copyright r. janow – spring 2015 1 physics 121 - electricity and magnetism lecture 2 - electric...

1Copyright R. Janow – Spring 2015

Physics 121 - Electricity and Magnetism Lecture 2 - Electric Charge Y&F Chapter 21, Sec. 1 - 3

• History• Electromagnetism• Electric Charge• Quantization of Charge• Structure of Matter• Conservation of Charge • Conductors and Insulators• Coulomb’s Law – Force• Shell Theorem• Examples• Summary

Copyright R. Janow – Spring 2015

• Ancients discover static electricity and natural magnets– Rubbed amber (electron), “lodestone”– Today rub balloons, comb & paper– Shocks from door knobs, pipes,…– Van de Graaf generator

• 17th Century: (Newton)– Action at a distance “field” concept for gravitation

• 18th Century: (Ben Franklin)– Two flavors of “Electric Fluid” (+ & -) – Lightning is electrical

• 19th Century: – 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 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

Short History - Charge Static Electricity through Electromagnetism

ELECTRIC & MAGNETIC EFFECTS ARE INTERCONNECTED


Fields are used to explain “Action at a Distance”

Field Type Source Acts on Definition(dimensions)

Strength

gravitational massanother

massForce per unit

mass at test pointag = Fg / m

electrostatic chargeanother

charge

Force per unit charge at test

pointE = F / q

magneticelectric

current .length

another current .length

Force per unit current.length B ~ F/qv or F/iL

• 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).


The Electromagnetic Spectrum

c f

Accelerated charges radiate


WHAT’S CHARGE?CHARGE IS A BASIC ATTRIBUTE OF MATTER, LIKE MASS• mass = source of gravitational field

strength of response to gravitational field

• charge = source of electrostatic (electromagnetic) field

strength of response to electromagnetic field

WEAK

STRONG

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 (+)

CHARGED ROD ATTRACTS NEUTRAL

CONDUCTOR BY INDUCTION.

+ + + + + +F

+ + + + + +

- F

CHARGES WITH SAME SIGN REPEL

F+ + + + + +

- - - - - - - -- F

CHARGES WITH OPPOSITE SIGN

ATTRACT

+ + + + + +

- F

- - - + + +F

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: 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.


MODERN ATOMIC PHYSICS (1900 – 1932): Rutherford, Bohr• Charge is quantized: e = 1.6x10-19 coulombs (small).

• Every charge is an exact multiple of +/- e. Milliken ~1900.

• 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,

• 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

The physical world we see is electrical

Proton

Electron

Neutron

2-1: Which type of charge is easiest to pull out of an atom?

A. ProtonB. Electron

The protons all repel each other Why don’t nuclei all fly apart?


What’s the Difference between Insulators & Conductors?

InsulatorsCharge not free to move(polarization not shown)

ConductorCharge free to move in copper rodPlastic rod repels electrons Charge separation induced in cooper rodF always attractive for plastic rod near either end of copper rodWhat if plastic rod touches copper rod?

inducedcharge

inducedcharge

netcharge


Charging a conducting sphere by induction

1.

neutral

2.

induce charges

- - - - - -

++

+-

-

- 3.

draw off negative charge

- - - - - -

++

+

groundwire

4.

remove wire

- - - - - -

++

+5.

uniformdistribution

+ ++

SOLID CONDUCTORS

LIQUID & GASEOUS CONDUCTORS

metals for example

• Regular lattice of fixed + ions• conduction band free electrons wander when E field is applied

sea water, humans,hot plasmas

+

+

+

+

+

+

+

+

+

+

+

+

- - - -- - - -

- - - -

+

++

++

+

+

+

+

++

+

- - - -- - - -

- - - -

• Electrons and ions both free to move independently• Normally random motion• Electrons & ions move in opposite directions in an E field.

Atomic view of Conductors: charges free to move


Insulators: Charges not free to move

Semiconductors: - Normally insulators, but can be weak conductors when voltage, doping, or heat is applied - Bands are formed by overlapping atomic energy levels

valence band

conduction band

EN

ER

GY

BA

ND

GA

P

Electrons are tightly bound to ions... ...BUT insulators can be induced to polarize by nearby charges

Polarization means charge separates but does not leave home

MOLECULES CAN HAVE PERMANENT

POLARIZATION... ...OR MAY BE INDUCED TO ROTATE OR

DISTORT WHEN CHARGE IS NEAR

+ -

Insulators can be solids, liquids, or gases

THE DIELECTRIC CONSTANT MEASURES MATERIALS’ ABILITY TO POLARIZE

negativepolarizationcharge

positivepolarizationcharge

field inside insulating material is smaller than it would be in vacuum

----

++++

----

++++

freecharge

freecharge


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?

Answer choicesA. 4 is attractive, 5 is repulsive

B. 4 is attractive, 5 is attractive

C. 4 is repulsive, 5 is repulsive

D. 4 is repulsive, 5 is attractive

E. not enough information

A

A

A B

B DD D

CC

1 2 3 4 5

? ?


ELECTROSTATIC FORCE LAW (coulomb, 1785)

q1

q2

r12

F12

F21

3rd law pair of forces

212

2112

r

qqkF

Force onq1 due to

q2

(magnitude)

Constantk=8.89x109 Nm2/coul2

repulsionshown

• 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

• 0= 8.85 x 10-12 Why this value? Units

9x10 k 9

04

1

Gravitation is weak

221110676

212

2112

kg/m.Nx.Gr

mmGF

Unit of charge = Coulomb1 Coulomb = charge passing through a cross section of a wire carrying 1 Ampere of current in 1 second1 Coulomb = 6.24x1018 electrons [ = 1/e]

dtdq

i current or... ...flow charge idtdq


Coulomb’s Law using vector notation

12F

Force on q1 due to q2

12r

Displacement to q1 from q2

12r Unit vector pointing radiallyaway from q2 at location of q1

Sketch shows repulsion:

12to parallel is r F12

For attraction (opposite charges):

12to direction opposite in r F12

122

12

21

012123

12

21

012 4

1

4

1r

r

qqF r

r

qqF or

cubed squared

q2

q1

x

y

12r

12r

12F

2112

1212 r

r

rr


Superposition of forces & fieldsThe 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)

-q4

+q1

x

y

12r

12F

+q3

+q2

13r

14r

13F

14F

Example: Find NET force on q1

Method: find forces for all pairs of charges involving q1, then add the forces vectorially

.....,,,

n

ii,1 on net

FFF

FF

413121

21

• F11 is meaningless• Use coulomb’s law to calculate individual forces• 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


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.

-q

-2q

+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

• 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

• Solving for y, ,

y is less than half-way to the top negative charge

0y

qQk

)yL(

qQ2kFFF 22q romfq2 romf

22 y

1

)yL(

2

2y

yL 2

y

)yL(2

2

L412.021

Ly

y

L

-q

-2q

+Q


How do we know that electrostatic force - not gravitational force - holds atoms & molecules together?

Example: Find the ratio of forces in Hydrogen

a

eFelec 2

0

2

04

1

2

0a

mmGF

pegrav

• 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.

• 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?

2

2

112

2

20

20

0 18331067618334

1

)(9.11x10

)(1.6x10

x.

10 x 9

m

e

a

a

GF

F31-

19-9

egrav

elec

10 x 3.1 F

F 39

grav

elec


Example: Forces on an electron and two protons along a line (1 D)

Show forces on each of the objects in free body diagrams for # 1, 2, & 3

• 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?

e p p

1 2 3r12 r23

F21

F23

FBD for 1

FBD for 2

FBD for 3

F12

F13

F31F32

212

912 109

r

e |F|

2

213

913 109

r

e |F|

2

1221 FF

223

923 109

r

e |F|

2

1331 FF

2332 FF

By symmetry, all forces lie along the horizontal axis


Find F12 - the Force on q1 due to q2

F21 has the same magnitude, opposite direction

EXAMPLE: Vector-based solution… … Start with two point charges on x-axis

q1 = +e = +1.6 x 10-19 Cq2 = +2e = +3.2 x 10-19 C.R = 0.02 m

29

22

012

020109

4

1

.

10 x 3.2 10 1.6

R

|q||q| F

-19-191 xx

x

x

Let:

Magnitude of force:

N10 1.15 F -24x12

small!

In vector notation:

q1 q2

RF12 F21

x

N i10 1.15 F N i10 1.15 F -24 -24xx 2112


EXAMPLE continued: add a 3rd charge on x-axis between q1 and q2

q3 = - 2e = -3.2 x 10-19 C. r13= ¾ R, r23= ¼ R

q1 q2

RF12 F21

xq3

r13r23

• Because force is pairwise: F12 & F21 not affected...but...• Four new forces appear, only two distinct new magnitudes F13 and F23 .... FBDs below.

Let:

What changes?

…see next page...

q1

q2

RF12

F21

q3

r13

r23

F13

F31

F32

F23


Find net forces on each of the three charges by applying superposition

EXAMPLE continued: calculate magnitudes

29

213

313

750109

)R.(

10 3.2 10 1.6

r

|q||q| KF

-19-191 xxx

x

x

N10 2.05 F -24x13

N i10 2.05 F- F N i10 2.05 F -2413

-24xx

3113

29

223

323

250109

)R.(

10 3.2 10 3.2

r

|q||q| KF

-19-192 xxx

x

x

N10 3.69 F -23x13

N i10 3.69 F- F N i10 3.69 F -2323

-23xx

3223

On 1:

On 2:

i10 2.05 i10 1.15 F F F -24 -2413,net xx

121 N. i10 9.0 F -25,net x1

i10 1.15 i10 3.69 F F F -24 -2321,net xx

232 N. i10 3.58- F -23

,net x2

On 3: i10 3.69 i10 2.05 F F F -23 -2432,net xx

313 N. i10 3.49- F -23

,net x3


Now move charge off-axis – Now 2 dimensional

unchanged

As before:q1 = +e = +1.6 x 10-19 C.q2 = +2e = +3.2 x 10-19 C.R = 0.02 mNow: q3 goneq4 = -2e = -3.2 x 10-19 C.cos(60o) = .5sin(60o) = .866

F F F 14,net

121


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 ......???

• Spheres repel each other. How big is the force?We applied shell theorem outside • Do the spheres approximate point charges? What about polarization?

B

Aa

+Q

B

Aa

+Q/2

+Q/2 B

Aa

+Q/2

+Q/2

2

02

2

16

12

a

Q

a

)/Q(KFAB

How do we know the final charges are

equal?Are they equal the

spheres are not identical?

Another Example: Find the final charges after the spheres below touch?

B

Aa

-50e

+20e B

Aa

-15e

-15eBA

-30e

What determines how much charge ends up on each?


What force does a fixed spherical charge distribution exert on a point test charge nearby?

Shell Theorem (Similar to gravitational Shell Theorem) • Holds for spherical symmetry only: shell or solid sphere• Uniform surface charge density - not free to move (insulator)• Shell has radius R, total charge Q•

2m / Coulombs are dimensions

2 RQ/4 Q/A

R

++

+

++

+

+

+

+

+

+

++

+++ +

r q

P

Proof by lengthy integration or symmetry + Gauss Law

Why does the shell theorem work? spherical symmetry cancellations

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.


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 spread themselves out uniformly because of symmetry• Sphere acts 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

The Shell Theorem works approximately for conductors if the test charge induces very little polarization; e.g., if q << Q and/or d >> R.

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

Q q Q q

d d

Q is a point charge here

2R


Example: three charges along x-axisq1 = + 8q q not statedq2 = -2qqproton = +e

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)

y

qprotq2q1

Lx

Region IRegion III Region II

Proton in Region I: Both forces to the rightNo equilibrium possible qprot

1q,protF

2q,protF

Proton in Region III:qprot

1q,protF

2q,protF

Forces opposed but 21 q q x L x and

Forces can never balance

Proton in Region II:qprot

2q,protF

1q,protF

Forces opposed 21 q q x L x and

So balance point exists here

2b,a

bab,a

r

qqKF

Did the test charge magnitude make any difference?

2q,prot

2q,prot

L)-(x

2qeK F

x

8qeK F

2

1Equilibrium x: 22 L)-(x

1

x

4 L 2 x

copyright r. janow – spring 2015 1 physics 121 - electricity and magnetism lecture 2 - electric...

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