Exam 2 covers Ch. 27-33, Lecture, Discussion, HW, Lab

1

Exam 2 covers Ch. 27-33,Lecture, Discussion, HW, Lab

Chapter 27: The Electric Field Chapter 29: Electric potential & work Chapter 30: Electric potential & field

(exclude 30.7) Chapter 31: Current & Resistance Chapter 32: Fundamentals of Circuits

(exclude 32.8) Chapter 33: The Magnetic Field

(exclude 33.5-33.6, 33.9-10, & Hall effect)

Exam 2 is Tue. Oct. 27, 5:30-7 pm, 145 Birge

2

Electric field lines

• Local electric field tangent to field line

• Density of lines proportional to electric field strength

• Fields lines can only start on + charge

• Can only end on - charge.

• Electric field lines can never cross

Oct. 22, 2009

Physics 208 Lecture 15

3

QuestionHere is a picture of electric field lines. Which choice most accurately ranks the magnitude of the electric field at the different points?

A) E1=E3>E2=E4

B) E1=E2>E3>E4

C) E4=E3>E1=E2

D) E4=E2>E1>E3

E) E4<E3<E1<E2

1

2

3

4

Oct. 22, 2009

Physics 208 Lecture 15

Oct. 22, 2009 Physics 208 Lecture 15 4

Charge Densities Volume charge density: when a charge is distributed

evenly throughout a volume = Q / V dq = dV

Surface charge density: when a charge is distributed evenly over a surface area = Q / A dq = dA

Linear charge density: when a charge is distributed along a line = Q / dq = d

Electric fields and potentials from these charge elements superimpose

Oct. 22, 2009 Physics 208 Lecture 15 5

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

Infinite line of charge, charge density λ

r

€

rE = 2ke

λ

r=

1

2πεo

λ

r

Infinite sheet of charge, charge density η

r

€

rE = 2π keη =

η

2εo

Oct. 22, 2009 Physics 208 Lecture 15 6

Ring of uniform positive charge

y

x

z

€

E z = kzQ

z2 + R2( )

3 / 2

€

E z

Which is the graph of on the z-axis?

A)

B)

C)

D)

E)

z

7

Properties of conductors

everywhere inside a conductor

Charge in conductor is only on the surface

surface of conductor

€

rE = 0

€

rE ⊥

----

-

-

+ +++++

8

Electric potential: general

Electric field usually created by some charge distribution. V(r) is electric potential of that charge distribution

V has units of Joules / Coulomb = Volts

€

ΔU =r F Coulomb • d

r s ∫ = q

r E • d

r s ∫ = q

r E • d

r s ∫

Electric potential energy difference ΔU

€

ΔU /q ≡ ΔV = Electric potential difference

Depends only on charges that create E-fields

€

= r

E • dr s ∫

9

Electric Potential

Q source of the electric potential, q ‘experiences’ it

Electric potential energy per unit chargeunits of Joules/Coulomb = Volts

Example: charge q interacting with charge Q

Electric potential energy

Electric potential of charge Q

€

=UQq = ke

Qq

r

€

=VQ r( ) =UQq

q= ke

Q

r

10

Example: Electric PotentialCalculate the electric potential at B

Calculate the work YOU must do to move a Q=+5 mC charge from A to B.

Calculate the electric potential at A

x

+-

B

A

d1=3 m 3 m

d2=4 m

3 m

y

-12 μC +12 μC

d

€

VB = kq

d−

q

d

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

€

VA = kq

d1

−q

3d1

⎛

⎝ ⎜

⎞

⎠ ⎟= k

2q

3d1

€

WYou = ΔU = UB −UA = Q(VB −VA ) = −k2qQ

3d1

€

=WE− field = −ΔUWork done by electric fields

11

Potential from electric field

Electric field can be used to find changes in potential

Potential changes largest in direction of E-field.

Smallest (zero) perpendicular to E-field

€

dV = −r E • d

r l

€

dV = −r E • d

r l

€

dr l

€

rE

V=Vo

€

V = Vo −r E d

r l

€

V = Vo +r E d

r l

€

dr l

€

dr l

€

V = Vo

12

Electric Potential and Field

Uniform electric field of

What is the electric potential difference VA-VB?

A) -12V

B) +12V

C) -24V

D) +24V

€

rE = 4 ˆ y N /C

A

x

y

2m

5m

2m

5m

B

Capacitors

Energy stored in a capacitor:

€

U =Q2

2C=

1

2CΔV 2 =

1

2QΔV

C = capacitance: depends on geometry of conductor(s)

Conductor: electric potential proportional to charge:

€

V = Q /C

Example: parallel plate capacitor

€

ΔV = E insided =Qd

εoA= Q /C ⇒

€

C =εoA

d

+Q -Q

d

Area A

€

ΔV

13

€

rE inside =

1

2εo

Q

A−

1

2εo

−Q

A=

Q

εoA

€

rE outside = 0

Question

What is the voltage across capacitor 1 after the two are connected?

14

C2=3µF

V1=1V

C1=1µF

V2=0V

A. 1V

B. 2V

C. 0V

D. 0.25V

E. 4V

€

Q1before = Q1

after + Q2after

C1V1before = C1V

after + C2Vafter

V after =C1

C1 + C2

V1before =

1μF

1μF + 3μF1V = 0.25V

15

Isolated charged capacitorPlate separation increased The stored energy 1) Increases2) Decreases3) Does not change

A)B)C)

Stored energy

q unchanged because C isolated

€

U =q2

2C

Cini =ε0A

d→ C fin =

ε0A

D⇒ C fin < Cini ⇒ U fin > U ini

q is the sameE is the same = q/(Aε0)ΔV increases = EdC decreasesU increases

16

Conductors, charges, electric fields Electrostatic equilibrium

No charges moving No electric fields inside conductor. Electric potential is constant everywhere Charges on surface of conductors.

Not equilibrium Charges moving (electric current) Electric fields inside conductors -> forces on charges. Electric potential decreases around ‘circuit’

17

Resistance and resistivity Ohm’s Law: ΔV = R I (J = σ E or E = ρ J ΔV = EL and E = J /A = ΔV/L R = ρL/A Resistance in ohms (Ω)

Question

5cm

2cm1cm

I

A block is made from a material with resistivity of 10-4Ω-m. It has 10 A of current flowing through it. What is the voltage across the block?

€

R = 10−4 Ω ⋅m( )0.05m

0.02m( ) 0.01m( )

= 0.025Ω

A. 0.1V

B. 0.25V

C. 0.5V

D. 1.0V

E. 5.0V

18

Current conservation

Iin

Iout

Iout = Iin

I1

I2

I3I1=I2+I3

I2

I3

I1

I1+I2=I3

19

Resistors in Series and parallel Series I1 = I2 = I Req = R1+R2

R1

R2

=R1+R2

2 resistors in series:R LLike summing lengths

R1R2

€

R = ρL

A€

1

R1

+1

R2

⎛

⎝ ⎜

⎞

⎠ ⎟

−1

=

I

I

II1 I2

I1+I2

Parallel V1 = V2 = V Req = (R1

-1+R2-1)-1

20

Quick Quiz

What happens to the brightness of bulb A when the switch is closed?

A. Gets dimmer

B. Gets brighter

C. Stays same

D. Something else

21

Quick QuizWhat is the current

through resistor R1?

R1=200Ω

R2=200Ω

R3=100Ω

R4=100Ω9V

Req=100Ω

Req=50Ω

9V

A. 5 mA

B. 10 mA

C. 20 mA

D. 30 mA

E. 60 mA

6V

3V

Oct. 22, 2009 Physics 208 Lecture 15 22

Power dissipation (Joule heating)

Charge loses energy from c to d.

Ohm’s law:

Energy dissipated in resistor as Heat (& light) in bulb

Power dissipated in resistor =

€

E lost = −ΔE = − ΔKE + ΔU( ) = 0 − q Vd −Vc( )

€

Vc −Vd( ) = IR

€

dE lost

dt=

dq

dtIR = I2R Joules / s = Watts

€

E lost = qIR

23

Capacitors as circuit elements

Voltage difference depends on charge Q=CV Current in circuit

Q on capacitor changes with time Voltage across cap changes with time

24

Capacitors in parallel and series

ΔV1 = ΔV2 = ΔV Qtotal = Q1 + Q2

Ceq = C1 + C2

Q1=Q2 =Q ΔV = ΔV1+ΔV2 1/Ceq = 1/C1 + 1/C2

Parallel Series

25

Example: Equivalent Capacitance

C1 = 30 μFC2 = 15 μFC3 = 15 μFC4 = 30 μF

C2V C3

C1

C4

Parallel combinationCeq=C1||C2

€

C23 = C2 + C3 =15μF +15μF = 30μF

€

C1, C23, C4 in series

€

⇒1

Ceq

=1

C1

+1

C23

+1

C4

€

1

Ceq

=1

30μF+

1

30μF+

1

30μF⇒ Ceq =10μF

26

RC CircuitsR

Cε

€

q(t) = Cε(1− e−t / RC )

I(t) =ε

Re−t / RC

R

C

€

Vcap t( ) = ε 1− e−t / RC( )

€

Vcap t( ) = qo /C( )e−t / RC

€

q t( ) = qoe−t / RC

I t( ) =qo /C

Re−t / RC

Start w/uncharged CClose switch at t=0

Start w/charged CClose switch at t=0

Time constant

€

τ =RC

Charge Discharge

27

Question

What is the current through R1 Immediately after the switch is closed?

10V

R1=100Ω

R2=100ΩC=1µF

A. 10A

B. 1 A

C. 0.1A

D. 0.05A

E. 0.01A

28

Question

What is the current through R1 a long time after the switch is closed?

10V

R1=100Ω

R2=100ΩC=1µF

A. 10A

B. 1 A

C. 0.1A

D. 0.05A

E. 0.01A

29

Question

What is the charge on the capacitor a long time after the switch is closed?

A. 0.05µC

B. 0.1µC

C. 1µC

D. 5µC

E. 10µC

10V

R1=100Ω

R2=100ΩC=1µF

30

RC Circuits

What is the value of the time constant of this circuit?

A) 6 msB) 12 msC) 25 msD) 30 ms

31

FB on a Charge Moving in a Magnetic Field, Formula

FB = q v x B FB is the magnetic force q is the charge v is the velocity of the

moving charge B is the magnetic field

SI unit of magnetic field: tesla (T)

CGS unit: gauss (G): 1 T = 104 G (Earth surface 0.5 G)

€

T =N

C ⋅m /s=

N

A ⋅m

32

Magnetic Force on a Current

S

N

I

Current

Magnetic field

Magnetic force

€

rF = IBL

€

qr v ×

r B Force on each charge

Force on length of wire

€

dr s

€

Idr s ×

r B

Force on straight section of wire, length L

Oct. 22, 2009 Physics 208 Lecture 15 33

Law of Biot-Savart

Each short length of current produces contribution to magnetic field.r

I in plane of pageds

€

dr B =

μo

4π

Idr s × ˆ r

r2

B out of page

ds

dB

r

€

μo = 4π ×10−7 N / A2= permeability of free space

r = distance from current element

Field from very short section of current

€

dr s

34

Magnetic field from long straight wire:Direction What direction

is the magnetic field from an infinitely-long straight wire? I

x

y

€

rB =

μoI

2π r

€

μo = 4π ×10−7 N / A2= permeability of free space

r = distance from wire

Oct. 22, 2009 Physics 208 Lecture 15 35

Magnetic field from loopWhich of these graphs

best represents the magnetic field on the axis of the loop?

Bz

Bz

Bz

Bz

A.

B.

C.

D.

z

z

z

z

z

x

y

I