1

Exam 2 covers Ch. 27-32,Lecture, Discussion, HW, Lab

Chapter 27: Electric flux & Gauss’ law Chapter 29: Electric potential & work Chapter 30: Electric potential & field Chapter 28: Current & Conductivity Chapter 31: Circuits Chapter 32: Magnetic fields & forces

(exclude 32.6,32.8,32.10)

Exam 2 is Wed. Mar. 26, 5:30-7 pm, 2103 Ch: Adam(301,310), Eli(302,311), Stephen(303,306), 180 Science Hall: Amanda(305,307), Mike(304,309), Ye(308)

2

Electric flux Suppose surface make angle surface normal

E = EA cos E =0 if E parallel A

E = EA (max) if E A

Flux SI units are N·m2/C

€

rE = E ||

ˆ s + E⊥ˆ n

€

ˆ n

€

ˆ s

€

rA = A ˆ n

Component || surface

Component surface

Only component‘goes through’ surface

€

E =r E •

r A

3

Gauss’ law

net electric flux through closed surface = charge enclosed /

€

E =r E • d

r A ∫ =

Qenclosed

εo

4

Field outside uniformly-charged sphere

Field direction: radially out from charge Gaussian surface:

Sphere of radius r

Surface area where

Value of on this area:

Flux thru Gaussian surface:

Charge enclosed:

€

rE • d

r A ≠ 0 :

€

4π r2

€

rE • d

r A

€

E

€

Q

€

E 4π r2

Gauss’ law:

€

E 4π r2 = Q /εo ⇒ E =1

4πεo

Q

r2

5

Electric potential energy

€

Whand = ΔU since they repel! potential energy increases

Work is Force x distance (taking into account cosθ between 2 vectors!)

>0

If opposte charges they attract => W <0 and potential energy decreases

6

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

7

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

8

A. W = +19.8 mJB. W = -19.8 mJC. = 0

Work and electrostatic potential energy

−μC

−μC

−μC m

m m

Question: How much work would it take YOU to assemble 3 negative charges?

Likes repel, so YOU will still do positive work!

q3

q2q1

€

W1 = 0

W2 = kq1q2

r12

= 9 ×109 −1×10−6 × −2 ×10−6

5= 3.6mJ

W3 = kq1q3

r13

+ kq2q3

r23

=16.2mJ

W tot = kq1q2

r12

+ kq1q3

r13

+ kq2q3

r23

= +19.8mJ

€

UE =19.8mJ electric potential energy of the system increases

9

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

10

Electric Potential and Field• Uniform electric field of E = 4i+3j N/C

• Points A at 2m and B at 5m on the x axis.

• What is the potential difference VA - VB?

B(5,0)

E = 4i N/C x(m)

A(2,0)0

A) -12VB) +12VC) -24VD) +24V

€

E = −∇V ⇒ Ex = −dV

dx⇒ dV

A

B

∫ = −Ex dxA

B

∫ ⇒

VA −VB = 4 × 3 =12V

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 = Q /C

€

C =εoA

d

+Q -Q

d

Area A

€

ΔV

12

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

13

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’

Electric current

Current density J= I/A = nqvd

(direction of + charge carriers)

L

SI unit: ampere 1 A = 1 C / s

Average current:

Instantaneous value:

n = number of electrons/volumen x AL electrons travel distance L = vd Δt

Iav = ΔQ/ Δt = neAL vd /L

15

Resistance and resistivity

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

16

Current conservation

Iin

Iout

Iout = Iin

I1

I2

I3I1=I2+I3

I2

I3

I1

I1+I2=I3

17

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

18

Quick Quiz

How does brightness of bulb B compare to that of A?

A. B brighter than A

B. B dimmer than A

C.Both the same

Battery maintain constant potential difference

Extra bulb makes extra resistance -> less current

19

Quick QuizWhat happens to the brightness of bulb B

when the switch is closed?

A. Gets dimmer

B. Gets brighter

C. Stays same

D. Something else

Battery is constant voltage,not constant current

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

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

22

RC CircuitsR

C

€

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

23

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

24

Calculate the equivalent Capacitance C1 = 10 μF

C2 = 20 μFC3 = 30 μFC4 = 40 μFV = 50 Volts

C2V C3

C1

C4

€

1

Ceq

=1

C1

+1

C2 + C3

+1

C4

⇒ Ceq = 6.9μF

€

V = V1 + V23 + V4 V23 = V2 = V3

Q = Q1 = Q23 = Q4 Q23 = Q2 + Q3

V =Q

Ceq

=Q

C1

+Q

C2 + C3

+Q

C4 parallel

C1, C23, C4 in series

25

RC Circuits

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

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

26

Magnetic fields and forces

€

FB = qvBsinθ

€

F = qv × B

I

B

€

FB = Ids × B Magnetic force

on current-carrying wire

Magnetic torque on current loop

Magnetic force on moving charged particle

I

B

27

Effect of uniform magnetic field

Effect of uniform B-field on charged particle If charged particle

is not moving - no effect If particle is moving:

force exerted perpendicularto both field and velocity

€

F = qv × B

vector ‘cross product’

12/09/2002 U. Wisconsin, Physics 208, Fall 2006 28

Lorentz force

Electron moves in plane of screen the page.

B- field is in the plane of screen to the right.

Direction of instantaneous magnetic force on electron is

A) toward the top of the page

B) into the page

C) toward the right edge of the page

D) out of the pagev

B

F

electron

29

Trajectory in Constant B Field

FFv

x x x x x x

x x x x x x

x x x x x xv B

q

x x x x x x

x x x x x x

x x x x x x

• Charge enters B field with velocity shown. (vB)

• Force is always to velocity and to B.

Path is a circle.

Radius determined by velocity:

€

R =mv

qB

30

Current loops & magnetic dipoles Current loop produces magnetic dipole field. Magnetic dipole moment:

€

rμ

€

rμ =IA

currentArea of loop

€

rμ

magnitude direction

Effect of uniform magnetic field

Magnetic field exerts torqueTorque rotates loop to align with

€

rτ =

rμ ×

rB ,

€

rμ

€

rB

€

rτ =

rμ

rB sinθ

12/09/2002 U. Wisconsin, Physics 208, Fall 2006 31

Which of these loop orientations has the largest magnitude torque?

(A) a (B) b (C) c

Question on torque

Answer: (c). all loops enclose same area and carry same current ⇒ magnitude of μ is the same for all.

(c) μ upwards, μ ⊥ B and τ = μB. (a), τ = 0 (b) τ = μBsin

a bc

μμ ττ