Physics 121: Electricity and Magnetism
Introduction
Syllabus, rules, assignments, exams, etc.


Text: Young & Friedman, University Physics
Homework & Tutorial System: Mastering Physics
Course Content:
•
•
•
•
•
•
5 Weeks: Stationary charges –
– Forces, fields, Electric flux, Gauss’ Law,
–
Potential, potential energy, capacitance
2 Weeks: Moving charges –
– Currents, resistance, circuits containing resistance and capacitance,
– Kirchoff’s Laws, multi-loop circuits, RC circuits
2 Weeks: Magnetic fields (static fields due to moving charges)
– Magnetic force on moving charges,
– Magnetic fields caused by currents (Biot-Savart’s and Ampere’e
Laws)
2 Weeks: Induction & Inductance
– Changing magnetic flux (field) produces currents (Faraday’s Law)
– Inductance, LR Circuits
2 – 3 Weeks: AC (LCR) circuits,
– Eelectromagnetic oscillations, Resonance
– Impedance, Phasors
Not covered:
– Maxwell’s Equations - unity of electromagnetism
– Electromagnetic Waves – light, radio, gamma rays,etc
Copyright R. Janow – Fall 2015
1
Physics 121 - Electricity and Magnetism
Lecture 01/02 – Vectors, Charge, Coulombs Law
Y&F Chapter 21, Sec. 1 - 3
Review of Vectors:
• Components in 3D. Right-hand rule
• Scalar multiplication, Dot product, vector
product
Charge:
• Basic properties of Electric Charge
• Quantization of Charge
• Structure of Matter
• Conservation of Charge
• Conductors and Insulators
• Coulomb’s Law – Force
• Shell Theorem
• Examples
Copyright R. Janow – Fall 2015
2
Definition: Right-Handed Coordinate Systems
•
•
•
We always use right-handed
coordinate systems.
In three-dimensions the righthand rule determines which way
the positive axes point.
Curl the fingers of your RIGHT
HAND so they go from x to y.
Your thumb will point in the
positive z direction.
z
y
x
This course uses several right hand rules related to this one!
Copyright R. Janow – Fall 2015
Vectors in 3 dimensions
•
Unit vector (Cartesian) notation:
•
Spherical polar coordinate representation:
1 magnitude and 2 directions
Rene Descartes
1596 - 1650

a  a x î  a y ĵ  a zk̂

a  (a, , )
•
z
Conversion into x, y, z components
a x  a sin  cos 
a y  a sin  sin 
a z  a cos 
•

a
az
Conversion from x, y, z components

a  a 2x  a 2y  a 2z
1
ax
  cos a z / a
  tan  1 a y / a x
x
Copyright R. Janow – Fall 2015
ay

a sin()
y
Surface Integral Example: Show that the surface area of a
sphere A= 4pR2 by integrating over the sphere’s surface
Find dA – an area segment on the surface of the sphere,
then integrate on angles  (azimuth) and  (co-latitude).
z
dA  dl  dh

dA  r̂ dA
Where:
• dl is a curved length segment of the circle around
the z-axis (along a constant latitude line)
• dh is a segment along the  direction (along a
constant longitude line)
dl  r sin() d

r

y
dh  r d
x
  [0, p ]
Angle range for a full sphere:

A   r̂.dA 
surface
  [0, 2p ]
Factors into 2 simple 1 dimensional integrations
2p
p
0
0
 
2p
p
0
0
r sin() dd  r {  d} {  sin() d }
2
2
p
p
 2pr 2  sin() d  2pr 2 [  cos()] 0  2pr 2 ()[ 1  1]
0
 A  4 pr 2
Copyright R. Janow – Fall 2015
There are 3 Kinds of Vector Multiplication
Multiplication of a vector by a scalar:

sA  sA x î  sA y ĵ

A

sA
vector times scalar  vector whose length is multiplied by the scalar
Dot product (or Scalar product or Inner product):

B


A
- vector times vector  scalar
- projection of A on B or B on A
- commutative
 
 
AoB  ABcos( )  BoA  A xBx  A yB y  A zBz
unit vectors measure
perpendicularity:
Copyright R. Janow – Fall 2015
î  ĵ  0,
ĵ  k̂  0, î  k̂  0
î  î  1,
ĵ  ĵ  1, k̂  k̂  1
Vector multiplication, continued
Cross product (or Vector product or Outer product):
- Vector times vector  another vector perpendicular to the
plane of A and B
- Draw A & B tail to tail: right hand rule shows direction of C
  
 
C  A  B  - B  A (not commutative)


magnitude : C  ABsin( )
B
A



where

is
the
smaller
angle
from
A
to
B

C
- If A and B are parallel or the same, A x B = 0
- If A and B are perpendicular, A x B = AB (max)

 
 
 
distributiv e rule : A (B
Algebra:
  C)   A  B  A  C
associative rules : sA
  (sB)
 B
  (sA)  B  A
( A  B)  C  A  (B  C)
î  ĵ  k̂, ĵ  k̂  î , î  k̂  - ĵ
Unit vector
representation:
î  î  0, ĵ  ĵ  0, k̂  k̂  0
 
A  B  (Ax î  Ay ĵ  A zk̂)  (B x î  By ĵ  Bzk̂)
 (A yBz - AzBy ) î  (A zBx - AxBz ) ĵ  (A xB y - AyBx )k̂


  
 

Applications:
L  r p
F  qE 
 2015
r F
Copyright R. Janow – Fall
i
j
k
 
qv  B
Fields are used to explain “Action at a Distance”
• Place a test mass, test charge, or test current at some test point in a field
• It feels a force due to the presence of remote sources of the field.
• The sources “alter space” at every possible test point.
• The forces (vectors) at a test point due to multiple sources add up
via superposition (the individual field vectors add & form the net field).
See also: Fields Examples – Gravitation posted on web page
Field Type
Source
Acts on
Definition
Strength
(dimensions)
mass
another
mass
Force per
unit mass at
test point
ag = F g / m
electrostatic
charge
another
charge
Force per
unit charge
at test point
E=F/q
magnetic
electric
current
.length
another
current
.length
Force per unit
current.length
B ~ F/qv or
F/iL
gravitational
Copyright R. Janow – Fall 2015
WHAT’S CHARGE?
CHARGE IS A BASIC ATTRIBUTE OF MATTER, LIKE MASS
WEAK
•
mass =
source of gravitational field
strength of response to gravitational field
•
charge = source of electrostatic (electromagnetic) field
strength of response to electromagnetic field
GENERATING STATIC CHARGE
•
•
Feet on nylon carpet, dry weather  sparks
Face of monitor or TV  dust, sizzling if touched
•
Rub glass rod (+) with silk (-) 
•
Rub plastic rod (-) with fur (+) 
+ + + + + +
+ + + + + +
-F
Copyright R. Janow – Fall 2015
CHARGED ROD
ATTRACTS NEUTRAL
CONDUCTOR
BY INDUCTION.
+ + + + + +
-F
CHARGES WITH
SAME SIGN REPEL
F
STRONG
F
-F
- - - - - - - -
F
+ + + + + +
CHARGES WITH
OPPOSITE SIGN
ATTRACT
- - -
+ + +
copper
CHARGE IS A “CONSERVED” QUANTITY
LIKE ENERGY, MOMENTUM, ANGULAR MOMENTUM, usually MASS
•
•
•
•
•
Net charge never just disappears or appears from nowhere.
Charges can be cancelled or screened by those of opposite sign
All of chemistry is about exchanging & conserving charge among atoms & molecules
Nuclear reactions conserve charge, e.g. radioactive decay: 92U238  90Th234 + 2He4
Pair Production & annihilation: g  e- + e+ (electron – positron pair)
ALL MATTER IS ELECTRICAL…
… BUT…matter is normally electrically neutral
(equal numbers of + and – charges).
(Why?)
OBJECTS WITH A NET CHARGE HAVE LOST OR GAINED SOME BASIC CHARGES
• Net + charge  deficit of electrons (excess of “holes”)
• Net – charge  excess electrons (deficit of “holes”).
• Why isn’t all matter a gas? Electrical forces hold matter together.
• Outside neutral objects electrical forces are “screened”, but not completely
• Weak “leftover” electrical force (Van der Waal’s) makes atoms stick together
& defines properties of solids, liquids, and gases.
Copyright R. Janow – Fall 2015
The physical world we see is electrical
MODERN ATOMIC PHYSICS (1900 – 1932): Rutherford, Bohr
•
Charge is quantized: e = 1.6x10-19 coulombs (small).
•
•
Electrons ( -e) are in stable orbital clouds with radius ~ 10-10 m.,
•
•
mass me = 9.1x10-31 kg. (small). Atoms are mostly space.
Protons ( +e) are in a small nucleus, r ~10-15 m , mp= 1.67x10-27 kg,
•
•
•
Every charge is an exact multiple of +/- e. Milliken ~1900.
Ze = nuclear charge, Z = atomic number. There are neutrons too.
A neutral atom has equal # of electrons and protons.
An ion is charged + or – and will attract opposite charges to become neutral
Electron
Proton
Neutron
Copyright R. Janow – Fall 2015
2-1: Which type of charge is
easiest to pull out of an
atom?
A.
Proton
B.
Electron
The protons all repel each other
Why don’t nuclei all fly apart?
What’s the Difference between Insulators & Conductors?
Insulators
Charge not free to move
(polarization not shown)
induced
charge
induced
charge
net
charge
Conductor
Copyright R. Janow – Fall 2015
Charge free to move in copper rod
Plastic rod repels electrons
Charge separation induced in cooper rod
F always attractive for plastic rod
near either end of copper rod
What if plastic rod touches copper rod?
Charging a conducting sphere by induction
- - - -
- - - -
1.
2.
neutral
- - - -
+
+
-
+
-
3.
-
+
+
+
ground
wire
draw off
negative charge
induce
charges
4.
+
+
5.
+
remove
wire
+
+
+
uniform
distribution
Atomic view of Conductors: charges free to move
SOLID CONDUCTORS
-
metals for example
LIQUID & GASEOUS
CONDUCTORS
sea water, humans,
hot plasmas
Copyright R. Janow – Fall 2015
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
• Regular lattice of fixed + ions
• conduction band free electrons
wander when E field is applied
• Electrons and ions both free
to move independently
• Normally random motion
• Electrons & ions move in opposite
directions in an E field.
Insulators: Charges not free to move
Electrons are tightly bound to ions...
...BUT insulators can be induced to polarize by nearby charges
Insulators can be solids, liquids, or gases
Polarization means charge
separates but does not leave home
free
charge
MOLECULES CAN HAVE
PERMANENT POLARIZATION...
...OR MAY BE INDUCED TO
ROTATE OR DISTORT WHEN
CHARGE IS NEAR
-
positive
polarization
charge
-
+
+
+
+
field inside insulating
material is smaller than it
would be in vacuum
+
+
+
+
negative
polarization
charge
THE DIELECTRIC CONSTANT MEASURES
MATERIALS’ ABILITY TO POLARIZE
Semiconductors:
- Normally insulators, but can be
weak conductors when voltage,
doping, or heat is applied
- Bands are formed by overlapping
atomicR. energy
levels
Copyright
Janow – Fall
2015
ENERGY
BAND GAP
+
-
free
charge
conduction band
valence band
ELECTROSTATIC FORCE LAW (coulomb, 1785)
3rd law pair of
forces
Constant
k=8.89x109 Nm2/coul2
F12
q1
Force on
q1 due to q2
r12
repulsion
shown
q2
F21
(magnitude)
• Force is between pairs of point charges
• An electric field transmits the force
(no contact, action at a distance)
• Symmetric in q1 & q2 so F12 = - F21
• Inverse square law
• Electrical forces are strong compared to gravitation
•
e0= 8.85 x 10-12
Why this value? Units
Unit of charge = Coulomb
1 Coulomb = charge passing through a cross section
of a wire carrying 1 Ampere of current in 1 second
1 Coulomb = 6.24x1018 electrons [ = 1/e]
charge flow
Copyright R. Janow – Fall 2015
dq  idt
F12 
k
k q1q2
2
r12
1
 9x109
4p e0
Gravitation is weak
G m 1m 2
F12 
2
r12

G 6.67x10 11 N.m2 / kg2
dq
... or... current  i 
dt
Coulomb’s Law using vector notation

F12  Force on q1 due to q2

r12  Displacement to q1 from q2
r̂12  Unit vector pointing radially
y
away from q2 at location of q1
Sketch shows repulsion:

F12 is parallel to r̂12

r12
q2
q1

F12
r̂12

r
r̂12  12  r̂21
r12
For attraction (opposite charges):
x

F12 in opposite direction to r̂12

F12 
1 q1q2 
r
3 12
4p e0
r12
Copyright R. Janow – Fall 2015
or
cubed

F12 
1 q1q2
r̂
2 12
4p e0
r12
squared
Superposition of forces & fields
The electrostatic force between a specific pair of point charges
does not depend on interaction with other charges that may be
nearby – there are no 3-body forces (same as gravitation)
Example: Find NET force on q1
y
Method: find forces for all pairs of
charges involving q1, then add the
forces vectorially
n 

Fnet on 1   F1,i
i 2


 F1,2  F1,3  F1,4  .....
•
•
•
•
•
+q2

r12

r13
+q1

F13


F12
F14

r14
+q3
x
F11 is meaningless
Use coulomb’s law to calculate individual forces
-q4
Keep track of direction, usually using unit vectors
Find the vector sum of individual forces at the point
For continuous charge distributions, integrate instead of summing
Copyright R. Janow – Fall 2015
Example: Charges in a Line
2-4: Where do I have to place the +Q charge in order
for the forces on it to balance, in the figure at right?
A. Cannot tell, because + charge value is not given.
B. Exactly in the middle between the two negative
charges.
C. On the line between the two negative charges, but
closer to the 2q charge.
D. On the line between the two negative charges,
but closer to the q charge.
E. There is no location that will balance the forces.
Copyright R. Janow – Fall 2015
-2q
+Q
-q
Calculate the Exact Location for the forces to balance
•
•
•
•
Force on test charge Q is attractive toward both negative
charges, hence they could cancel.
By symmetry, position for +Q in equilibrium is along y-axis
Coordinate system: call the total distance L and call y the
position of charge +Q from charge q.
Net force is sum of the two force vectors, and has to be
zero, so
F  Ffrom 2q  Ffrom q  k
•
2qQ
qQ

k
0
2
2
(L  y)
y
k, q, and Q all cancel due to zero on the right, so our
answer does not depend on knowing the charge values.
We end up with
2
1

(L  y)2 y2
(L  y)2
2 
2
y
•
Solving for y,
y
Ly
 2
y
L
,
 0.412L
1 2
y is less than half-way to the top negative charge
Copyright R. Janow – Fall 2015
-2q
+Q
L
y
-q
How do we know that electrostatic force - not gravitational
force - holds atoms & molecules together?
Example: Find the ratio of forces in Hydrogen
• 1 proton in nucleus, neutral Hydrogen atom has 1 electron
• Both particles have charge e = 1.6 x 10-19 C.
• me = electron mass = 9.1 x 10-31 kg.
• mp = proton mass = 1833 me
• a0 = radius of electron orbit = 10-8 cm = 10-10 m.
Felec
1 e2

4p e0 a02
Fgrav  G
memp
a 02
Felec
a 02
1
e2
9 x 109
(1.6x10 -19 )2


Fgrav 4p e0G a 02 1833 me2
6.67x10 11 1833 (9.11x10 - 31)2

Felec
 3.1 x 1039
Fgrav
• Electrostatics dominates atomic structure by a factor of ~ 1039
• So why is gravitation important at all? Matter is normally Neutral!
• Why don’t atomic nuclei with several protons break apart?
Copyright R. Janow – Fall 2015
Example: Forces on an electron and two protons along a line (1D)
1
2
r12
r23
p
e
3
p
By symmetry, all forces lie along the horizontal axis
Show forces on each of the objects in free body diagrams for # 1, 2, & 3
F12
FBD for 1
F13
F21
FBD for 2
FBD for 3
•
•
•
•
2

9 e
| F12 |  9  10 2
r12
2

9 e
| F13 | 9  10 2
r13


F21  F12
2

9 e
| F23 | 9  10 2
r23
F23
F31
F32


F31  F13


F32  F23
There are 3 action-reaction pairs of forces
Which forces are attractive/repulsive?
Could net force on any individual charge be zero above?
What if r12 = r23?
Copyright R. Janow – Fall 2015
EXAMPLE: Vector-based solution…
… Start with two point charges on x-axis
Find F12 - the Force on q1 due to q2
F21 has the same magnitude, opposite direction
F12
F21
R
q1
Let:
q2
x
q1 = +e = +1.6 x 10-19 C
q2 = +2e = +3.2 x 10-19 C.
R = 0.02 m
Magnitude of force:

F12 
-19
-19
1 | q1 | x | q2 |
1.6
x
10
x
3.2
x
10
9

9
 10
x
2
4p e0
R
0.02 2

F12  1.15 x 10-24 N
small!
In vector notation:

F12   1.15 x 10-24 î N
Copyright R. Janow – Fall 2015

F21   1.15 x 10-24 î N
EXAMPLE continued: add a 3rd charge on x-axis between q1 and q2
R
F12
q3
q1
Let:
q3 = - 2e = -3.2 x 10-19 C.
F21
r23
r13
r13= ¾ R,
x
q2
r23= ¼ R
What changes?
• Because force is pairwise: F12 & F21 not affected...but...
• Four new forces appear, only two distinct new magnitudes
F13 and F23 .... FBDs below.
R
F12
q1
F13
r13
F31
q3
…see next page...
Copyright R. Janow – Fall 2015
F32
r23
F23
F21
q2
EXAMPLE continued: calculate magnitudes
F13  K
| q1 | x | q3 |
r13 2
 9  10
9
1.6 x 10-19 x 3.2 x 10-19
x

F13   2.05 x 10-24 î N
F23  K
| q2 | x | q3 |
r23 2
 9  10
9

F23   3.69 x 10-23 î N
(0.75R ) 2


F31  - F13   2.05 x 10-24 î N
3.2 x 10-19 x 3.2 x 10-19
x
F13  2.05 x 10-24 N
F23  3.69 x 10-23 N
(0.25R ) 2


F32  - F23   3.69 x 10-23 î N
Find net forces on each of the three charges by applying superposition
On 1:



Fnet ,1  F12  F13   1.15 x 10-24 î  2.05 x 10-24 î

Fnet ,1  9.0 x 10-25 î N.
On 2:



Fnet ,2  F23  F21   3.69 x 10-23 î  1.15 x 10-24 î

Fnet ,2  - 3.58 x 10-23 î N.
On 3:



Fnet ,3  F31  F32   2.05 x 10-24 î  3.69 x 10-23 î

Fnet ,3  - 3.49 x 10-23 î N.
Copyright R. Janow – Fall 2015
Example: Movement of Charge in Conductors
• Two identical conducting spheres. Radii small compared to a.
• Sphere A starts with charge +Q, sphere B is neutral
• Connect them by a wire. What happens and why?
Charge redistributes until ......???
a
B
A
a
+Q
B
+Q/2
A
+Q/2
a
B
+Q/2
A
+Q/2
How do we know the
final charges are
equal?
Are they equal if the
spheres are not
identical?
• Spheres repel each other. How big is the force?
We applied shell theorem outside
• Do the spheres approximate point
charges? What about polarization?
FAB  K
(Q / 2)2
a2
1 Q

 
16p e0  a 
2
Another Example: Find the final charges after the spheres below touch?
B
a
+20e
B
A
-50e
ACopyright
R. Janow – Fall 2015
-30e
a
B
-15e
A
-15e
What determines
how much charge
ends up on each?
What force does a fixed spherical charge distribution
exert on a point test charge nearby?
+
+
+
+
+
+
+
+
+
q
r
+
R
+
+
+
+
+
P
+
Proof by lengthy integration
or symmetry + Gauss Law
+
Shell Theorem (Similar to gravitational Shell Theorem)
• Holds for spherical symmetry only: shell or solid sphere
• Uniform surface charge density s - not free to move (insulator)
• Shell has radius R, total charge Q
•
s  Q/A  Q/4 pR 2
2
dimensions are Coulombs / m
Outside a shell, r > R: Uniform shell responds to charged
particle q as if all the shell’s charge is a point charge at
the center of the shell.
Inside a shell, r < R: Charged particle q inside uniform hollow
shell of charge feels zero net electrostatic force from the shell.
Why Copyright
does the
shell
theorem
work? spherical symmetry  cancellations
R. Janow
– Fall
2015
Charge on a spherical conductor – charge free to move
•
•
Net charge Q on a spherical shell spreads out everywhere over the
surface.
Tiny free charges inside the conductor push on the others and move
as far apart as they can go. When do they stop moving?
If no other charges are nearby (“isolated” system)
•
•
Charges would spread themselves out uniformly because of symmetry
Sphere would act as if all of it’s charge is concentrated at the center
(for points outside the sphere) as in the Shell Theorem.
Bring test charge q<<Q near the sphere
q
Q
2R
d
Q is a point charge here
q
Q
d
The test charge induces polarization charge on the sphere
• Charge distribution will NOT be uniform any more
• Extra surface charge is induced on the near and far surfaces of the sphere
• The net force will be slightly more attractive than for point charges
The Shell Theorem works approximately for conductors if the test charge
induces very little polarization; e.g., if q << Q and/or d >> R.
Copyright R. Janow – Fall 2015
Example: three charges along x-axis
y
q1
Region III
q1 = + 8q
q not stated
q2 = -2q
q q
qproton = +e
Fa,b  K a 2b
x
L
qprot
q2
Region I
ra,b
Region II
For q1 & q2 as in the sketch where should a proton (test charge) be placed to
be in equilibrium (Fnet = 0, use “free body diagrams” & symmetry)
Proton in Region I:
qprot
Proton in Region III:

Fprot ,q2

F
 prot ,q1
Fprot ,q1
Forces opposed but q1  q2

Fprot ,q2
Proton in Region II:
Forces opposed
Equilibrium
x:

Fprot ,q2
qprot
and x  L  x

Fprot ,q1
qprot
q1
Fprot ,q1  K
 Fprot ,q2
Both forces to the right
No equilibrium possible
 q2 and x  L  x
Forces can never balance
So balance point exists here
8qe

x2
2qe
 K
(x - L)2
Did theR.test
charge
magnitude
Copyright
Janow
– Fall 2015
4
x2

1
(x - L)2
make any difference?
x  2L