initially the flux is
and after the field changes the flux is
2 20.30 0.30m =0.085 T mi T
2 20.20 0.30m =-0.057 T mf T
The magnitude of the average induced emf is:
2 20.085 T m -0.057 T m0.095
1.5f iemf V
t t s
369
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 . 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 :
V1.4s/m.T1.4s80.0
)m.T0m.T0162.0(200
t
N 222
B
370
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.
371
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 :
This expression indicates that the induced emf decays exponentially in time.
Note that the maximum emf occurs at t=0, where max = aABmax.
atmax
atmax
B eaABedt
dAB
dt
d
372
BlvElV
BvF qB
As the wire moves,
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,
vBEorqEqvBorFF EB
Motional EMF
373
Motional EMF in a Circuit
BlxBAB
dt
dxBlBlx
dt
d
dt
d B
E
BlvER
Blv
RI
E
If the bar is moved with constant velocity,
IlBFF Bapp
RR
vlBvIlBvFapp
2222 EP
374
Example: Motional emf Induced in a Rotating BarA 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.
375
Solution
Consider a segment of the bar of length dr having a velocity v.
The magnitude of the emf induced in this segment is :
Because every segment of the bar is moving perpendicular to B, an emf d 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 :
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 :
Bvdrd
Bvdr
20 B
2
1rdrBvdrB
376
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.
377
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 :
We know that I=B v/R, and so we can write this expression as :
BIdt
dvmmaFx
vR
B
dt
dvm
22
dtmR
B
v
dv 22
378
Integrating this equation using the initial condition that v=vi at t=0, we find that:
where the constant
From this result, we see that the velocity can be expressed in the exponential form :
This expression indicates that the velocity of the bar decreases exponentially with time under the action of the magnetic retarding force.
t0
22vv dt
mR
B
v
dvi
tt
mR
B
v
vln
22
i
22B
mR
/tievv
379
Lenz’s LawThe 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
380
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
381
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.
382
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.
383
A Loop Moving Through a Magnetic Field
384
Induced EMF and Electric Fields
Changing Magnetic Flux EMF Electric Field Inside a Conductor
This induced electric field is non-conservative and time-varying
dt
d BE
rFqW E 2 E
rqEq 2E
rE
2
E
dt
dBrE
Brdt
d
rdt
d
rE B
2
2
1
2
1 2
dt
dd B
sE. General Form of Faraday’s Law
385
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?
386
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 = BR2; hence :
The magnetic field inside a long solenoid is given by , B = onI.
Substitute I = Imax cos t into this equation and then substitute the result into Eq. (1), we find that :
dt
dBRrEsdE
dt
dBRRB
dt
dsdE
2
22
)2(
)(
(1)
387
tsinnIR
)t(cosdt
dnIR)r2(E
maxo2
maxo2
tsinr2
RnIE
2maxo
(2)(for r > R)
388
Solution for (b)
For an interior point (r < R), the flux threading an integration loop is given by Br2.
Using the same procedure as in part (a), we find that :
tsinnIrdt
dBr)r2(E maxo
22
tsinr2
nIE maxo
(3) (for r < R)
389
Generators and Motors
tBABAB coscos
NAB
tNABtdt
dNAB
dt
dN B
max
sincos
E
E
390
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.
VNAB
sf
136
3772
max
1
)B (What is the maximum induced current when the output terminals are connected to a low-resistance conductor?
AR
I 3.11maxmax
391
Maxwell’s Equations
dt
dd B
sE.
dt
dId E
000. sB
0. AB d
0
.Q
d AE Gauss’ Law
Gauss’ Law for Magnetism no magnetic monopoles
Faraday’s Law
Ampère-Maxwell Law
BvEF qq Lorentz Force Law
392
Lecture 20
Discussion
393
[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.
394
AR
VIb
Vdt
dN
mTa
125
60]
60]2
8.0)[150(
.8.0]400)[2.0)(1.0(] 2
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-to-right across the front of the loop .
395
NFapp 0.1 smv /0.2
B
0.8R
]2[ A conducting rod of length l moves on two (frictionless) horizontal rails, as shown to the right. A
constant force of magnitude moves the bar at a constant speed of
through a magnetic field directed into the page. The resistor has
a value
(a) What is the current through the resistor R?
(b) What is the mechanical power delivered by theconstant force?
396
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:
R
Blv I IR BlvΔV
397
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
Plug that into the first equation:
Il
F B FBIlF app
appm
R
vF I
IR
vF
IlR
lvF
R
Blv
R
VI appappapp
2
You should get I=0.5 A.
398
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
You should get 2 W.
Alternatively, you note that power delivered by a force is
where θ is the angle between the force and velocity. In this case, θ = 0, so
RIRF2PP
cosFvF P
WFvF 2P
399
[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?
400
[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.
401
[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.
402
[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,
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