1

Exam 2 covers Ch. 28-33,Lecture, Discussion, HW, Lab

Chapter 28: Electric flux & Gauss’ law Chapter 29: Electric potential & work Chapter 30: Electric potential & field

(exclude 30.7) Chapter 31: Current & Conductivity Chapter 32: Circuits

(exclude 32.8) Chapter 33: Magnetic fields & forces

(exclude 33.3, 33.6, 32.10, Hall effect)

Exam 2 is Tue. Oct. 28, 5:30-7 pm, 2103 Ch

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

Properties of conductors

everywhere inside a conductor

Charge in conductor is only on the surface

surface of conductor

€

rE = 0

€

rE ⊥

----

-

-

+ +++++

5

Gauss’ law example: Charges on parallel-plate capacitor

Determine fields by superposition

Q -Q

Area A=Length X Width

€

E =Q

Aεo

€

E = 0

€

E = 0

Apply Gauss’ law: E=0 inside metal E=0 to left No charge on outer surface

€

⇒ E = 0, Qencl = 0

Apply Gauss’ law: E=0 inside metal E=Q/Ao in middle

€

E =Q

Aεo

Asurf , Qencl = η inner Asurf

⇒ η inner = Q / A

6

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 proportional to charge q that work is done on

€

ΔU /q ≡ ΔV = Electric potential difference

Depends only on charges that create E-fields

€

= r

E • dr s ∫

7

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

8

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

9

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

10

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

11

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 = Q /C

€

C =εoA

d

+Q -Q

d

Area A

€

ΔV

13

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

14

Spherical capacitor

Charge Q moved from outer to inner sphere

Gauss’ law says E=kQ/r2 until second sphere

Potential difference

+ ++

++ +

++

+

€

ΔV = E • dsa

b

∫

Along path shown

€

ΔV =kQ

r2a

b

∫ = −kQ1

r a

b

= kQ1

a−

1

b

⎛

⎝ ⎜

⎞

⎠ ⎟

€

C =Q

ΔV= k

1

a−

1

b

⎛

⎝ ⎜

⎞

⎠ ⎟−1

€

C =Q

ΔV= k

1

a−

1

b

⎛

⎝ ⎜

⎞

⎠ ⎟−1

Gaussian surface to find E

Path to find ΔV

15

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

17

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)

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

R1=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

22

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

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

Parallel Series

24

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

25

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

Time constant

€

τ =RC

Charge Discharge

26

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

27

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

28

RC Circuits

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

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

29

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

30

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

31

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

32

Current loops & magnetic dipoles

• Current loop produces magnetic dipole field.

• Magnetic dipole moment:

€

rμ

€

rμ =IA

currentArea of loop

€

rμ

magnitude direction

In a uniform magnetic field

Magnetic field exerts torqueTorque rotates loop to align with

€

rτ =

rμ ×

rB ,

€

rμ

€

rB

€

rτ =

rμ

rB sinθ