R3-1

Physics IReview 3

Review NotesExam 3

R3-2

Newton’s Law ofUniversal Gravitation

rrmm

GF 221

The meaning of each term:

F: Gravitational force on object 1 from object 2.G: Universal gravitational constant = 6.673 x 10–11 N m2/kg2.

1m: Mass of object 1.

2m:Mass of object 2.2r: Center distance from object 1 to object 2, squared.r: Unit vector from object 1 to object 2.

R3-3

If Gravity Varies As 1/r2,Where Does g = 9.8 m/s2 Fit In?

C o n s i d e r t h e f o r c e o n a n o b j e c t n e a r t h e s u r f a c e o f t h e e a r t h .( A s s u m e t h e e a r t h i s a s p h e r e a n d i g n o r e r o t a t i o n e f f e c t s . )R = r a d i u s o f t h e e a r t h .M = m a s s o f t h e e a r t h .m = m a s s o f t h e o b j e c t .

gmrR

MGmr

RMm

GF 22

( W h a t i s t h e d i r e c t i o n ? )

g = 9 . 8 m / s 2 o n l y s e e m s c o n s t a n t b e c a u s e w e d o n ’ t g o v e r y f a rf r o m t h e s u r f a c e o f t h e e a r t h .

R3-4

Class #17Take-Away Concepts

1 . F o u r f u n d a m e n t a l f o r c e s k n o w n t o p h y s i c s : G r a v i t a t i o n a l F o r c e E l e c t r o m a g n e t i c F o r c e W e a k N u c l e a r F o r c e S t r o n g N u c l e a r F o r c e

2 . N e w t o n ’ s L a w o f U n i v e r s a l G r a v i t a t i o n

rr

mmGF 2

21

3 . G r a v i t a t i o n a l P o t e n t i a l E n e r g y ( l o n g - r a n g e f o r m )

rMmG

)r(U g

R3-5

Coulomb’s Law ofElectrostatic Force

)r(rqq

41

F 221

0

(Prof. B’s version.)

The meaning of each term:

F: Electrostatic force on charge 1 from charge 2.

041:Electrostatic force constant = 8.98755 × 10+9 N m2/C2

1q: Value of charge 1, positive or negative.

2q: Value of charge 2, positive or negative.2r: Center distance from point charge 1 to point charge 2.r: Unit vector from charge 1 to charge 2.

R3-6

Direction of Electrostatic Force“Opposites Attract”

R3-7

Superposition ofElectrostatic Forces

)r(r

qq4

1F

N

2ii2

i

i1

01on

(find and add X and Y components)

resultant

+1.0 C+5.0 C

resultant -3.0 C

+5.0 C

X

Y

1

R3-8

How to Calculate aUnit Direction Vector

X8

Y

0

6

0

F r o m b l u e ( x 0 , y 0 ) t o r e d ( x , y ) :1 . F i n d t h e d i s p l a c e m e n t i n X , Y c o m p o n e n t s .

jˆ6i8jˆ)yy(i)xx(d 00

2 . F i n d t h e l e n g t h o f t h i s v e c t o r .

1068)yy()xx(|d|r 2220

20

3 . D i v i d e b y t h e l e n g t h t o g e t a u n i t v e c t o r .

jˆ6.0i8.010jˆ6i8rdr

R3-9

Class #18Take-Away Concepts

1 . C h a r g e s c o m e i n p o s i t i v e a n d n e g a t i v e .

2 . O p p o s i t e s c h a r g e s a t t r a c t , l i k e c h a r g e s r e p e l .

3 . C o u l o m b ’ s L a w o f E l e c t r o s t a t i c F o r c e

)r(r

qq4

1F 2

21

0

4 . S i m i l a r i t i e s a n d d i f f e r e n c e s w i t h g r a v i t y .

5 . T h e p r i n c i p l e o f s u p e r p o s i t i o n .

)r(r

qq4

1F

N

2ii2

i

i1

01on

R3-10

The Idea of Electric Field

T h e e l e c t r i c f i e l d i s a v e c t o r f i e l d t h a t d e p e n d s o n t h e l o c a t i o n o f t h e p o i n t w h e r e w e c a l c u l a t e i t . d e p e n d s o n t h e u n i t v e c t o r s t o t h e o t h e r c h a r g e s . d e p e n d s o n t h e d i s t a n c e s t o t h e o t h e r c h a r g e s . d e p e n d s o n t h e v a l u e s o f t h e o t h e r c h a r g e s .

I t d o e s n o t d e p e n d o n t h e v a l u e o f a n y c h a r g e a t t h a t p o i n t .I n f a c t , i t c a n b e c a l c u l a t e d e v e n w h e n t h e r e i s n o c h a r g e t h e r e !

)r(r

q4

1)1#point(E

N

2ii2

i

i

0

)1#point(EqF 11on

Electric Field

Force / Field Relationship

R3-11

-+

The Electric Field of a Point Charge (as a Source)

The electric field is a vector field, meaning at each point inspace the electric field has a magnitude and a direction. Weshow that by drawing arrows at representative points in thecorrect directions with lengths proportional to the magnitudes.

Away from positiveToward negative

Just because we don’t draw an electric field vector ata point doesn’t mean there is no electric field there.

R3-12

Calculating the Electric Field of a Charge at a Point (Example)

X8 cm

Y

0

6 cm

0

)r(rq

41

E 20

F i r s t c a l c u l a t e t h e u n i t v e c t o r . T h i s v e c t o r i s f r o m t h e p o i n t ( X )t o t h e c h a r g e . U s e t h e m e t h o d o n s i d e R 3 - 8 :

jˆ)06.0(i)08.0(d

1.0)06.0()08.0(r 22 jˆ)6.0(i)8.0(1.0]jˆ)06.0(i)08.0[(r

C100.1q 9

R3-13

Calculating the Electric Field of a Charge at a Point (Example)

X8 cm

Y

0

6 cm

0

)r(rq

41

E 20

N o w c a l c u l a t e t h e n u m b e r i n ( ) :

900)1.0()100.1)(100.9(r

q4

1 2992

0

F i n a l a n s w e r :

jˆ)540(i)720(]}jˆ)6.0(i)8.0[(){900(E

N / C

C100.1q 9

R3-14

Class #19Take-Away Concepts

1 . E lec tric fie ld fro m po in t charge sources:(T o ta l fie ld is the superposition o f po in t source fie ld s.)

)r(r

q4

1E i2

i

i

0

2 . F orce on a charge in an e lec tric fie ld :

EqF

3 . E lec tric fie ld po in ts aw ay from + source charges.4 . E lec tric fie ld po in ts to w ard – source charges.

R3-15

Electric Potential

J u s t a s w e n o t i c e d w h e n w e d e f i n e d t h e e l e c t r i c f i e l d , w e s e e t h a tt h e r e i s a c e r t a i n q u a n t i t y t h a t i s i n d e p e n d e n t o f t h e c h a r g ee x p e r i e n c i n g t h e e l e c t r o s t a t i c f o r c e . T h i s i s c a l l e d t h e e l e c t r i cp o t e n t i a l a n d i t i s n o r m a l l y r e p r e s e n t e d b y t h e s y m b o l V :

A

0

xdE)A(V

T h e r e l a t i o n s h i p b e t w e e n U a n d V i s

VqU V e x i s t s e v e r y w h e r e t h a t a n e l e c t r o s t a t i c f i e l d e x i s t s . U e x i s t s o n l yw h e r e t h e r e a r e c h a r g e s i n t h a t f i e l d .

R3-16

Calculating Electric Potentialfrom Point Source Charges

1 m

1 m

C1C2

3

669

i

i

0i

i

0

107.3

2102

1101

109

rq

41

rq

41

V

R3-17

Equipotential Linesand Field Lines

Equipotential Lines – Lines connecting points with equal electricpotential. (In 3D, these are equipotential surfaces.)

Electric Field Lines – Lines formed by taking tiny electric fieldvector arrows and connecting them “tail to head.” Electric field lines are always perpendicular to equipotential lines. Electric field strength is higher where electric field lines are closer

together. Electric field strength is higher where equipotential lines are closer

together (assuming equal increments of potential per line). Electric field lines point in the direction of decreasing potential.

R3-18

Equipotential Linesand Field Lines

+

equipotential

field line

R3-19

Calculating Electric Fieldfrom Electric Potential

X

Y

V = 50X = 2

V = 0X = 4

V = 100X = 0

m/V254

100xV

xdVd

Ex

0ydVd

Ey

R3-20

Calculating Change in K.E.from Electric Potential

V = 50

V = 0

V = 100

initial

final0UK o r UK

V)e(VqU o r V)e(U

J106.1

)100()106.1()V)(e(K17

19

e

e

R3-21

Class #20Take-Away Concepts

Physicist General’s Health Warning:Confusing electric potential (V) with electric potential energy(U) will be hazardous to your score on the next exam.

1. Electric potential (V) and electric potential energy (U):VqU

2. Electric potential from point source charges:

i

i

0 rq

41

V

3. Relationship of electric field and electric potential.4. Electric field lines and equipotential lines.5. Calculating change in kinetic energy using electric potential.

R3-22

How to Solve Problems Asking“Where is Net Electric Field = 0?”

Many things to consider: The number of regions (N+1) is one more than number of charges (N). The total electric field is the sum of the individual fields from each charge.

(Principle of Superposition) In each region, the direction of electric field from each charge is constant in that region.

(Away from +, toward –.) The magnitude of electric field depends on the value of charge and inverse distance squared

to the charge. In any region, if you get close enough to one of the charges at the end points, the inverse

distance squared dependence will make the field from that charge much larger than thefields from all of the other charges.

If you get far enough away from all the charges in Region I or Region N+1, the inversedistance term is about the same for all points, so the relative sizes of the electric fields fromeach charge will be determined by the sizes of the charges.

In a region where the total electric field is zero at some location, it will point in one directionat one end of the region and the other direction somewhere else in the region – the middle orthe opposite end.

+XRegion IIIRegion I Region II

- charge+ charge

R3-23

How to Solve Problems Asking“Where is Net Electric Field = 0?”

+XRegion IIIRegion I Region II

-4 C+7 CRegion I:E field from +7C direction is to the left.E field from –4C direction is to the right.Every point is closer to +7C than to –4C, so net field is always to the left.No points where E = 0.

Region II:E field from +7C direction is to the right.E field from –4C direction is to the right.Net field is always to the right.No points where E = 0.

Region III:E field from +7C direction is to the right.E field from –4C direction is to the left.Close to –4C, net E field is to the left.Far from –4C, net E field is to the right because 7 > 4 and far enough away the inversedistance squared will be about the same for both.There is one point where E = 0.

R3-24

How to Solve Problems Asking“Where is Net Electric Field = 0?”

+XRegion IIIRegion I

x-d-4 C+7 C

Region II

x

d

d097.4d27

7x

x2)dx(7

dx2

x7

)dx(4

x7

0)dx(

44

1x7

41

22

20

20

R3-25

How to Solve Problems Asking“Where is Electric Potential = 0?”

+XRegion IIIRegion I Region II

- charge+ chargeThis is similar to finding zero electric field, but V is a scalar, not vector! The total electric potential is the sum of the individual potentials from each charge.

(Principle of Superposition) The sign of an individual contribution is always the same as the sign of charge. The magnitude of potential depends on the value of charge and inverse distance to the charge –

not squared in this case. If you get close enough to one of the charges, the inverse distance dependence will make the

potential from that charge much larger (in magnitude) than the potentials from all of the othercharges.

If you get far enough away from all the charges, the inverse distance term is about the same forall, so the relative sizes of the electric potentials from each charge will be determined by therelative sizes of the charges.

On any line or curve segment, not passing through a charge, where the electric potential isnegative at one end and positive at the other, there will be at least one point on that segmentwhere the electric potential is zero.

The electric potential can be zero at a point not at infinity only if there are at least two charges inthe problem, one positive and one negative. In that case, there will be infinitely many points notat infinity where the electric potential is zero.

R3-26

How to Solve Problems Asking“Where is Electric Potential = 0?”

+XRegion IIIRegion I Region II

-4 C+7 CRegion I:Every point is closer to +7C than to –4C, so net V isalways positive.No points where V = 0.

Region II:On the line from +7C to –4C, V switches sign.There is a point where V = 0.

Region III:Close to –4C, net V is negative.Far from –4C, net V is positive.There is a point where V = 0.

R3-27

How to Solve Problems Asking“Where is Electric Potential = 0?”

d333.2d47

7x

x4)dx(7

)dx(

4

x

7

0)dx(

4

4

1

x

7

4

1

IIIRegion

00

+XRegion IIIRegion I Region II

-4 C+7 C

d6364.0d47

7x

x4)xd(7

)xd(

4

x

7

0)xd(

4

4

1

x

7

4

1

IIRegion

00

R3-28

Class #21Take-Away Concepts

To find points where E = 0, consider the following:

1. Consider N+1 regions for N charges on a line.2. Superposition Principle: Add E from each charge.3. E points away from + charge, toward – charge.4. E depends on charge value and inverse distance squared.5. Very close to a charge, E from that charge dominates.6. Far away right or left, all distances are about the same.7. Look for reversal of total E direction in a region.

Finding where V = 0 is similar but easier, since V is a scalar.Look for regions where V changes sign.

R3-29

Magnetic Field Lines

From N to S.

In direction ofcompassneedle.

Try this web site:http://home.a-city.de/walter.fendt/phe/mfbar.htm

R3-30

Class #22Take-Away Concepts

1. The magnetic force was known in ancient times.

2. Magnets have two poles, N and S.

3. Opposite poles attract, like poles repel.

4. N and S poles are always in pairs, never alone.

5. Electric currents also cause magnetic fields.

6. Magnetic field lines start at N and go to S,following the direction of a compass needle.

7. Units of tesla (T) and gauss. 1 T = 10,000 gauss.

R3-31

Magnetic Force on aMoving Charge

BvqF

q: charge of the particle (C; + or –)v:velocity of the particle (m/s)B:magnetic field (T) Force is at a right angle to velocity. Force is at a right angle to magnetic field.

Important: If q is negative, that reverses the direction of force.

R3-32

Analysis of the Magnetic Force

X

Z

Y

F

BvqF

We will evaluate this expression beforethe electron starts turning.

First, evaluate Bv. In this case, they are 90° apart, so all we

need is the direction. v is +X, B

is –Z, so Bv

is +Y.

Next, we need to account for q. This is an electron, so q isnegative. Therefore, the magnitude of the force is (e v B) and thedirection is –Y.

R3-33

The Radius of the Circle

v

r

F

A l t h o u g h t h e d i r e c t i o n s o f t h e v e c t o r s a r ec h a n g i n g , t h e m a g n i t u d e s s t a y t h e s a m e .

rv

mamF2

BvqF

rv

mBvq2

Bqvm

Bvqv

mr2

R3-34

Class #23Take-Away Concepts

1. The Lorentz (magnetic) force on a moving, charged particle:

BvqF

2. The Lorentz force cannot change a particle’s speed, only thedirection of its velocity.

3. Radius and angular frequency of a charged particle in uniformcircular motion in a magnetic field:

Bqvm

r

mBq

R3-35

Apparatus for Measuring e/m

We will set Potential in tube (V). Current in coils (I).

We will observe Radius of circular

electron path (r).

We will calculate e/m.

R3-36

Analysis of Electron Acceleration

Electron Stream

pot = -V pot = 0

cathode anode UK V)e(U

Ve)]V(0[e

]V)[e(K

E l e c t r o n s h a v e v e r y l o w k i n e t i ce n e r g y w h e n t h e y l e a v e t h ec a t h o d e – e s s e n t i a l l y z e r o .

Vevm 22

1 o r

mVe2

v

R3-37

Derivation of e/m Formula

Be

vmr

Br

v

m

e

m

Ve2v

Br

V2

m

e

Br

V2m

e

Brm

Ve2

m

e

Br

V2

m

e o r22 Br

V2

m

e We set or observe all of the variableson the right side of the equation.

R3-38

Mass Spectrometer Diagram

Adjusting the acceleratingpotential (V) and/or themagnetic field (B) “tunes”the spectrometer to detectonly ions with a specificratio of charge to mass.

R3-39

Class #24Take-Away Concepts

1. The ratio of e/m is determined by accelerating a streamof electrons through an electric potential difference andobserving their path in a magnetic field.

2. The formula for e/m (know how to derive it):

22 Br

V2

m

e

3. Mass spectrometers work by the same principle.

R3-40

Relationship of Energy and Electric Potential Changes

Memorize these tables or know how to derive them!K = –UU = q V or U = q V

Potential Energy Change

V > 0 (+) V < 0 (–)+ charge + (increase) – (decrease)– charge – (decrease) + (increase)

Kinetic Energy Change

V > 0 (+) V < 0 (–)+ charge – (decrease) + (increase)– charge + (increase) – (decrease)