Physics 2020 Lectures and Clicker Quizzes

Weeks 3-4

M. GoldmanSpring, 2011

1

2/6/11

Upcoming exam

• Concepts versus facts = thinking versus rememberingo You do need to remember some facts but we are not so

interested in that in this course. (Paper with formulas that you are allowed to bring to exams and formulas given in exam)

o In physics you mainly have to think to apply concepts to solve problems you haven't seen before. If you deeply understand concepts you will have a better chance of solving problems.

o Seeing analogies is an important ingredient in solving new problems.

o Math skills are also important

• Knowledge of facts, understanding of concepts and ability to solve unexpected problems are essential for workers in the health/medical industry 2

Does the force, F on a particle of charge, q, due to an electric field, E point in the same direction as E?

3

F = qE

E

positive charge, q

F+++++

E++++

F

negative charge, q

A positively charged solid metal sphere in equilibrium

A metal in electrostatic equilibrium

E = 0 inside

net charge on surface only

E-field surface

Why does the electric field vector at the surface of a metal point have no component on the surface?

5

The E-field must be perpendicular to the surface (in electrostatic

equilibrium), otherwise the component of the E-field along the

surface would push electrons along the surface causing movement

of charges (and we would not be in equilibrium).

metal

E

Ex

Ey

On the surface of metal, if Ex was

not zero, there would be a force

Fx = q Ex = –e Ex on electrons in the

metal pushing them along the

surface.

How can a uniform electric field be produced?

• The electric field between two parallel metal slabs with equal and opposite charges will be almost uniform

• This is a capacitor!

Put a conducting slab in an electric field, E. Electronsin slab will distribute themselves so that there

is zero net field inside the slab

7

Faraday Cage• A metal-screen cage in an external E-field will also

distribute its electrons to eliminate any net E-field inside the cage

• http://www.youtube.com/watch?v=WqvImbn9GG4

8

Would you try this?

9

Van de Graaff demos of "lightning" and of "tassel spreading"

10

Positively unmanageable hair!

11

Are you safe in your car during lightning?

12

URBAN LEGEND:

The insulating rubber tires on a car protect you from lightning.

REAL PHYSICS:

The metal body of the car keeps the charges on the outside, away from you! It is essentially a Faraday charge

Explanation

13

Remember that Electric Fields inside conductors (metals) are zero when in electrostatic equilibrium.

Inside a hollow metal object (like a car) that has excess charge put on it (from lightning) the charge will be on the outside surface of the object!

However, this is not a case of electrostatic equilibrium. The charges are moving, but by similar principle they stay along the outside conductor and then try to leave via the easiest path (in this case to the ground).

Clicker Question 1, Monday, Jan 24

14

-|Q|

+|Q| -|Q|

+|Q|

P

What is the magnitude of the E-field at point P?

A) |E| = 2k|Q |/ s2

B) |E| = sqrt(2) k|Q| / s2

C) |E| = k|Q| / ( sqrt(2) s)2

D) zero

E) none of the above

s

s

s

s

Now begin subject matter of Chapter 17:Potentials and voltage

15

Beware of high voltage

You have already studied gravitational potential energy

o When an object has positive gravitational potential energy the force of gravity can do work on it.

o Conservation of energy says that as work is done on the body by gravity its potential energy, PE, will decrease and its kinetic energy, KE, will increase but the sum of the two is constant.

o Near the surface of Earth the gravitational force is approximately constant and equal to mg, pointing down.

o The gravitational potential energy of an object a distance h above theground is the work youdo to lift it there PE = F·h = mgh.

16

Review of relationship between constant force, work and P.E.

17

m

Fgravity

m

Fgravity

Fhandm

Fgravity

h

P.E. = (mg)h Conservative force independent of path. No work to move m horizontally. Only verticalcomponent of Fhand has to balance gravity.

Electric forces and fields can be understood in a similar way by defining electric potential energy

• The constant electric force, F, experienced by a positive charge, q, at height h inside a capacitor with downward field E is analogous to the "constant gravitational force"

18

qh E

F

Electricpotentialenergy =F·h =q·|E|·h

Clicker question 2, Jan 24, 2011

• Neglecting gravity, what is the kinetic energy of the particle (with +charge, q) put down at rest when it hits the bottom of the capacitor?A) It cannot be determined without knowing its mass

B) It is Eqh

C) It is mgh

19

Clicker question 3, Jan 24, 2011

• What happens if the electric charge put down at rest at the same position is negative, = -|q|? Once again, you can neglect gravity.A) The charge moves the same way as the positive charge

B) The charge will have negative kinetic energy

C) The potential energy of the charge will decrease as it moves

20

Explanation

21

Electricpotentialenergy =F·h =q·|E|·h

21

-|q|

h E

P.E. is negative. Becomes more negative as chargemoves up from h. More negative means decrease!

E.g. -6V < -4V

22

Electrical P.E. depends on the sign of q. Define a quantity independent of q, called the electric

potential, V

• Electric potential (voltage), V, is related to electric field, E• The potential (voltage) V is not a vector. It is a ± scalar.• The potential energy (P.E.) of a (test) charge, q, at a point

where potential = V is always given by P.E. = qV• For the constant downward electric field, E, inside the

capacitor, as drawn, the electric potential (voltage), V, iso V = P.E./q = (Eqh)/q = Eho V does not depend on the (test) charge, q, which feels

force, F, due to E when placed down as shown.o F can be up or down, depending on sign of q

How to visualize electric potential (voltage)

• "Connect the dots" to visualize voltageo A curve through all points with the same value of V is called an

equipotential curve.o We can draw several of these curves – one for each different

value of V (label them by the value of V)o The equipotential lines will always be perpendicular to the

electric field lines o For the uniform field lines inside the capacitor the equipotential

lines appear as straight lines perpendicular to the field lines

23

Compare four different ways of understanding the electric force on a test charge, q:

Vector or scalar

Units Depends on test charge?

Electric force, F

vector Newtons, N Yes

Electric poten-tial energy,

(P.E.)

± scalar Joules, J,or eV (elec-tron volts)

Yes

Electric field, E=F/q

vector N/C=Volts/m

No

Electric poten-tial (voltage),

V = (P.E.)/q

± scalar J/C = Volts, V No

24

SAT-type analogy question

• Electric force is to electric field as electric P.E. is to _______

A) Charge

B) Field line

C) Voltage

D) Gravitational P.E.

E) None of the above

25

Comments about electrical units

• Why are electric field units N/C the same as V/m?o Answer: Because volt has units V =

work/charge(Nm)/C (also same as J/C)

• What is the energy unit eV (electron volts)?o Answer: K.E. gained by electron in electric field

when it loses P.E. associated with 1 Volt.

26

eV = charge on electron( )×(1 volt) = 1.6 10−19CV =1.6 10−19 J

Use equipotential lines to visualize how a test charge, q, set down at rest will move

• A charged particle, q, set down at rest will always move from higher P.E. to lower P.E. as it gains kinetic energy, K.E. (Doesn't matter if q = + or -)

• If q is positive it will move from higher V to lower V (V = voltage = electric potential)

• If q is negative it will move from lower V to higher V

27

6 volts0 volts

You can infer the strength of the electric field from the equipotential lines

28

6 volts0 volts10-3 m V = Eh

E = V/h = 6V/10-3m

Ey =−ΔVΔy

≡−Vf −Viyf −yi

=− 0−60−10−3 =−6000V /m

More generally,

y-direction

= -6 kV/m

There are some conventions for equipotential lines or surfaces

29

6 volts0 volts

12 volts18 volts

10-3 m

Ey =−ΔVΔy

≡−Vf −Viyf −yi

=− 0−60−10−3 =−6000V /m

The capacitor field, E, is the same betweenany two equipotential lines. It is uniform in this case.

y-direction

10-3 m10-3 m

The voltage difference between any two drawn equipotentiallines is the same unless they are labeled otherwise

equally spaced

Some set of charges produce the Voltages (electric potentials) shown below. How much work is required to move the particle from a to b?

A)2 CoulombsB)20 VoltsC)40 JoulesC)10 JoulesE) None of the Abovea

Clicker question 1, Wed. Jan 26, 2011

30

10V 20V 30V 40V 50V

bQ = 2 C

E =−ΔVs

=−Vlarger circle−Vsmaller circlerlarger circle−rsmaller circle

What do equipotential curves look like when E has a different magnitude/direction at each point in space (is

spatially nonuniform). Consider point (+)charge, Q:

31

Red dashed circles are equipotential curves

They are still perpendicular to the electric field lines

The voltage difference is the same between any two equipotential circles

5 V

10 V

15 V

The spatial separation, s = rlarger - rsmaller between circles shrinks as V increases

ss The electric field is givenmore and more accurately as s decreases by

Equipotential curves are like topo maps used for hiking in the mountains

32

Relief (3D) map vs topo (2D) map

33

The equipotential lines are like constant altitude lines on a topo map (but voltage is the altitude)

34

Consider the equipotential lines of a dipole

This is a topo map of the following potential terrain

(+)charges create ‘mountains’, (-)charges create ‘valleys.’Potentials from each charge add as ±numbers to get net V.Test (+)charges go ‘downhill’, test ( -)charges go ‘uphill’.Crowded lines indicate higher E-field.

Clicker question 1, Fri., Jan 28

35

The equipotential surfaces around a line of charge (viewed end-on) are shown. Each equipotential line is 2 meters from the nearest-neighbor equipotential. What is the approximate magnitude of the electric field at point A?

A

-1.0V-1.4V-1.8V

-2.2V

A)0.1 Volts/mB) 0.2 Volts/mC) 0.4 Volts/mD) 0.7 Volts/mE) None of the above

P

Clicker Question 2, Fri, Jan 28, 2011

The equipotential lines are labeled in volts. Spatial distances can be determined from the x and y axes. What is the approximate magnitude and direction of the electric field, E, at point P?

A) 2.7 V/m, in the x-direction

B) 3.2 V/m, in the y-direction

C) 1 V/m, in the negative y-direction

D) .5 V/m, in the y-direction

E) None of the above

36Correct answer is: .25 V/m up (negative y direction

P

Further questions to nail down your understanding of the meaning of potential

37

Additional questions: •How can you visualize the electric field lines and vectors? •What are the signs of the two point charges which create the electric field. •Which charge has the greater magnitude? •What is the P.E. of a charge q = 3 C put down at P?•What is the force on it? •What is the P.E. of a charge of q = -3 C put down at P?•What is the force on it?

Clicker Question 3, Friday, Jan 28, 2011

38

Voltage is like altitude, and E-field is like slope

If you know the voltage at a single point only, can you determine the electric field there?

(a)No(b)Yes(c)Magnitude, but not direction(d)Direction, but not magnitude

P.E. = Work = - F(x)dx∞

r

∫ =− kQqx2

⎡⎣⎢

⎤⎦⎥dx

∞

r

∫ =kQqr

Voltage = V = P.E./q = kQr

What is the formula for voltage, V (electric potential) surrounding a point (+)charge, Q?

39

First we find the potential energy, P.E. of another positivecharge, q, put down at the tip of the position vector, r.

Q q

P.E. = Work to bring charge q from infinity, where P.E. is defined = 0 to position r. Path doesn't matter for conservative force.

r

Work NOT EQUAL to force times distance because this force depends on position.

x

y

But circle of radius, r, is equipotential (voltage is same everywhere on circle)

40

Q r

Voltage caused by point charge, Q, anywhere on circle of

radius r is V = kQ/r

41- 1.0 - 0.5 0.0 0.5 1.0

2

4

6

8

- 1.0 - 0.5 0.0 0.5 1.0

- 1.0

- 0.5

0.0

0.5

1.0x

y

y

x

V

y

r = √(x2+y2)

V

If charge Q = -|Q| is negative the formula for voltage is still correct but values of V are negative.

42

V = -k|Q|/r

42

- 1.0 - 0.5 0.0 0.5 1.0

- 1.0

- 0.5

0.0

0.5

1.0

x

y

x

V

y

r = √(x2+y2)

x

V

43

Equipotential surfaces areconcentric spheres ratherthan circles

xy

z

r = √(x2+y2+z2)

Can the electric field from a source charge or charges be "blocked?"

• The field goes right through any charges put in its way• But you must add the field of the charges in its way to get the

net field on the other side of the charges in its way.

44

(+) chargeE of (+)charge here (-) charge

E of (-)charge

Net E

What can you learn about the electric field if you know the equipotentials (equal voltage curves)

• Step 1: visualize the mountains and valleys of voltage; imagine you are standing on skis at some point on voltage terrain

• Step 2: electric field, E, is like downhill slope of mountain or valley at that point. Steeper slope means stronger E.

• Step 4: Your ski trail along the steepest slope (no turns) becomes an electric field line when projected onto the contour map

45Contour map Relief map

Clicker Question 1, Monday, Jan 31, 2011

46

A set of equipotential lines are shown below. Each equipotential line is 2 meters from the nearest-neighbor equipotential. What is the approximate electric field at point A?

A10V

8V6V4V

A)1.0 V/m in –x directionB) 1.0 V/m in +y directionC) 2.0 V/m in +x directionD) 2.0 V/m in –y directionE) None of the above

x

y

1 V/m in +x direction

Clicker Question 2, Monday, Jan 31, 2011

47

An electron of mass, m, is fired away (to the left) from a large positive charged point starting at a distance r. Its initial velocity (to the left) is v

+Q-e

The electron will escape (go out to -infinity) if:

A)

B)

mr

kQev

2||

mr

kQev ||

C)

D) It can never escape.

r

kQv ||

r

48

A = 84-96A- = 80B+ = 76B = 72B- = 68C+ = 64C = 56-60C- = 52D = 40-48F = 24-36

Roughguide togrades

Questions 20-25

49

18

18

19.519.5

21 21

26

26

32

32

38

38

- 2 - 1 0 1 2- 1.5

- 1.0

- 0.5

0.0

0.5

1.0

1.5

Clicker question 3

• An electron is set down with an initial velocity, v, near a proton What determines the trajectory (path the electron follows at later times)A) The initial velocity, v together with the distance from the

proton

B) The initial velocity alone

C) The force due to the proton alone

D) The distance from the proton alone

E) It cannot be determined

50

Explanation

• Set down an electron at a position relative to a proton with different initial velocity vectors, v.

• Position determines the force and its directiono Does a particle always move in the direction of the force

on it?o No, initial velocity also determines its path

• http://www.Colorado.edu/physics/2000• YouTube

51

• Add voltage created by charge A to voltage created by charge B at any point to get net voltage at that point. Scalars, not vectors

• Remember that voltages can be either positive or negative.• Find net voltage at point P below

What is the (net) voltage created by two charges?

52

++++++

+++

QA = - 2C QB = +2Cd

PdA dB

VA =kQA

dA=−k 2

dA, VB =k

QB

dB=+k 2

dB

Net VP =VA +VB =2k −1dA

+ 1dB

⎛

⎝⎜

⎞

⎠⎟

Potential of a dipole at large distances, dB >> d.Definition of dipole moment.

53

++++++

+++

QA = - |Q| QB = +|Q|d

PdA = dB + a

a ≈ d·cosθ

dB

dB

aθ

VA =kQA

dA=−k|Q |

dA, VB =k

QB

dB=k|Q |

dA

Net VP =VA +VB =k|Q | − 1dA

+ 1dB

⎛

⎝⎜

⎞

⎠⎟=

k|Q | − 1dB +dcosθ

+ 1dB

⎛

⎝⎜

⎞

⎠⎟≈k|Q | dcosθ

dB2

⎛

⎝⎜

⎞

⎠⎟=k

pcosθr2

• Consider three (+)charges of equal magnitude below.• P.E. of system of three charges = sum of work done

to bring each of them in from infinity to position showno No work to bring the first charge to its present position

o Work on second charge (say, Q2) brought to its position was calculated earlier

o Q3 can be brought in to its positionfrom infinity along any convenient path Can break up work into sum of

works against Q1 and Q2 separatelyand then add.

Potential energy of an assembly of charges

541 2

Q3

Gauss's Law

• Gauss law is a mathematical expression of Coulomb's lawo If you had calculus it is an integral form of Coulomb's lawo There is also a differential eqn form of Coulomb's law called Poisson's equation

• It often enables you to find electric field, E, when it is not due to a few point charges but due to a distribution of charges (fluid).

• Non-calculus statement of Gauss's law: o Draw an imaginary closed surface around a distribution of source charges. o Field lines from the source charges will pierce the surface.

o If E⊥ is the perpendicular component of E on a small piece of surface with area ∆A, then the total charge, Qencl inside the closed surface is given by

• Calculus statement of Gauss's law:

55

E⊥ΔA=Qenclosed

ε0closed surface∑ , k= 1

4πε0

Examples of using Gauss's law to find electric fields from distributed charges

• In the simplest examples, E is everywhere perpendicular to the surface and of constant magnitude

• Example 1: Charged spherical conductor• Example 2: Close to a wide charged flat conducting slab• Example 3: Inside a capacitor

56

+ + + + + + + + + + + +

_ _ _ _ _ _ _ _ _ _ _ _

E· 4πr2⎡⎣ ⎤⎦=Qon sphere

ε0,

E =kQon sphere

r2

rQon sphere

E = E⊥

+ + + + + + + + + + + + +

∆A

E = E⊥

∆Q

E =Q / A( )ε0

=σε0

Capacitor: Any two pieces of metal (conductor) brought near each other.

57

Parallel plate capacitor:L

W

d

Area A = L x W

We charge the capacitor by (somehow) putting charge onto the plates (i.e. excess electrons or deficit of electrons).

“Charged Capacitor” has no net charge.

58

+++++++++++++++++++++++

-----------------------------------------

+Q

-Q

E=0

E=0

ECharges are on the inside surfaces of the conductors due to attraction.

Electric Field (treating plates as infinite) and having area A.

d

A

QEE

00

1

εεσ

E = 0

E = 0

Q and σ are assumed to be positive

Voltage and charge-carrying capacity

59

EdrEVV ΔΔ +++++++++++++++++++++++

-----------------------------------------

+Q

-Q

Ed

A

QE

00

1

εεσ

A

QdV

0ε

C ≡QV

For parallel plate capacitor d

AC 0ε

Q and V are assumed to be positive

Clicker Question

60

An electric field of 500 V/m is desired between two parallel plates 10.0 mm apart. How large a Voltage should be applied?

A) 500 VoltsB) 0.5 VoltsC) 2.0 VoltsD) 5.0 VoltsE) None of the above answers

xx

VE ˆ

rEV

ΔΔ

Clicker Question

61

LL

d

A parallel-plate capacitor has square plates of edge length L, separated by a distance d. If we doubled the dimension L and halve the dimension d, by what factor have we changed the capacitance?

A) C is unchangedB) 0.5xC) 2xD) 4xE) 8x

22 2oo o o

2LA L LC 8

d d d / 2 d

εε ε ε