Physics 121 - Electricity and Magnetism
Lecture 2 - Electric Charge
Y&F Chapter 21, Sec. 1 - 3
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History
Electromagnetism
Electric Charge
Quantization of Charge
Structure of Matter
Conservation of Charge
Conductors and Insulators
Coulomb’s Law – Force
Shell Theorem
Examples
Summary
1
Copyright R. Janow – Spring 2016
A Short History of Charge
Static Electricity through Electromagnetism
•
Ancients discover static electricity and natural magnets
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•
17th Century: (Newton)
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Action at a distance “field” concept for gravitation
18th Century: (Ben Franklin)
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•
Rubbed amber (electron), “lodestone”
Today rub balloons, comb & paper
Shocks from door knobs, pipes,…
Van de Graaf generator
Two flavors of “Electric Fluid” (+ & -)
Lightning is electrical
19th Century:
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Currents produce magnetic fields (Ampere) & deflect a compass needle (Oersted 1820)
Generator principle: changing magnetic fields produce electric fields (Faraday)
Action at a distance “fields applied to E&M.
Maxwell’s Equations unify Electricity & Magnetism (James Clerk Maxwell, ~ 1865)
• Predicted accelerated (e.g., vibrating) charges cause radiation
• Electromagnetic waves as linked, oscillating electric and magnetic fields
– Electromagnetic (radio) waves created in lab using LC circuit (Hertz, 1887)
• EM SPECTRUM: Radio, Infra-red, Visible light ,Ultra-violet, X-Rays, Gamma rays
ELECTRIC & MAGNETIC EFFECTS ARE INTERCONNECTED
Copyright R. Janow – Spring 2016
The Electromagnetic Spectrum
Accelerated charges radiate
f c
Copyright R. Janow – Spring 2016
WHAT’S CHARGE?
CHARGE IS A BASIC ATTRIBUTE OF MATTER, LIKE MASS
WEAK
•
mass =
•
charge = source of electrostatic (electromagnetic) field
strength of response to electromagnetic field
source of gravitational field
strength of response to gravitational field
CHARGED ROD
ATTRACTS NEUTRAL
CONDUCTOR
BY INDUCTION.
HOW TO GENERATE STATIC CHARGE
•
•
Feet on nylon carpet, dry weather  sparks
Face of monitor or TV  dust, sizzling if touched
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Rub glass rod (+) with silk (-) 
•
Rub plastic rod (-) with fur (+) 
+ + + + + +
-F
CHARGES WITH
SAME SIGN REPEL
F
+ + + + + +
+ + + + + +
-F
STRONG
F
-F
- - -
+ + +
copper
- - - - - - - -
F
+ + + + + +
CHARGES WITH
OPPOSITE SIGN
ATTRACT
Copyright R. Janow – Spring 2016
CHARGE IS “CONSERVED”
AS IS ENERGY, MOMENTUM, ANGULAR MOMENTUM, MASS*
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Net charge never just disappears or appears from nowhere.
Charges can be cancelled or screened by charges 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 – Spring 2016
The physical world we see is electrical
MODERN ATOMIC PHYSICS (1900 – 1932): Rutherford, Bohr
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Charge is quantized: e = 1.6x10-19 coulombs (small).
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Electrons ( -e) are in stable orbital clouds with radius ~ 10-10 m.,
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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,
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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
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?
Neutron
WHAT”S WRONG WITH THIS PICTURE?
Copyright R. Janow – Spring 2016
What’s the Difference between Insulators & Conductors?
Insulators (polarization)
Charge not free to move
(polarization not shown)
induced
charge
induced
charge
net
charge
Conductor (induction, screening)
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 Copyright
rod touches
rod?
R. Janowcopper
– Spring 2016
Conductors: Charge a metal 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
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-
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-
-
-
-
-
-
-
-
-
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• Regular lattice of fixed + ions
• Free electrons in conduction band
wander when E field is applied
• Electrons and ions both free
to move independently
• Normally in random motion
• Electrons & ions move in opposite
Copyright R. Janow – Spring 2016
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
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
+
+
+
+
free
charge
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
atomic energy levels
ENERGY
BAND GAP
Polarization means charge
separates but does not leave home
conduction band
valence band
Copyright R. Janow – Spring 2016
Insulators and Conductors
2-2: Which of the following are good conductors of electricity?
A.
A plastic rod.
B.
A glass rod.
C.
A rock.
D.
A wooden stick.
E.
A metal rod.
2-3: Balls A, B, and D are charged plastic. Ball C is made of metal and has zero net
charge on it. The forces between pairs are as shown in sketches 1, 2, 3. In sketches 4
and 5 are the forces between balls attractive or repulsive?
A
C
A
B
A
D
1
2
3
C
B
?
?
D
D
4
5
Answer choices
A.
B.
C.
D.
E.
4 is attractive, 5 is repulsive
4 is attractive, 5 is attractive
4 is repulsive, 5 is repulsive
4 is repulsive, 5 is attractive
not enough information
Copyright R. Janow – Spring 2016
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
dq  idt
F12 
k
k q1q2
2
r12
1
 9x109
4 e0
Gravitation is weak
G m 1m 2
F12 
2
r12
G 6.67x10 11 N.m2 / kg2
dq
... or... current  i 
Copyright R. Janow – Spring
dt 2016
Coulomb’s Law using Vector Notation

F12  Force on q1 due to q2

r12  Displacement to q1 from q2
y

r12
r̂12  Unit vector pointing radially
away from q2 at location of q1
Sketch shows repulsion:

F12 is parallel to r̂12
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
4 e0
r12
or
cubed

F12 
1 q1q2
r̂
2 12
4 e0
r12
squared
Copyright R. Janow – Spring 2016
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
+q2
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  .....
•
•
•
•
•

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 – Spring 2016
Example: Charges in a Line
2-4: Where do I have to place the charge +Q 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.
-2q
+Q
-q
Copyright R. Janow – Spring 2016
Calculate the exact location for the forces to balance
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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
(L - y)
2
-k
qQ
y
2
0
-2q
+Q
L
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:
-q
2
1

(L - y)2 y2
(L - y)2
2 
y2
•
Solving for y,
y
L-y
 2
y
L
, 0.412L
1 2
Copyright R. Janow – Spring 2016
How do we know that electrostatic force – not the 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

4 e0 a02
Fgrav  G
memp
a 02
Felec
a 02
1
e2
9 x 109
(1.6x10 -19 )2


Fgrav 4 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 – Spring 2016
Example: Forces on an electron and two protons along a line (1D)
1
2
r12
3
r23
p
e
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 – Spring 2016
NEXT 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
q1 = +e = +1.6 x 10-19 C
q2
q2 = +2e = +3.2 x 10-19 C.
x
R = 0.02 m
Magnitude of force:

F12 
-19
-19
1 | q1 | x | q2 |
x 3.2 x 10
9 1.6 x 10
 9  10 x
2
4 e0
R
0.02 2

F12  1.15 x 10-24 N
small!
In vector notation:

F12  - 1.15 x 10-24 î N

F21   1.15 x 10-24 î N
Copyright R. Janow – Spring 2016
EXAMPLE continued: add a 3rd charge on x-axis between q1 and q2
R
F12
q3
q1
q3 = - 2e = -3.2 x 10-19 C.
F21
r23
r13
x
r13= ¾ R,
q2
r23= ¼ R
What changes?
• Because force is pairwise F12 & F21 not affected...but...
• Four new forces appear,having only two distinct new magnitudes
F13 and F23 .... FBDs below.
R
F12
q1
F13
r13
F31
q3
…see next page...
F32
r23
F23
F21
q2
Copyright R. Janow – Spring 2016
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 – Spring 2016
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

 
16 e0  a 
2
Another Example: Find the final charges after the spheres below touch?
B
a
A
+20e
-50e
B
A
-30e
a
B
-15e
A
-15e
Copyright R. Janow – Spring 2016
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 (for gravitational Shell Theorem see Lecture 1)
• 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 R2
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 does the shell theorem work? spherical symmetry
Copyright
R.cancellations
Janow – Spring 2016
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 – Spring 2016
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
qpro
q2
Region I
ra,b
Region II t
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 q1
Equilibrium
x:
Fprot ,q1  K
 Fprot ,q2
Both forces to the right
No equilibrium possible

Fprot ,q2
qprot
and x  L  x Forces can never balance

Fprot ,q1
qprot
 q2 and x - L  x
So balance point exists here
8qe

x2
2qe
 K
(x - L)2
4
x2

1
(x - L)2
x  2L
Copyright R. Janow – Spring 2016
Did the test charge magnitude make any difference?