physics 212 lecture 29, slide 1 physics 212 lecture 29 course review the topics for today –...

Physics 212 Lecture 29, Slide Physics 212 Lecture 29, Slide 11

Physics 212Physics 212Lecture 29Lecture 29

Course ReviewCourse Review• The Topics For Today

– Electric Fields/Gauss’ Law/Potential– Faraday’s Law– RC/RL Circuits– AC Circuits

Physics 212 Lecture 29Physics 212 Lecture 29

MusicWho is the Artist?Who is the Artist?

A)A) Ray CharlesRay CharlesB)B) Solomon BurkeSolomon BurkeC)C) Henry ButlerHenry ButlerD)D) Johnny AdamsJohnny AdamsE)E) Otis ReddingOtis Redding

“Rediscovered” Soul singer from 60’sAbsolutely Beautiful Voice

Highly Recommended DVD:Highly Recommended DVD:““One Night History of the Blues”One Night History of the Blues”

Why?Why?

Great singer to end the courseGreat singer to end the course

Electric Field


Horizontal components cancelHorizontal components cancel E from top arc points downE from top arc points downE from bottom arc points upE from bottom arc points up

Top arc produces smaller horizontal componentsTop arc produces smaller horizontal components EEtotaltotal points down points down

sinsin

top

toptop

rQE

sin

4 20

Calculation:Calculation:

cos4 2

0rdqEtop

toprQ

2

top

rdrQ

rE

toptop

02

0cos2

241

Two curved rods each have charge +Q uniformly distributed over their lengthWhich statement best describes the electric field due to these two rods at the midpointbetween the two rods (marked by an X)A. E points up B. E points down C. E = 0 48%

Conductors, Potential Energy


Potential Energy is a measure of work done by E fieldPotential Energy is a measure of work done by E field

Spherical symmetry & Gauss’ law Spherical symmetry & Gauss’ law E = 0 inside shellE = 0 inside shell

E = 0 inside shell E = 0 inside shell No work done to move qNo work done to move q

No change in potential energy !No change in potential energy !

A thin non-conducting spherical shell carries a positive charge Q spread uniformlyover the surface of the shell. A positive test charge q is located inside the shell.

Compare UA, the potential energy of the test charge in position A (near the shell)with UB, the potential energy of the test charge in position B (center of the shell)A. UA > UB B. UA = UB C. UA < UB 53%

Electric Potential, Gauss’ Law


bQ

aQ 32

41

0(A)

aQ

bQ 23

41

0(B)

0(C)

aQ

bQ 22

41

0(E)

ALWAYS START FROM ALWAYS START FROM DEFINITION OF DEFINITION OF POTENTIALPOTENTIAL

a

brdEV

Spherical symmetry & Spherical symmetry & Gauss’ law determines Gauss’ law determines EEa < r < a < r < b:b: 20

241

rQE

a

b rdrQV 204

2

baQV 11

42

0

A thin non-conducting spherical shell of radius a carries a uniformly distributed net surface charge 2Q. A second thin non-conducting shell of radius b carries a uniformly distributed net surface charge 3Q. The two shells are concentric. Calculate the potential difference V = V(a) – V(b) between the inner and outershells. 37%

bQ

aQ 22

41

0(D)

Electric Potential, Gauss’ Law


Spherical symmetry & Spherical symmetry & Gauss’ law determines Gauss’ law determines EE

r < a:r < a: 0E0

0

ardEV

bQ

aQ 32

41

0(A)

aQ

bQ 23

41

0(B)

0(C)

aQ

bQ 22

41

0(E)

A thin non-conducting spherical shell of radius a carries a uniformly distributed net surface charge 2Q. A second thin non-conducting shell of radius b carries a uniformly distributed net surface charge 3Q. The two shells are concentric.

Calculate the potential difference V = V(a) – V(0) between the inner shell and the origin. 45%

bQ

aQ 22

41

0(D)


)(4 2212

abQQ

(A)

21

4 bQ

(B)

(D)

(E)

21

4 bQ

21

4 aQ

21

4 aQ

Charge must be Charge must be induced to insure induced to insure

E = 0 within E = 0 within conducting shellconducting shell

1 1( )enclosed a bQ Q Q Q Q 21

4 bQ

0

enclosedQAdE

Spherical symmetry & Spherical symmetry & Gauss’ law determines Gauss’ law determines EE

04 2 rE

(C)

A solid conducting sphere of radius a is centered on the origin, and carries a total charge Q1. Concentric with this sphere is a conducting spherical shell of inner radius b and outer radius c, which carries a total charge Q2. What is the surface charge density, b, on the inner surface of the outer spherical shell (r = b)? 59%

Gauss’ Law, Conductors


r < a: E = 0r < a: E = 0 b < r < c: E = 0b < r < c: E = 0

Eliminate (e)Eliminate (e)Eliminate (a) and (c)Eliminate (a) and (c)

QQ11 = -3 = -3CCQQ11 + Q + Q22 = +2 = +2CC

a < r < b: E = kQa < r < b: E = kQ11/r/r22

r > c: E = k(Qr > c: E = k(Q11+Q+Q22)/r)/r22

For r > c, For r > c, E must be less than the E must be less than the

continuation of E from a to continuation of E from a to bb

A solid conducting sphere of radius a is centered on the origin, and carries a total charge Q1. Concentric with this sphere is a conducting spherical shell of inner radius b and outer radius c, which carries a total charge Q2. Which of the following graphs best represents the magnitude of the electric field at points along the positive x axis? 70%

Q1 = - 3 CQ2 = 5 C

Gauss’ Law

Electric Potential


ALWAYS START FROM ALWAYS START FROM DEFINITION OF DEFINITION OF POTENTIALPOTENTIAL

b

rdEV0

Break integral into two Break integral into two piecespieces

b

a

ardErdEV

0

same for same for conductor conductor & insulator& insulator

conductor: = 0conductor: = 0insulator: insulator: 0 0

A solid conducting sphere of radius a is centered on the origin, and carries a total charge Q1. Concentric with this sphere is a conducting spherical shell of inner radius b and outer radius c, which carries a total charge Q2.

If the inner conducting sphere were replaced with an insulating sphere having the same charge, Q1, distributed uniformly throughout its volume, the magnitude of the potential difference |Vb – V0| would 53%

A. increase B. decrease C. stay the same

Q1 = - 3 CQ2 = 5 C

RL Circuits


Strategy: Back to First PrinciplesStrategy: Back to First Principles• The time constant is determined The time constant is determined from a differential equation for the from a differential equation for the current through the inductor.current through the inductor.• Equation for current through Equation for current through inductor obtained from Kirchhoff’s inductor obtained from Kirchhoff’s RulesRules

II I – II – I11

II11

01 VIRdtdIL11

22 0)( 11 RIIdtdIL

11 dtdILVIR 1

22 0111 RIdtdILV

dtdIL

VRIdtdIL 112

RL

RL 2""""

In the circuit below, V = 6 Volts, R = 10 Ohms, L = 100 mH. The switch has been open for a long time. Then, at t = 0, the switch is closed. What is the time constant for the current through the inductor? 30%A. R/L B. R/2L C. L/RD. 2L/R E. L/2R

Faraday’s Law


A rectangular wire loop travels to the right with constant velocity, starting in a region of no magnetic field, moving into a region with a constant field pointing into the page (shaded rectangle below), and continuing into a region of no magnetic field. Which plot below best represents the induced current in the loop (Iloop) as it travels from the left through these three regions? Note: a positive Iloop corresponds to a counter-clockwise current.

Current induced Current induced onlyonly when flux is when flux is changing.changing.Flux is changing because loop is moving.Flux is changing because loop is moving.As loop enters field, current will be As loop enters field, current will be induced to reduce the increase in flux induced to reduce the increase in flux (Lenz’ law): counterclockwise current (Lenz’ law): counterclockwise current generates B field pointing out of the generates B field pointing out of the page.page.When loop is completely inside field, flux When loop is completely inside field, flux is constant, therefore, current is is constant, therefore, current is zerozeroAs loop leaves field current will be As loop leaves field current will be induced to produce B field pointing into induced to produce B field pointing into the page.the page.

Faraday’s Law


R = 20 Ohms

v = 6 m/s

R = 20 Ohms

v

R = 20 Ohms

v = 6 m/s

R = 20 Ohms

v

Two fixed conductors are connected by a resistor R = 20 Ohms. The two fixed conductors are separated by L = 2.5 m and lie horizontally. A moving conductor of mass m slides on them at a constant speed, v, producing a current of 3.75 A. A magnetic field with magnitude 5 T points out of the page. In which direction does the current flow through the moving conductor when the bar is sliding in the direction shown?

A. to the right B. to the left

L

HFlux is changing because area of the Flux is changing because area of the loop is changing.loop is changing.As the moving conductor slides As the moving conductor slides downward the area and therefore the downward the area and therefore the flux increases.flux increases.Therefore a current must flow in the Therefore a current must flow in the clockwise direction to generate a clockwise direction to generate a magnetic field pointing into the pagemagnetic field pointing into the page

Faraday’s Law


R = 20 Ohms

v = 6 m/s

R = 20 Ohms

v

R = 20 Ohms

v = 6 m/s

R = 20 Ohms

v

Two fixed conductors are connected by a resistor R = 20 Ohms. The two fixed conductors are separated by L = 2.5 m and lie horizontally. A moving conductor of mass m slides on them at a constant speed, v, producing a current of 3.75 A. A magnetic field with magnitude 5 T points out of the page. At what speed is the conductor moving?

A. 1 m/s B. 3 m/s C. 5 m/s D. 6 m/s E. 9 m/s

As the moving conductor slides As the moving conductor slides downward the area of the loop, and downward the area of the loop, and therefore the flux, increases.therefore the flux, increases.

d dA dHIR B BL BLvdt dt dt

3.75 20 6 /5 2.5

IR Av m sBL T m

Faraday’s Law: Faraday’s Law: d dE d IRdt dt

Phasers


Phasor diagram at t = Phasor diagram at t = 00

What is VWhat is VCC at t = at t = /(2/(2))

(A)

(B)

(C)

(D)

sinmaxCV

sinmaxCV

cosmaxCV

cosmaxCV

VC

VR

VL

Phasor diagram at t = Phasor diagram at t = /(2/(2))

Voltage is equal to Voltage is equal to projection of projection of phasor along phasor along vertical axisvertical axis

Ampere’s Law Integrals


.

Two infinitely long wires carrying current run into the page as indicated. Consider a closed triangular path that runs from point 1 to point 2 to point 3 and back to point 1 as shown. Which of the following plots best shows B·dl as a function of position along the closed path?

The magnetic field points in the azimuthal direction and is oriented The magnetic field points in the azimuthal direction and is oriented clockwise.clockwise.

B B

No current is contained within the loop 1-2-3-1, therefore NOT (A)No current is contained within the loop 1-2-3-1, therefore NOT (A)The magnetic field falls off like 1/r (r is the distance from the wire).The magnetic field falls off like 1/r (r is the distance from the wire).From 3 to 1, From 3 to 1, BB··dldl is < 0, largest magnitude near wires. is < 0, largest magnitude near wires.From 1 to 2, From 1 to 2, BB··dldl is > 0 is > 0


62%62%

BwhBAAdB

Flux definition:Flux definition:

Faraday’s Faraday’s law:law: dt

dBwhdtd

sTdtdB /2 A

RI 012.

1505/9


65%65%

Current is determined by time rate of change of the Current is determined by time rate of change of the fluxfluxdd/dt is determined by dB/dt/dt is determined by dB/dtdB/dt (6s) = dB/dt (5 s) = -2 T/sdB/dt (6s) = dB/dt (5 s) = -2 T/sThe induced currents at t = 5s and t = 6s are The induced currents at t = 5s and t = 6s are equalequal

Faraday’s Law


76%76%

Current induced because flux is changingCurrent induced because flux is changingFlux is changing beause B is changingFlux is changing beause B is changingAt t = 5 seconds, B is positive, but decreasingAt t = 5 seconds, B is positive, but decreasingLenz’ law: emf induced to oppose change that brought it into Lenz’ law: emf induced to oppose change that brought it into beingbeing

Induced current must produce positive B fieldInduced current must produce positive B fieldPositive B field produced by counterclockwise Positive B field produced by counterclockwise currentcurrent

physics 212 lecture 29, slide 1 physics 212 lecture 29 course review the topics for today –...

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