1 / 61

# Physics 2020 Lectures and Clicker Quizzes Weeks 3-4 - PowerPoint PPT Presentation

2/6/11. Physics 2020 Lectures and Clicker Quizzes Weeks 3-4. M. Goldman Spring, 2011. Upcoming exam. Concepts versus facts = thinking versus remembering

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' Physics 2020 Lectures and Clicker Quizzes Weeks 3-4' - arella

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Physics 2020 Lectures and Clicker Quizzes Weeks 3-4

M. Goldman

Spring, 2011

• Concepts versus facts = thinking versus remembering

• 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)

• 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.

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

• 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

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

-

positive charge, q

++

F = qE

F

+

++

E

E

F

negative charge, q

A positively charged solid metal sphere in equilibrium

net charge on surface only

E-field  surface

E = 0 inside

A metal in electrostatic equilibrium

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

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).

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.

E

Ey

Ex

metal

• 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 thereis zero net field inside the slab

• A metal-screen cage in an external E-field will also distribute its electrons to eliminate any net E-field inside the cage

Positively unmanageable hair!

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

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).

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

-|Q|

+|Q|

s

s

s

P

s

-|Q|

+|Q|

Now begin subject matter of Chapter 17:Potentials and voltage

Beware of high voltage

You have already studied gravitational potential energy

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

• 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.

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

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

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

m

Fhand

h

m

m

Choose zeroof P.E. here

Fgravity

Fgravity

Fgravity

P.E. = minimum work done by hand (against gravity) to bring m up a vertical distance, h = Fhandh. Minimum |Fhand| = |Fgravity|.

P.E. = (mg)hConservative 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"

Electricpotentialenergy =F·h =q·|E|·h

q

h

E

F

• 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

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

• The charge moves the same way as the positive charge

• The charge will have negative kinetic energy

• The potential energy of the charge will decrease as it moves

-

Electricpotentialenergy =F·h =q·|E|·h

-|q|

E

h

21

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

E.g. -6V < -4V

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, is

• V = P.E./q = (Eqh)/q = Eh

• V does not depend on the (test) charge, q, which feels force, F, due to E when placed down as shown.

• F can be up or down, depending on sign of q

• "Connect the dots" to visualize voltage

• A curve through all points with the same value of V is called an equipotential curve.

• We can draw several of these curves – one for each different value of V (label them by the value of V)

• The equipotential lines will always be perpendicular to the electric field lines

• For the uniform field lines inside the capacitor the equipotential lines appear as straight lines perpendicular to the field lines

SAT-type analogy question force on a test charge, q:

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

• Charge

• Field line

• Voltage

• Gravitational P.E.

• None of the above

• Why are electric field units N/C the same as V/m?

• Answer: Because volt has units V = work/charge(Nm)/C (also same as J/C)

• What is the energy unit eV (electron volts)?

• Answer: K.E. gained by electron in electric field when it loses P.E. associated with 1 Volt.

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

6 volts

0 volts

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

y-direction

10-3 m

V = Eh

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

More generally,

6 volts

= -6 kV/m

0 volts

There are some equipotential linesconventions for equipotential lines or surfaces

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

y-direction

10-3 m

10-3 m

10-3 m

equally spaced

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

18 volts

12 volts

6 volts

0 volts

Clicker question 1, Wed. Jan 26, 2011 equipotential lines

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

• from a to b?

• 2 Coulombs

• 20 Volts

• 40 Joules

• 10 Joules

• E) None of the Above

b

Q = 2 C

a

10V

20V

30V

40V

50V

What do equipotential curves look like when equipotential linesE has a different magnitude/direction at each point in space (is spatially nonuniform). Consider point (+)charge, Q:

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

Thespatial separation, s = rlarger - rsmallerbetween circlesshrinks as V increases

5 V

s

s

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

10 V

15 V

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

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 a topo map (but

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?

• 0.1 Volts/m

• B) 0.2 Volts/m

• C) 0.4 Volts/m

• D) 0.7 Volts/m

• E) None of the above

A

-1.0V

-2.2V

-1.4V

-1.8V

Clicker Question 2, Fri, Jan 28, 2011 a topo map (but

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?

• 2.7 V/m, in the x-direction

• 3.2 V/m, in the y-direction

• 1 V/m, in the negative y-direction

• .5 V/m, in the y-direction

• None of the above

P

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

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

• 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?

P

Clicker Question 3, Friday, Jan 28, 2011 meaning of potential

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

• No

• Yes

• Magnitude, but not direction

• Direction, but not magnitude

Voltage is like altitude, and

E-field is like slope

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

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

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.

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

y

q

Q

r

x

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

Voltage caused by point charge, Q, anywhere on circle of radius r is V = kQ/r

Q

r

Visualizations meaning of potentialof voltage V = kQ/rsurroun-ding (+) point charge, Q > 0 at different x and y

r = √(x2+y2)

y

V

y

x

V

y

x

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

V = -k|Q|/r

r = √(x2+y2)

V

y

x

x

y

V

42

x

Visualization meaning of potentialof voltage V = kQ/rsurroun-ding (+) point charge, Q > 0 at different x, y, and z

r = √(x2+y2+z2)

Equipotential surfaces areconcentric spheres ratherthan circles

z

y

x

• 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.

E of (+)charge here

(-) charge

Net E

(+) charge

E of (-)charge

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

Contour map

Relief map

Clicker Question 1, Monday, Jan 31, 2011 equipotentials (equal voltage curves)

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?

• 1.0 V/m in –x direction

• B) 1.0 V/m in +y direction

• C) 2.0 V/m in +x direction

• D) 2.0 V/m in –y direction

• E) None of the above

10V

A

y

8V

6V

x

4V

1 V/m in +x direction

Clicker Question 2, Monday, Jan 31, 2011 equipotentials (equal voltage curves)

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

r

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

A)

B)

C)

D) It can never escape.

Rough equipotentials (equal voltage curves)guide togrades

A = 84-96

A- = 80

B+ = 76

B = 72

B- = 68

C+ = 64

C = 56-60

C- = 52

D = 40-48

F = 24-36

Questions 20-25 equipotentials (equal voltage curves)

Clicker question 3 equipotentials (equal voltage curves)

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

• The initial velocity, v together with the distance from the proton

• The initial velocity alone

• The force due to the proton alone

• The distance from the proton alone

• It cannot be determined

Explanation equipotentials (equal voltage curves)

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

• Position determines the force and its direction

• Does a particle always move in the direction of the force on it?

• No, initial velocity also determines its path

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

• 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

-

++

+

++

P

dA

dB

d

QB = +2C

QA = - 2C

Potential of a dipole at large distances, d equipotentials (equal voltage curves)B >> d.Definition of dipole moment.

P

-

++

dA = dB + a

a ≈ d·cosθ

+

++

dB

r

dB

p = |Q|d = dipole momentr ≈ dB and θ are "polar coordinates"

a

θ

θ

d

QB = +|Q|

QA = - |Q|

Potential energy of an assembly of charges equipotentials (equal voltage curves)

• 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 shown

• No work to bring the first charge to its present position

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

• Q3 can be brought in to its positionfrom infinity along any convenient path

• Can break up work into sum ofworks against Q1 and Q2 separatelyand then add.

Q3

1

2

Gauss's Law equipotentials (equal voltage curves)

• Gauss law is a mathematical expression of Coulomb's law

• If you had calculus it is an integral form of Coulomb's law

• 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:

• Draw an imaginary closed surface around a distribution of source charges.

• Field lines from the source charges will pierce the surface.

• 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:

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

∆A

E = E⊥

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

σ = ∆Q/∆A =charge per unit area

∆Q

E = E⊥

r

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

Qon sphere

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

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

Parallel plate capacitor:

L

Area A = L x W

W

d

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

“Charged Capacitor” has no distributed chargesnet charge.

E=0

E = 0

+Q

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

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

d

E

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

-Q

E = 0

E=0

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

Q and s are assumed to be positive

Voltage distributed chargesand charge-carrying capacity

+Q

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

d

E

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

-Q

Define property of capacitorcalled Capacitance:

Q and V are assumed to be positive

For parallel plate capacitor

Clicker Question distributed charges

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 Volts

B) 0.5 Volts

C) 2.0 Volts

D) 5.0 Volts

E) None of the above answers

Clicker Question distributed charges

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?

L

L

• C is unchanged

• 0.5x

• 2x

• 4x

• 8x

d