Chapter 15 Serway Physics

8/9/2019 Chapter 15 Serway Physics

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2/102

Concept: Electricity (Cont.)

Electricity occurs due to several types of physics:—Electric charge : charged particles interacting each other—Electric current : a movement of charges particles—Electric field : electromagnetic field (produced

with/without charge

particles)― Electric potential : the capacity of an electric filed to

do work on an electric charge

—Electromagnets : electrical currents generate magnetic fields, and changing magnetic fields generate electrical current.

Section 15.1


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Concept: Electricity (Cont.)

Electricity gives a wide variety of well ‐known

electrical effects :— lightning— static electricity

— electromagnetic induction — the flow of electrical current in an electrical wire — electricity permits the creation and reception of

electromagnetic radiations such as radio waves

Section 15.1


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Concept: Charge

Electric charge is a property of certain subatomic particles which gives rise to and interacts with the

electromagnetic force , one of the four fundamental force of nature.

Section 15.1

Charge generates in the atom, in which its most familiar carriers are the electrons and protons .


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Section 15.1

Properties of Electric Charges• Two types of charges exist

— they are called positive and negative

• Nature’s basic carrier of positive charge —proton (do not move from one material to

another because they are held firmly in the nucleus).

• Nature’s basic carrier of negative charge — electron (far lighter than protons and hence more

easily accelerated by forces).

• Gaining or losing electrons is how an object becomes

charged .


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Section 15.1

Properties of Electric Charges (cont.)

• Interaction between charges― Like charges repel one another ― Unlike charges attract one another


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Section 15.1


• Electric charge is always conserved .— Charge can not be created , only

exchanged— Objects become charged because negative charge is transferred from

one object to another.—Objects loss negative charges (i.e., electrons) and have a net positive

charges.—Objects gain electrons and have a net ___?____charges.


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• Charge is quantized (Millikan Oil‐Drop Experiment 15.7). –

All charge is a multiple of a fundamental unit of charge , symbolized by e (i.e. q=ne )• Quarks are the exception (fractional values of e )

– Electrons have a charge of ‐e – Protons have a charge of +e –

The SI (System International) unit of charge is the Coulomb (C)

• e = 1.6 x 10 ‐19 C

Section 15.1



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Conductors

• Conductors are materials in which the electric charges move freely in response to an electric force.

– Copper, aluminum and silver are good conductors.

– When a conductor is charged in a small region, the charge readily distributes itself over the

entire surface

of

the

material.

Section 15.2


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Insulators• Insulators are materials in which electric charges

do not move freely. – Glass and rubber are examples of insulators. – When insulators are charged by rubbing, only

the rubbed area becomes charged. – There is no tendency for the charge to move

into other regions of the material.

Section 15.2


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Section 15.2

Semiconductors

• The characteristics of semiconductors are between those of insulators and conductors.

— Silicon and germanium are examples of semiconductors.


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Charging by Conduction

• A charged object is placed in contactwith another object.

• When the rod is removed , the sphere is left with a net negative charge. (why?)

•

The object

being

charged

is

always

left with a charge having the same sign as the object doing the charging. (why?)

Section 15.2


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Charging by Induction

• A neutral conducting sphere has

equal number

of

electrons

and

protons. (why?)

• The positive and negative charges are uniformly distributed . (why?)

Section 15.2


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Charging by Induction (cont.)

• A negatively charged rubber rod is brought near an uncharged

sphere. The charges in the sphere are redistributed . (why?)

Section 15.2

• Think about if the sphere is an insulator , can electrons redistribute when a charged rod is brought close? (why?)


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• The region of the sphere nearestthe negatively charged rod has an excess of positive charge because of the migration of electrons away from this location. (why?)

Section 15.2

• A grounded conducting wire is connected to the sphere.

–

Allows some of the electrons to move from the sphere to the ground

(why?)


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• When an object is connected to a conducting wire or pipe

buried in

the

earth,

it

is

said

to

be grounded .

•

The Earth can be considered an infinite reservoir for electrons ; in effect, it can

accept or support an unlimited number of electrons.


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• The wire to ground is removed , the sphere is left with an excessof induced positive charge (why?)

• Initially, the positive charge on the sphere is nonuniformlydistributed. (why?)

Section 15.2


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• Eventually , the excess positive charge becomes evenly distributed

due to

the

repulsion between

the

positive charges. (why?)

Section 15.2

•

This process

is

known

as

induction .—Charging by induction requires no

contact with the object inducing the

charge.• Does the charged rubber rod lose its

negative charge ? Why?


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Polarization

• In the presence of a charged object , these centers

may separate slightly. – This results in more positive charge on one side of

the molecule than on the other side

Section 15.2

• In most neutral atoms or molecules, the center of positive charge coincideswith the center of negativecharge.


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Polarization

• This realignment of charge on the surface of an insulator is known as polarization .

• The charged object induces charge on the surface of the insulator .

Section 15.2


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Tips for Solving Physical Problems

Read the problem and form a quick picture.Read the problem again, slowly, and look for hidden

information, as

well

as

what

is

really

being

asked.

Write down all given information.

Draw a sketch .Figure out the important information for getting from

what you know to what you want to know.

Attack!

Check your answer for reasonability .


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Coulomb

s Law (Cont.)• Coulomb shows that an electric

force has the following properties:

Section 15.3

q1

q2F

2

1ˆeF r r ∝

r

– It is inversely proportional to the square of the separation distance , r, between them

r̂ – It is directed along the line

joining the two particles


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Coulomb

s Law (Cont.)

Section 15.3

– It is proportional to the product of the magnitudes of the charges , |q 1|and |q 2|on the two particles

1 22

ˆeq qF r

r ∝

1 21 2

2 ˆ 0e

q qF r for q q

r ∝ >

1 21 22

ˆ 0eq q

F r for q qr

∝ − <

— It is attractive if the charges are of opposite signs and repulsive if the charges have the same signs


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Coulomb

s Law (Cont.)• Mathematically,

• k e is called the Coulomb Constant – k e = 8.9875 x 10 9 N m2/C2

(SI)

•

Typical charges (q1 and q2) can be in the µC (10‐6C) range

– Remember, Coulombs must be used in the equation

• Remember that force is a vector quantity

Section 15.3

1 22

ˆe eq q

F k r r

=


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Coulomb

s Law (Cont.)

• Electrostatic forces: electric forces between unmoving charges

• Moving charges ― create magnetic forces (Ch. 19)

Section 15.3

• Applies only to point charges and sphericaldistributions of charges

—r is the distance between the two centers of

charge1 2

2 ˆe e

q qF k r

r =


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Characteristics

of

Particles

Section 15.3


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Vector Nature of Electric Forces• Two point charges are

separated by a distance r • The like charges produce a

repulsive force between them

• The force on q1 is equal in

magnitude and opposite in direction to the force on q2(why?)

Section 15.3

12 21F F = −

1 212 2 ˆe

q qF k r

r =

r̂


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Vector Nature of Forces (Cont.)

• The unlike charges produce an attractive force between

them• The force on q1 is equal in

magnitude and opposite in

direction to the force on q2

Section 15.3

12 21F F = −

1 2

12 2 ˆ

e

q qF k r

r = −

r̂


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30/102

Electrical Force Compared to Gravitational Force

•

Both are

inverse

square

laws .• The mathematical form of both laws is the same.

– Masses (m 1, m 2) replaced by charges (q1, q2) – G replaced by k e

• Electrical forces can be either attractive or repulsive .• Gravitational forces are always attractive .• Electrostatic force is stronger than the gravitational

force.

Section 15.3

1 22

12

ˆe e q qF k r r = 1 2

212

ˆg m mF G r r = −


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The Superposition Principle

• When a number of separate charges act on the charge of interest, each exerts an electric force.

• These electric forces can all be computed separately , one at a time, then added as vectors. This is another

example of the superposition principle .

Section 15.3


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The Superposition Principle (Cont.)

• The resultant force on any one charge equals the vector sum of the forces exerted by the other individual charges that are present.

– Find the electrical forces between pairs of charges separately

– Then add the vectors• Remember to add the forces as vectors

Section 15.3

12 13 14 ...totalF F F F = + + +


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Superposition Principle Example

• The force exerted by q1 on q3 is

•

The force exerted by q2• ( ) on q3 is• The total force exerted on

q3 is the vector sum of and

Section 15.3

13F

23F

23F 13F

13 23F F F = +

2 1q q<


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Superposition Principle Example (Cont.)

Section 15.3

1 313 2

13e q qF k r =

13 13 13ˆ ˆ

x yF F i F j= +

13 13

13 13

cos36.9

sin36.9 x

y

F F

F F

=

=

Step 1:

2 323 2

23

e

q qF k

r =

23 23 23ˆ ˆ

yF F i F j= +23 23

23 23

cos180sin180

x

y

F F F F

==

Step 2:

r13

r23


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Superposition Principle Example

Section 15.3

( ) ( )( ) ( )

13 23 13 13 23 23

13 23 13 23

ˆ ˆ ˆ ˆ

ˆ ˆ ˆ ˆ

x y x y

x x y y x y

F F F F i F j F i F j

F F i F F j F i F j

= + = + + +

= + + + = +

2 2 x yF F F = +

1tan y

x

F

F θ −

⎛ ⎞= ⎜ ⎟

⎝ ⎠

Step 3:

F

θF y

F x


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Problem Solving Strategy for Electric Force•

Draw a diagram

of

the

charges

in

the

problem.

• Identify the charge of interest. – You may want to circle it

• Units – Convert all units to SI. – Need to be consistent with k e

Section 15.3

1 22

ˆe eq q

F k r r

=

—k e: 8.9875 x 10 9 N m2/C2 (SI)

—q1, q2: Coulomb (C)—r : meter (m)—F : Newton (N)


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Problem Solving Strategy for Electric Force

(Cont.)• Apply Coulomb

s Law. –

For each

charge,

find

the

force

on

the

charge

of

interest.

– Determine the direction of the force. – The direction is always along the line of the two

charges.• Sum all the x ‐ and y ‐ components.

– This gives the x‐ and y‐components of the resultant force

• Find the resultant force by using the Pythagorean theorem and trigonometry.

Section 15.3

1 22

ˆe eq q

F k r r

=


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Electrical Field• Faraday developed an approach to discussing the

electrical field .•

An electric field is said to exist in the whole spacecentered at a charged object. – When another charged object enters this electric

field, the field exerts a force on the second charged object.

– Electric field exist independent of the existence of the second charge.

Section 15.4


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Electric Field (Cont.)

• A charged particle, with charge Q, called the source

charge , produces an electric field E in the region of space around it.

Section 15.4

• A small test charge , qo, placed in the field, will experience a force.


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Electric Field (Cont.)

• Mathematically,

•

SI unit : N / C• Use this for the magnitude of the field• The electric field is a vector quantity• The direction of the field is defined to be the

direction of the electric force that would be exerted on a small positive test charge placed at that point.

• Electric force and electric field,

Section 15.4

02 2

0 0

1 ˆ ˆe eQqF Q

E k r k r q q r r

⎛ ⎞≡ = =⎜ ⎟⎝ ⎠

00 02 2ˆ ˆe e

Qq QF k r q k r q E

r r

⎛ ⎞= = =⎜ ⎟⎝ ⎠


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Direction of Electric Field

• The electric field produced

by a

negative charge

is

directed toward the charge.

– A positive test charge would be attracted to the negative source

charge.

Section 15.4


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Direction of Electric Field (Cont.)

• The electric field produced

by a

positive charge

is

directed away from the charge.

– A positive test charge would be repelled from the positive source

charge.

Section 15.4


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Electric Field, Direction Summary

Section 15.4


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More About a Test Charge and The Electric Field

• The test charge is required to be a small charge . –

It can cause no rearrangement of the charges on the source charge. – Mathematically, the size of the test charge makes

no difference.• Using qo = 1 C is convenient

•

The electric field exists whether or not there is a test charge present.

Section 15.4


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Problem Solving Strategy for Electric Fields

• Calculate Electric Fields of point charges. – Use the equation to find the electric field due to

the source charges . – The direction is given by the direction of the force

on a positive test charge .

Section 15.4

r̂ r̂

20

ˆeF Q

E k r q r

≡ =

2e

Q E k r =

Direction:along for Q>0

opposite for Q


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Electric Field Lines

• A convenient aid for visualizing electric field patterns is to draw lines pointing in the direction of the field

vector at any point .

Section 15.5

+Q

Source charge

p1•1 E p2

•2 E

p3• 3

E

p4• 4

E 2 ˆe

Q E k r

r =

• These lines are called electric field lines and were introduced by Michael Faraday.


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Electric Fields and Superposition Principle•

The superposition principle holds when calculating the electric field due to a group of charges.—Draw a diagram of the charges in the problem.

—Identify the charge of interest.—Find the fields due to the individual charges .― Add them as vectors:

—Use symmetry whenever possible to simplify the problem.

Section 15.4

( ) ( )1 2 3

1 2 3 1 2 3

...

ˆ ˆ... ...

total

x x x y y y

E E E E

E E E i E E E j

= + + +

= + + + + + + +


48/102

Electric Field Due to Two Point Charges

Section 15.4

• Find the magnitude and the direction of the electric

field at point P produced by two charges, q1>0, q2


49/102

Electric Field Due to Two Point Charges

Section 15.4

1 11 1 1 12 2

1 1

ˆ ;e eq q

E k r E E k r r

= = =

• Use the equation for the electric field due to individual charge q1and q2 :

222 2 2 2 22 2

2 2

ˆ ( 0);e eqq

E k r q E E k r r

= < = =

1r̂ 2r̂

r2r1

Direction: along

(i.e.

y

‐direction)1̂r

Direction: along the opposite direction of 2̂r


50/102

1r̂ 2r̂

r2r1

Electric Field Due to Two Point Charges (Cont.)

Section 15.4

1 1

1 1 1

cos(90 ) 0

sin(90 )

x

y

E E

E E E

= =

= =

E 1y

• Find the x‐ and y‐components of E 1and E2 using the triangle in the figure

(black ‐dash lines):

2 2 2

2 2 2

cos( ) cos

sin( ) sin x

y

E E E

E E E

θ θ

θ θ

= − == − = −

E 2y

E 2x

0.300 0.400cos 0.600; sin 0.800

0.500 0.500m mm m

θ θ = = = =


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Electric Field Due to Two Point Charges (Cont.)

Section 15.4

2 2 x y E E E = +

• Use the Pythagorean theorem to find the magnitude of the resultant field:

r2r1

• Use the inverse tangent function

yields the direction of the resultant field: 1tan y

x

E

E φ −

⎛ ⎞= ⎜ ⎟

⎝ ⎠

1r̂ 2

r̂

• Sum

the x‐ and y‐components of the resultant field:

Ex

Ey

1 2

1 2

x x x

y y y

E E E

E E E

= += +


52/102

Electric

Field

Lines• A convenient aid for visualizing electric field

patterns is to draw lines pointing in the direction of the field vector at any point.

• These lines are called electric field lines and were introduced by Michael Faraday.

Section 15.5


53/102

• The field lines are related to the field in the following manners:

– The electric field vector, , is tangent to the

electric field lines at each point. – 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.

– is large/strong when the filed lines are closetogether and small/weak when the lines are farapart.

Section 15.5

Rules for Drawing Electric Field Lines


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Electric Field Line Patterns: Point Charge

― For a positive source charge, the lines will radiate outward (why?).

― Note that This two ‐dimensionaldrawing contains only the filed lines that lie in the plane

containing the point charge.― The lines are actually directed

radically outward from the charge in all directions.

Section 15.5

q0>0

0F q E =


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Electric Field Line Patterns: Point Charge (Cont.)

― For a negative source charge, the lines will point inward .

― The lines are radial and

extend all the way to infinity.

Section 15.5


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Rules for Drawing Electric Field Lines (Cont.)

• The lines for a group of charges must begin onpositive charges and end on negative charges.

–

In the case of an excess of charge, some lines will begin or end infinitely far away (why?).• The number of lines drawn leaving a positive charge

or ending on a negative charge is proportional to the magnitude of the charge (why?).

• No two field lines can cross each other (why?).

Section 15.5

20

ˆeF Q

E k r q r

≡ =


57/102

Electric Field Line Patterns: Two Charges

• An electric dipole consists of two equal and opposite

charges.• Use the superposition

principle to obtain the total

electric field of the dipole at any point in the space.

Section 15.5

• Draw lines starting from the positive charge and ending to the negative charge, keeping the electric field being the tangent of the lines.

E +E ‐

E =E ++E ‐

r‐

r+

•P

P’•

El t i Fi ld Li P tt T Ch


58/102

Electric Field Line Patterns: Two Charges

• The high density of lines between the charges indicates the strong electric field in this

region.

• The number of lines that begin at the positive charge must equal the number that terminate at the negative

charge (why?)

Section 15.5


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Electric Field Line Patterns: Two Charges (Cont.)

• Two equal but like point charges

• Use the superposition principleto obtain the total electric field of the dipole at any point in the space.

Section 15.5

• Draw lines starting from the positive charge and

ending to the infinity, keeping the electric fieldbeing the tangent of the lines.

r2r1

El t i Fi ld Li P tt T Ch g (C t )


60/102

Electric Field Line Patterns: Two Charges (Cont.)• At a great distance from the

charges, the field would be approximately that of a single charge of 2q (why?)

• The bulging out of the field lines

between the

charges

indicates

the repulsion between the charges.

• The low field lines between the charges indicates a weak field in

this region.Section 15.5

Electric Field Line P tterns T o Ch rges (Cont )


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• Unequal and unlike charges• The number of lines leaving

charge +2q is twice the number

terminating on charge –q (why?).

• Hence only half of the lines that leave the positive charge end at the negative charge (why?)

• In the case of an excess of charge, some lines will begin or end infinitely far away.

Section 15.5


Electric Field Line Patterns: Two Charges (Cont )


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• At a great distance from the charges, the field would be equivalent to that of a singlecharge +q (why?).

Section 15.5


Answer the question: No two field lines can cross each other.

•P 1 E

2 E


63/102

Electric Field Lines, Final

• The electric field lines are not material objects.

• They are used only as a pictorial representation of the electric field at various locations.

Section 15.5

Conductors in Electrostatic Equilibrium


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Conductors in Electrostatic Equilibrium•

When no net motion of charge occurs within a conductor, the conductor is said to be in electrostatic equilibrium .

Section 15.6

• An isolated conductor has the following properties:

• Conductors: electrons move freely in response to an electric force.

• Questions ask ourselves: how the electrons distributed? How the electric field inside/outside the conductor?

P t 1


65/102

Property 1•

The electric field is zero everywhere inside the conducting material.

Section 15.6

– Consider if this were not true• If there were an electric field

inside the conductor, the free charge there would move and there would be a flow of charge (why?) .

•

If there

were

a

movement

of

charge, the conductor would not be in equilibrium .

P t 2


66/102

Property 2•

Any exces s charge on an isolated conductor resides entirely on its surface .

Section 15.6

– A direct result of the 1/r 2 repulsionbetween like charges in Coulomb ’ s

Law – If some excess of charge could be

placed inside the conductor, the repulsive forces would push them as far apart as possible, causing them to

migrate to the surface.


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Property 3• The electric field just outside a

charged conductor is perpendicularto the conductor ’ s surface.

Section 15.6

– Consider what would happen if this was not true.

– The component along the surf ace would cause the charge to move .

–

It would not be in equilibrium .

Property 4


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Property 4•

On an irregularly shaped conductor, the charge accumulates at locations where the radius of curvature of the surface is smallest or the curvature

is largest (that is, at sharp points).

Section 15.6


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Property 4 (Cont.)

― Consider a conductor that is fairly flat at one end A and relatively pointed at the other end B.

― Any excess charge placed on the conductor moves to its surface.

Section 15.6

Propert 4 (Cont )


70/102

• The forces between two such charges at the flatterend are predominantly directed parallel to the surface, so the charges move apart until repulsive forces from other nearby charges establish an equilibrium.

Property 4 (Cont.)

Section 15.6

Property 4 (Cont )


71/102

• At the sharp end, the repulsive forces are directed predominantly away from the surface. As the result, there is less tendency for the charges to move apart

along the surface.

Property 4 (Cont.)

Section 15.6

Property 4 (Cont )


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• The cumulative effect of many such outward forces from nearby charges at the sharp end produce a larger resultant force away from the surface that

can be great enough to cause charges to leap from the surface into the surrounding air.

Property 4 (Cont.)

Section 15.6

Faraday Cage –property 1 ‐ 2


73/102

y g p p y

• A Faraday cage or Faraday shield is an enclosure formed by conducting material or by a mesh of such material. Such an enclosure blocks external static and non ‐static electric fields.

Section 15.6

• Driver safety during electricalstorm due to the fact that charges on the metal shell of the car will reside on the outer surface of the car.

Faraday cage

Lightning rod ‐ property 4


74/102

Storm clouds

•

If the

lightning

rod

having

sharp

points

is

attached

to

a

house, most of any charge on the house passes through these points, eliminating the induced charge on the house produced by storm clouds.

Section 15.6

Lightning rod property 4

Lightning rod

Li ht i d t 4 (C t )


75/102

•

A lightning discharge striking the house passes through the metal rod and is safely carried to the ground through wires leading from the rod to the

Earth.

Section 15.6

Lightning rod ‐ property 4 (Cont.)

Storm clouds

Sears building in Chicago

Millikan Oil ‐ Drop Experiment


76/102

p p

• Oil droplets are charged by friction in an atomizer• A horizontal light beam is used to illuminate the

droplets which are viewed by a telescope.• The droplets appear as shining stars against a dark

background, and the rate of fall of individual drops can be determined

Section 15.7

E

Millik Oil D E i t (C t )


77/102

Millikan Oil ‐ Drop Experiment (Cont.)

Section 15.7

• A drop with mass m and negative charge q• if E =0, there two forces acting on the drop: the downward gravity , and the

upward viscous drag force• When the drop reaches its terminal speed ,

0F mg D= + =∑ v

D bv= −

Millikan Oil Drop Experiment (Cont )


78/102


Section 15.7

• When E is on , there three forces acting on the drop: the downward gravity , the viscous drag force (upward or downward?), and the electric

force (upward or downward?)• When the drop reaches its new terminal speed ,

' 0F mg D qE = + + =∑ 'v

' ' D bv= −



79/102

• After making measurements on thousands of droplets,

Millikan and

his

co

‐workers found that

every

drop

had

a charge equal to some integral multiple of the elementary charge e:

q = ne , n=0, ±

1, ±

2, ±

3, …e=1.60x10 ‐19 C.

• Charge is quantized . Millikan was awarded the Nobel

Prize in Physics in 1923Section 15.7

• The drop can be followed for hours as it alternately rises and falls, simply by turning the electric field on and off.

Van de Graaff Generator


80/102


• An electrostatic generator designed and built by Robert J. Van de Graaff in 1929

• Extensively used in

nuclear physics research• Charge is transferred to

the dome by means of a rotating belt.

• An electrostatic field is

established.Section 15.8

Motor ‐driven pulleyPositively charged metallic needles

attract electrons from the belt

Positively charged belt attract electrons from the dome

Excess positive

charge on the dome



81/102

• Can we accumulate charges on the dome surface indefinitely ?

Section 15.8

discharge

― The strength of the field becomes

greater enough

to

partially

ionize

the air near the surface, increasing the conductivity of the air. Charges on the dome now have a pathway to leak off into the air.

—Eventually an electrostatic discharge takes place, producing some spectacular ‘lightning bolts’.



82/102

Section 15.8

discharge• How to inhibit the electric

discharge ?

― Place the entire system in a

container filled with a high ‐pressure gas, which is significantly more difficult to ionize.

― Charge find is easier to leap off a surface at points where the curvature is great. We can increase its radius to reduce the curvature and to store more charges

Electric Flux


83/102

Section 15.9

S1

• There are numbers of field lines passing out of the closed surface S1

S2

• There are numbers of field lines passing into the closed surface S2

S3

• The number of field lines passing into the closed surface S3 equal to the number of lines passing out of S3

S4

• The number of field lines passing into the closed surface S3 is less than the number of lines passing out of S4

For a closed surface , the number of field lines pass into or pass out of the closed surface only depends on the charges inside the

surface, and is independent of charges outside of the surface.


84/102

Electric Flux (Cont.)

• Electric flux Φ E is a measure of how much the electric field vectors penetratethrough a given surface .

• Field lines penetrating an area A perpendicular to the field,

• The product of EA is the flux, Φ E=EA

Section 15.9

/ ; E E N A N EA= ∝ ∝



85/102


• In general: – Φ E = E A cos θ – The perpendicular to the

area A is at an angle θ to the field. θ is the angle

between the field and the normal of the area A

Section 15.9

E



86/102


– When the area is constructed such that a closed surface is formed, use the convention that flux lines passing into the interior of the volume are negative and those passing out of the interior of the volume are positive .

– SI unit: N . m² / C

Section 15.9

Gauss’s Law


87/102

Gauss s Law•

Gauss’s law is essentially a technique for calculating the electric field on a closed surface with high symmetry

• Consider a point charge q surrounded by a spherical surface of radius r centered on the charge

• The magnitude of the electric field everywhere on the surface

2eq

E k r

=

Section 15.9

E

Gauss’s Law


88/102

• The electric flux through the surface is

( )22

cos cos(0 )

4 4

E dA

e e

EdA EdA

q E dA k r k q

r

θ

π π

Φ = =

= = =∑ ∑

∑• This results can be proven for any closed

surface that surrounds the charge q .

• ε o is the permittivity of free space and

equals 8.85 x 10‐12 C2/Nm 2, i.e.,

12 2 20

18.85 10 /

4 eC N m

k ε

π −= = × ⋅

0

4 E e qk qπ ε Φ = =

Section 15.9

E dA

G ’ L (C )


89/102

Gauss’s Law (Cont.)

• Gauss ’ Law states that the electric flux through any closed surface is equal to the net charge Q inside the

surface divided by ε o

– The area in Φ is an imaginary surface , a Gaussian surface , it does not have to coincide with the

surface of a physical object.

Section 15.9


90/102

Applications

of

Gauss’

Law

Gauss’s law is essentially a technique for

calculating the

electric

field

on

a

closed surface with high symmetry

Section 15.9

0

inside E

Gaussiansurface

Q E dA

ε

Φ = ⋅ =∑

Electric Field of a Charged Thin Spherical Shell


91/102


Section 15.9

(a) Find the electric field in the interiorof the conducting shell for r < a

• Apply Gauss’s law to a problem with spherical symmetry.

(b) Find the electric field outside the shell for r > b

(d) What is the distribution of charge on the sphere in this case?

(c) If an additional charge of ‐2Q is placed at the

center, find the electric field for r > b

‐2Q



92/102

( )2cos0

4 0 0

inside inside inside insidesphere sphere sphere

inside inside

E dA E dA E dA

E r E π

Φ = ⋅ = =

= = → =

∑ ∑ ∑

(a) Find the electric field in the interiorof the conducting shell for r < a

(Cont.)

Section 15.9

•

The electric

field

inside the

shell is

zero .

• Use Gauss law:0

0insideinsideQ

ε Φ = =

0insideQ =

dA

• Assume on the surface, is radially outward (spherical symmetry):

inside E

inside E • Select a Gaussian surface : a sphere inside the shell with r < a < b (dashed circle).

Electric Field of a Charged Thin Spherical


93/102

• Select a Gaussian surface: a sphere

outside the shell with r > b (dashed circle). Use Gauss law:0 0

outsideoutside

Q Qε ε

Φ = =

Q

Shell (Cont.)

Section 15.9

(b) Find the electric field outside the shell for r > b

( )20

20

cos0

4

4

outside outside outsidesphere sphere

outside outsidesphere

outside

E dA E dA

Q E dA E r

Q E r

π ε

πε

Φ = ⋅ =

= = =

→ =

∑ ∑∑

dAoutside E

• The field outside the shell is identical to that of a

point charge and points outward .

Electric Field of a Charged Thin Spherical


94/102

Shell (Cont.)

Section 15.9

Why no charge on the inner surface?

• Consider a Gaussian surface inside the hollow sphere:

0

0 0

0

inner E inside

sphere sphere

inner

Q E dA dA

Q

ε Φ = ⋅ = ⋅ = =

→ =

∑ ∑

• No charge on the inner surface!

‐‐

‐ ‐

‐

‐‐

• Suppose there are negative charges on the inner surface, Q inner

Q inner

0inside E =

Property 1: at the electrostatic equilibrium, no electric field inside the conductor



95/102

g p(Cont.)

(c) Now additional charge of ‐2Q is placed at the centerof the sphere.

2

0 0 0

20 0

cos0

24

4

outside outside outsideGauss surface Gauss surface

total center outside

outside

E dA E dA

Q Q Q Q Q E r

Q Q E

r

π

ε ε ε

ε πε

Φ = ⋅ =

+ − += = = =

= − ⇒ = −

∑ ∑

Section 15.9

+

•‐2Q r

a

b

+

+

+

+ +

+

+

+

Q dA

outside E

Gauss surface

•

Applying the Gauss’s law and assuming the field directed outward , the electricfield outside the shell is

Direction: radially inward



96/102

(Cont.)(d) What is the distribution of charge on the sphere in this

case?

0

0 0

0

0

( 2 ) 2

inside shell inside shell

center inner surface

center inner surface

inner surface center

E dA dA

Q Q

Q Q

Q Q Q Q

ε

Φ = ⋅ = =

+= =

⇒ + =

⇒ = − = − − =

∑ ∑

2outer surface inner surface shell

outer surface inner surface

Q Q Q Q

Q Q Q Q Q Q

+ = =

⇒ = − = − = −

Section 15.9

•‐2Q r

a

b

‐‐

‐

‐

‐‐

‐

‐

‐

Q ++

+

++

++++

+

++

+

+

inner surfaceQouter surfaceQ

• Applying Gauss’s law for the shell a < r < b (note: ):

Gauss surface0inside shell E =

Electric Field of a Charged (Uniform) Circular Ring


97/102

Electric Field of a Charged (Uniform) Circular Ring

A circular ring of charge of radius r has a total charge q(>0) uniformly distributed around it. Find the electric field of inside the ring.

•

dq 1

dE1

dq 2

dE2

r

0

0

0

E i isphere

i

Q E dA

E

ε Φ = ⋅ = =

⇒ ≠

∑

http://www.youtube.com/watch?v=BF3wEV4tWq8

dq 3

dq 4

Electric Field of a Nonconducting Plane Sheet of


98/102

gCharge

• Apply Gauss’s law to a problem with plane symmetry.

Section 15.9

• Find the electric field above and below a nonconducting infinite plane sheet of charge with uniform positive charge per unit area σ .



99/102

Charge (Cont.)

• The field must be perpendicular to the sheet.

Section 15.9

• The field is uniform .

• The field is directed either toward or away from the sheet.

Symmetry analysis:—

• Consider a Gaussian surface with cylindrical

symmetry.



100/102

Charge (Cont.)• Use a cylindrical Gaussian surf ace

Section 15.9

00

c o s 0 c o s 0

c o s 9 0 2

E to p b o tto m

ins ides ide

E A E A

Q E A E A

ε

Φ = +

+ = =

top A

bottom A

side A side A

• The field is directed away from the sheet.

0

E to p to p b o tto m b o tto m s id e s id e

in s ide

E A E A E A

Qε

Φ = ⋅ + ⋅ + ⋅

=

•

There is

no

field

through

the

curved

part

of

the

surface.

0side E Φ =

0ins ideQ Aσ =

• The flux through the ends is EA0

( )0

0 0 02 2 A

E Aσ σ

ε ε = =


Charge (Cont )


101/102

Charge (Cont.)• Find the field above and below a

nonconducting infinite plane sheet of charge with uniform negative charge

per unit area σ .

Section 15.9( )

00

0

0

0 0 0

2

2 2

i n s i d e E

i n s i d e

Q E A

Q A

A

E A

ε

σ

σ σ

ε ε

Φ = == = −

− −= =

Q=‐σ A0

• Use a cylindrical Gaussian surface• Since the field is uniform , the flux through the ends is

EA0, there is no field through the curved part of the surface.

• The field is directed toward from the sheet.

Parallel Plate Capacitor


102/102

• The total electric field between

the plates is given by

• The device consists of plates of positive and negative

charge.

Parallel Plate Capacitor

0 0

0 0 0

02 2

2 2

total E E E

outside

inside

σ σ ε ε

σ σ σ

ε ε ε

+ −= +

⎧ − =⎪⎪= ⎨⎪ + =⎪⎩

E +

E +

E +

E −

E −

E −

• The field outside the plates is

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Chapter 15 Serway Physics