thur. oct. 8, 2009physics 208 lecture 111 last time… equipotential lines capacitance and...

Thur. Oct. 8, 2009 Physics 208 Lecture 11 1

Last time…

Equipotential lines

Capacitance and capacitors

€

ΔV =1

CQ


Parallel plate capacitor

+Q

-Q

d

€

ΔV = Q /C

€

C =εoA

d Geometrical factor determined from electric fields

€

U =1

2C ΔV( )

2=

1

2

εoA

dEd( )

2=

1

2Ad( )εoE 2

Energy stored in parallel-plate capacitor

€

U / Ad( ) =1

2εoE 2

Energy density

AreaA

AreaA


An isolated parallel plate capacitor has charge Q and potential V. The plates are pulled apart. Which describes the situation afterwards?

A) Charge Q has decreased

B) Capacitance C has increased

C) Electric field E has increased

D) Voltage difference V between plates has increased

E) None of these

+Q-Q+

+

+

+

-

-

-

-

dpullpull

E = (Q/A)/0 E constant

V= Ed V increases

C = 0A/d C decreases

Quick Quiz

Cap. isolated Q constant


An isolated parallel plate capacitor has a charge q. The plates are then pulled further apart. What happens to the energy stored in the capacitor?

1) Increases

2) Decreases

3) Stays the same

+q

-q+

+

+

+

-

-

-

-

dpullpull

Quick Quiz


Different geometries of capacitors

Parallel plate capacitor

Spherical capacitor

Cylindrical capacitor

+Q

-QL

€

C =Q

ΔV=

2πεoL

ln b /a( )

€

C =Q

ΔV=

2πεoL

ln b /a( )

€

C =Q

ΔV=

εoA

d

€

C =Q

ΔV=

εoA

d

+Q

-Q

d

A

€

C =Q

ΔV= 4πεo

1

a−

1

b

⎛

⎝ ⎜

⎞

⎠ ⎟−1

€

C =Q

ΔV= 4πεo

1

a−

1

b

⎛

⎝ ⎜

⎞

⎠ ⎟−1


Combining Capacitors — Parallel Connect capacitors together with metal wire

C1 C2Ceq

Both have same ΔV

Need different charge

€

Q1 = C1 /ΔV

€

Q2 = C2 /ΔV

“Equivalent” capacitorPotential difference ΔVTotal charge

€

Qeq = Q1 + Q2

€

Ceq =Qeq

ΔV=

Q1 + Q2

ΔV= C1 + C2 = Ceq


Combining Capacitors — Series

21

111CCCeq

+=21

111CCCeq

+=

C1

C2

Ceq

VA

VB

Vm

€

ΔV1 = VA −Vm = Q /C1

VA

VB

€

ΔV2 = Vm −VB = Q /C2

€

ΔV = VA −VB

= ΔV1 + ΔV2

Q

Q

-Q

-QQ

-Q

€

ΔV =Q

C1

+Q

C1

=Q

CeqQ on each is same


Current in a wire: not electrostatic equilibrium

Battery produces E-field in wire

Charge moves in response to E-field


Electric Current

Electric current = I = amount of charge per unit time flowing through a plane perpendicular to charge motion

SI unit: ampere 1 A = 1 C / s

Depends on sign of charge: + charge particles:

current in direction of particle motion is positive - charge particles:

current in direction of particle motion is negative


Quick Quiz An infinite number of positively charged particles are

uniformly distributed throughout an otherwise empty infinite space. A spatially uniform positive electric field is applied. The current due to the charge motion

A. increases with time

B. decreases with time

C. is constant in time

D. Depends on field

Constant force qE

Produces constant accel. qE/m

Velocity increases v(t)=qEt/m

Charge / time crossing plane increases with time


But experiment says…

Current constant in time Proportional to voltage

R = resistance (unit Ohm = )

Also written

J = current density = I / (cross-section area) = resistivity = R x (cross-section area) / (length)

Resistivity is independent of shape

€

I =1

RV

€

J =1

ρV


Charge motion with collisions Wire not empty space, has various fixed objects. Charge carriers accelerate, then collide. After collision, charged particle reaccelerates. Result: average “drift” velocity vd


Current and drift velocity

This average velocity called drift velocity

€

I = −ene A( )vd = −ne

e2τ

mA

⎛

⎝ ⎜

⎞

⎠ ⎟E

€

vd =eτ

m

⎛

⎝ ⎜

⎞

⎠ ⎟E

Current density J

€

J = I / A = −nee

2τ

mE = σ E

This drift leads to a current

Conductivity

Electric field


What about Ohm’s law?

Current density proportional to electric field

€

J = σ E

€

I = JA = σAE =σA

LEL( ) = V /R

Current proportional to current density through geometrical factor

Electric field proportional to electric potential through geometrical factor

€

R =L

σA= ρ

L

A


Resistivity

Resistivity

SI units Ω-m

€

=RA

LIndependent of sample geometry


Resistors

Schematic layout

Circuits

Physical layout


Quick Quiz

Which bulb is brighter?

A. A

B. B

C.Both the same

Current through each must be same

Conservation of current (Kirchoff’s current law)

Charge that goes in must come out


Current conservation

Iin

Iout

Iout = Iin

I1

I2

I3I1=I2+I3

I2

I3

I1

I1+I2=I3


Quick QuizHow 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


Resistors in Series I1 = I2 = I Potentials add

ΔV = ΔV1 + ΔV2 = IR1 + IR2 =

= I (R1+R2) The equivalent resistance

Req = R1+R2

R

R=

2R

2 resistors in series:R LLike summing lengths

€

R = ρL

A


Quick Quiz

What happens to the brightness of the 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


Resistors in Parallel ΔV = ΔV1 = ΔV2

I = I 1 + I 2 (lower resistance path has higher current)

Equivalent Resistance

R/2

R R

Add areas

€

R = ρL

A


Quick Quiz

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

A. Gets dimmer

B. Gets brighter

C. Stays same

D. Something else

thur. oct. 8, 2009physics 208 lecture 111 last time… equipotential lines capacitance and...

Documents

charge q

isolated q constant

time slide

charge motion

wire charge

efield slide

sign of charge

charge carriers