Tues. Oct. 11, 2007Physics 208 Lecture 121 Last time… Potential and electric field Capacitors...

Tues. Oct. 11, 2007

Physics 208 Lecture 12 1

Last time…

Potential and electric field

Capacitors

€

dr l

V=Vo

€

dr l

€

dr l

Friday Honors Lecture: Prof. R. Moss, Physiology

Detecting the body’s electrical signals

Friday Honors Lecture: Prof. R. Moss, Physiology

Detecting the body’s electrical signals

€

ΔV =1

CQ

Tues. Oct. 11, 2007


Parallel plate capacitor

Charge Q moved from right conductor to left conductor

Each plate has size Length x Width = Area = A

Plate surfaces behave as sheets of charge, each producing E-field

+Q -Q

d

innerouter

Tues. Oct. 11, 2007


Parallel plate capacitor

Charge only on inner surfaces of plates.

E-field inside superposition of E-field from each plate.

Constant E-field inside capacitor.

+

-

d

Tues. Oct. 11, 2007


What is the potential difference?

Electric field between plates

Uniform electric field

-Q+Q

d

+++

+

+

+++

+

++

+++

+

---

-

-

---

-

--

---

-

Etotal

€

E left + E right = η /2εo + η /2εo = η /εo

Potential difference = V+-V-

= (1/q)x(- work to move + charge from + to minus plate)

€

= 1/q( ) × − −qEd( )( )

ΔV = Ed = ηd /εo = Qd

εoA

⎛

⎝ ⎜

⎞

⎠ ⎟

Tues. Oct. 11, 2007


What is the capacitance?

+Q

-Q

d

€

ΔV = V+ −V− = Qd

εoA

⎛

⎝ ⎜

⎞

⎠ ⎟

€

ΔV = Q /C

€

C =εoA

d

This is a geometrical factor

Tues. Oct. 11, 2007


An isolated parallel plate capacitor with spacing d has charge q.

The plates are pulled a small distance further apart. Which of the following describe situation after motion?A) The charge decreases

B) The capacitance increases

C) The electric field increases

D) The voltage between the plates increases

E) None of these

+q

-q+

+

+

+

-

-

-

-

dpullpull

E= q/(0A) E constant

V= Ed V increases

C = 0A/d C decreases!

Quick Quiz

Tues. Oct. 11, 2007


Requires work to transfer charge dq from one plate:

Total work required = sum of incremental work:

Work done to charge a capacitor

€

dW = ΔVdq =q

Cdq

€

dW = ΔVdq =q

Cdq

€

W =q

C0

Q

∫ dq =Q2

2C

€

W =q

C0

Q

∫ dq =Q2

2C

Tues. Oct. 11, 2007


Energy stored = work done

Example: parallel plate capacitor

Energy stored in a capacitor

€

U =Q2

2C=

1

2QΔV =

1

2C ΔV( )

2

€

U =Q2

2C=

1

2QΔV =

1

2C ΔV( )

2

€

U =1

2Ad( )εoE 2

€

U =1

2Ad( )εoE 2

Energy density depends only on field strength

€

U / Ad( ) =1

2εoE 2

Tues. Oct. 11, 2007


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

Tues. Oct. 11, 2007


Human capacitors Cell membrane:

‘Empty space’ separating charged fluids (conductors)

~ 7 - 8 nm thick In combination w/fluids, acts as

parallel-plate capacitor

Cytoplasm

Extracellular fluid

Plasma membrane

100 µm

Tues. Oct. 11, 2007


Modeling a cell membrane

Charges are +/- ions instead of electrons

Charge motion is through cell membrane (ion channels) rather than through wire

Otherwise, acts as a capacitor ~0.1 V ‘resting’ potential

Cytoplasm

Extracellular fluid

Plasma membrane

Na+ Cl-

K+ A-

- - - - - -

+ + + + + +

7-8 nm

ΔV~0.1 V

Ionic charge at surfaces of conducting fluids

Capacitance:

€

oA

d=

8.85 ×10−12 F /m( )4π 50 ×10−6 m( )2

8 ×10−9 m= 3.5 ×10−11F = 35 pF

100 µm sphere surface area

~ 3x10-4 cm2

~0.1µF/cm2

Tues. Oct. 11, 2007


Cell membrane depolarization

Cell membrane can reverse potential by opening ion channels.

Potential change ~ 0.12 V

Ions flow through ion channels Channel spacing ~ 10x membrane thickness (~ 100 channels / µm2 )

Cytoplasm

Extracellular fluid

Plasma membrane

Na+ Cl-

K+ A-

- - - - - -+ + + + + +

7-8 nm

ΔV~0.1 V

+ + + + + +- - - - - -

ΔV~-0.02 V

Charge xfer required ΔQ=CΔV=(35 pF)(0.12V) =(35x10-12 C/V)(0.12V)

= 4.2x10-12 Coulombs1.6x10-19 C/ion -> 2.6x107 ions flow

This is an electric current!This is an electric current!

Tues. Oct. 11, 2007


Charge motion

Cell membrane capacitor: ~70 ions flow through each ion channel to depolarize membrane

Occurs in ~ 1 ms = 0.001 sec. This is a current, units of Coulombs / sec 1 Coulomb / sec = 1 Amp

Tues. Oct. 11, 2007


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

Tues. Oct. 11, 2007


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

Tues. Oct. 11, 2007


Current in a wire

Battery produces E-field in wire

Current flows in response to E-field

Tues. Oct. 11, 2007


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

Tues. Oct. 11, 2007


Charge motion with collisions

Suppose space not empty, but has various fixed objects. The charged particles would then collide with these objects.

Assumption: after collision, charged particle equally likely to move in any direction.

x

x

x

x

Before collision After collision

x

x

x

x

Tues. Oct. 11, 2007


Quick Quiz

Two cylindrical conductors are made from the same material. They are of equal length but one has twice the diameter of the other.

A. R1 < R2

B. R1 = R2

C. R1 > R2

21

Current flow ~ uniform. More cross-sectional area means more current flowing -less resistance.

€

R = ρA

L

Tues. Oct. 11, 2007


Resistors

Schematic layout

Circuits

Physical layout

Tues. Oct. 11, 2007


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

Tues. Oct. 11, 2007


Current conservation

Iin

Iout

Iout = Iin

I1

I2

I3I1=I2+I3

I2

I3

I1

I1+I2=I3

Tues. Oct. 11, 2007


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

Tues. Oct. 11, 2007


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

Tues. Oct. 11, 2007


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

Tues. Oct. 11, 2007


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

Tues. Oct. 11, 2007


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

Tues. Oct. 11, 2007


Quick QuizCalculate the voltage across each resistor if the battery has potential

V0= 22 V.

•R12 = R1 + R2

•V12 = V1 + V2

•I12 = I1 = I2

= 11

V0R12

= V0 = 22 Volts = V12/R12 = 2 Amps

•V1 = I1R1

•V2 = I2R2

= 2 x 1 = 2 Volts

= 2 x 10 = 20 Volts

R1=1

V0

R2=10

Check: V1 + V2 = V12

R1 and R2 are in series

Tues. Oct. 11, 2007


Quick Quiz

What happens to the current through resistor 2 when the switch is closed?

1. Increases2. Remains Same3. Decreases

Follow Up:What happens to the current through the battery?

1. Increases

2. Remains Same

3. Decreases

Switch closed:Ibattery = /Req = (1/R2+1/R3)

Ibattery = I2 + I3

Switch open:Ibattery = /R2 =

Tues. Oct. 11, 2007


How did the charge get transferred? Battery has fixed electric potential difference

across its terminals Conducting plates connected to battery

terminals by conducting wires.

ΔVplates = ΔVbattery across plates

Electrons move from negative battery terminal to -Q plate from +Q plate to positive battery terminal

This charge motion requires work The battery supplies the work

ΔV

€

ΔV =1

CQ

Tues. Oct. 11, 2007


Capacitors in Parallel C =

ε0Ad

Both ends connected together by wire

C1 C2 Ceq

15 V

10 V

15 V

10 V

15 V

10 V

Add Areas: Ceq = C1+C2

Share Charge: Qeq = Q1+Q2

= Veq • Same voltage: V1 = V2

Tues. Oct. 11, 2007


Capacitors in Series

Same Charge: Q1 = Q2 = Qeq

Share Voltage:V1+V2=Veq

Add d:

Ceq

C1

C2

++++

++++

++++

+

-+-

+

-+-

+

-

+Q

-Q +Q

-Q

+Q

-Q

21

111CCCeq

+= C =

ε0Ad

Tues. Oct. 11, 2007


Energy density

The energy stored per unit volume is

U/(Ad) = 1/2 oAdE2

This is a fundamental relationship for the energy stored in an electric field valid for any geometry and not restricted to capacitors

Tues. Oct. 11, 2007


Current Density Alternative expression of Ohm’s law

J = current density, = conductivity

Independent of sample geometry Local relation between J and E

€

I = nqvd A = nqqEτ

m

⎛

⎝ ⎜

⎞

⎠ ⎟A =

nq2τ

mAE

€

I

A=

nq2τ

mE ⇒ J = σE

Tues. Oct. 11, 2007


What is the drift velocity?

Copper wire: A = 3.31 x 10-6 m2 and = 8.95 g/cm3

Copper molar mass m = 63.5 g/mol

1 mol contains NA = 6.02 x 1023 atoms

Hence for I = 10 A

Even if the drift velocity is so low the effect of flipping a switch is

instantaneous since electrons do not have to travel from the light

switch to the light bulb in order for the light to operate but they are

already in the light filament

The volume occupied bya mol V=m/ = 7.09 cm3

Density of electronsn = NA/V = 8.49 x 1028 m-3

€

vd =I

nqA= 2.22 ×10−4 m / s

Tues. Oct. 11, 2007


Electric Current Electric current

rate of flow of charge through some region of space Charge moves in response to a force

Usually electric field, corresponding potential difference

SI unit: ampere 1 A = 1 C / s Average current:ΔQ charges moving perpendicular to a surface of area A cross it in time Δt

Instantaneous value Convention:

the current has the direction of the positive charge carriers

Tues. Oct. 11, 2007


Charge Carrier Motion in a Conductor

Electric field should produce acceleration. Experiment: I = V/R (Ohm’s law) Despite the collisions with

the atoms of the conductor the electrons drift in the opposite direction of the field with a small net velocity

Tues. Oct. 11, 2007


Current Density Ohm’s Law: for many materials, the current density is

proportional to the electric field producing the current J = E = conductivity of the conductor Materials that obey Ohm’s law are said to be ohmic When an electric field is applied free electrons experience an acceleration

(opposite to E): a = F/m = eE/m Calling t the mean time between 2 collisions electron-atoms electrons

achieve a drift velocity:

Hence we obtain for the conductivity:

€

vd = at =eE

mτ =

eVτ

mL

€

I =ΔQ

Δt=

nALe

L / vd

=nAe2τ

m

⎛

⎝ ⎜

⎞

⎠ ⎟V

L= σE

Tues. Oct. 11, 2007


Resistance The potential difference maintained across the wire of

length L creates an electric field ΔV = EL Hence

J = E = ΔV/L = I/A

And

The constant of proportionality is called the resistance of the conductor and SI units of resistance are ohms () 1 = 1 V / A

Resistance in a circuit arises due to collisions between the electrons carrying the current with fixed atoms

€

ΔV =L

σA

⎛

⎝ ⎜

⎞

⎠ ⎟I = RI

Tues. Oct. 11, 2007


Quick Quiz 1

Two cylindrical resistors are made from the same material. They

are of equal length but one has twice the diameter of the other.

1. R1 > R2

2. R1 = R2

3. R1 < R2

21

Smaller diameter not as easy for carriers to flow through

Tues. Oct. 11, 2007


Electrical Power A battery is used to establish an

electric current in a conductor As a charge moves from a to b, the

electric potential energy of the system increases by ΔU=QΔV The chemical energy of the

battery decreases by the same amount

As the charge moves through the resistor (c to d), the system loses the same amount of energy due collisions with atoms of the resistor

This energy goes into vibrational motion of the atoms in the resistor and energy irradiated and heat transfer to air

Tues. Oct. 11, 2007


Cell membrane as dielectric

Membrane is not really empty

It has molecules inside that respond to electric field.

The molecules in the membrane can be polarized

Cytoplasm

Extracellular fluid

Plasma membrane

Na+ Cl-

K+ A-

- - - - - -

+ + + + + +

7-8 nm

Dielectric: insulating materials can respond to an electric field by generating an opposing field.

Tues. Oct. 11, 2007


Effect of E-field on insulators If the molecules of the dielectric are non-polar

molecules, the electric field produces some charge separation

This produces an induced dipole moment

+

-

+

-

E=0E

Tues. Oct. 11, 2007


Dielectrics in a capacitor An external field can polarize

the dielectric

The induced electric field is opposite to the original field

The total field and the potential are lower than w/o dielectric E = E0/ and V = V0/

The capacitance increases C = C0

E0

Eind

Tues. Oct. 11, 2007


Cell membrane as dielectric

Without dielectric, we found 7 ions/channel were needed to depolarize the membrane. Suppose lipid bilayer has dielectric constant of 10. How may ions / channel needed?

Cytoplasm

Extracellular fluid

Plasma membrane

Na+ Cl-

K+ A-

- - - - - -

+ + + + + +

7-8 nm

C increases by factor of 1010 times as much charged needed to reach potential

A. 70

B. 7

C. 0.7

Tues. Oct. 11, 2007


Charge distributions

-Q arranged on inner/outer surfaces of outer sphere.

Charge enclosed by Guassian surface = -Q+Q=0

Flux through Gaussian surface=0, -> E-field=0 outside

Another Gaussian surface: E-field zero inside outer cond. E-field zero outside outer cond. No flux ->

no charge on outer surface!

+ ++

+++

++

+

-Q

+Q

- - -

-

-

-

---

-

-

-

-

Tues. Oct. 11, 2007


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

Tues. Oct. 11, 2007


Tues. Oct. 11, 2007


The electric dipole moment (p) along line

joining the charges from –q to +q Magnitude: p = aq The dipole makes an angle with a uniform external field E The forces F=qE produce a net torque: = p x E

of magnitude = Fa sin pE sin The potential energy is = work done by the torque to rotate

dipole: dW = dU = - pE cos = - p · E

When the dipole is aligned to

the field it is minimum

U = -pE equilibrium!

Electric Dipole alignment

Tues. Oct. 11, 2007


Polar Molecules Molecules are said to be polarized when a separation

exists between the average position of the negative charges and the average position of the positive charges

Polar molecules are those in which this condition is always present (e.g. water)

Tues. Oct. 11, 2007


How to build Capacitors Roll metallic foil interlaced with

thin sheets of paper or Mylar

Interwoven metallic plates are immersed in silicon oil

Electrolitic capacitors: electrolyte is a solution that conducts electricity by virtue of motion of ions contained in the solution

Tues. Oct. 11, 2007


Capacitance of Parallel Plate Capacitor

The electric field from a charged plane of charge per unit area

= Q/A is E = /20

For 2 planes of opposite charge

E= /0 = Q/(0A)

A

d

AE+

-

ΔV

0=1/(4ke)=8.85x10-12 C2/Nm2

+ -E+

E-E-

E+

E-

E+

Tues. Oct. 11, 2007


Spherical capacitor

Capacitance ofSpherical Capacitor:

Capacitance ofCylindrical Capacitor:

Tues. Oct. 11, 2007


Charge, Field, Potential Difference Capacitors are devices to store electric charge and energy

They are used in radio receivers, filters in power supplies, electronic flashes

V =VA – VB = +E0 d VA – VB = +2E0 d

Potential difference is proportional to charge: Double Q Double V

EQ, VE, QV

Charge Q on plates Charge 2Q on plates

+++++

-----

d

E=E0

+++++++++

---------d

E=2E0

Tues. Oct. 11, 2007


Human capacitors: cell membranes Membranes contain lipids and proteins Lipid bilayers of cell membranes can be modeled as a

conductor with plates made of polar lipid heads separated by a dielectric layer of hydrocarbon tails

Due to the ion distribution between the inside and outside of living cells there is a potential difference called resting potential

http://www.cytochemistry.net/Cell-biology/membrane.htm

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

Tues. Oct. 11, 2007


Human capacitors: cell membranes

The inside of cells is always negative with respect to the outside and the DV ≈ 100 V and 0.1 V

Cells (eg. nerve and muscle cells) respond to electrical stimuli with a transient change in the membrane potential (depolarization of the membrane) followed by a restoration of the resting potential.

Remember EKG! The Nobel Prize in Chemistry (2003) for fundamental

discoveries on how water and ions move through cell membranes.

- Peter Agre discovered and characterized the water channel protein

- Roderick MacKinnon has elucidated the structural and mechanistic basis for ion channel function.

http://nobelprize.org/nobel_prizes/chemistry/laureates/2003/chemadv03.pdf#search=%22membrane%20channels%22

Tues. Oct. 11, 2007


Ion channels Membrane channels are protein/sugar/fatty complexes that act as

pores designed to transport ions across a biological membrane In neurons and muscle cells they control the generation of

electrical signals They exist in a open or closed state when ions can pass through

the channel gate or not Voltage-gated channels in nerves and muscles open due to a

stimulus detected by a sensor Eg: in muscles there are 50-500 Na channels per m2 on

membrane surface that can be opened by a change in electric potential of membrane for ~1 ms during which about 103 Na+ ions flow into the cell through each channel from the intracellular medium. The gate is selective: K+ ions are 11 times less likely to cross than Na+

Na channel dimension and the interaction with negative O charges in its interior selects Na+ ions

Tues. Oct. 11, 2007


How much charge flow?

How much charge (monovalent ions) flow through each open channel making a membrane current?

Data: Resting potential = 0.1 V Surface charge density: Q0/A = 0.1 C/cm2

surface density of channels = C = 10 channels/m2 = 109 channels/cm2

1 mole of a monovalent ion corresponds to the charge

F = Faraday Constant = NA e = 6.02 x 1023 x 1.6 x 10-19 ≈ 105 C/mole

NA = Avogadro’s number = number of ions in a mole

Hence surface charge density = (Q0/A)/F = (10-7 C/cm2)/(105 C/mole) = 1 picomoles/cm2

Current/area =I/A = = (10-7 C/cm2)/(10-3 s) = 100 A/cm2

Current/channel = IC = (I/A)/C = = (10-4 A/cm2)/(109 channel/cm2) = = 0.1 pA/channel

(10-13 C/s/channel)/(105 C/mole) = 10-18 moles of ions/s in a channel (10-18 moles of ions/s)/(6.02 x 1023 ions/mole)= 6 x 105 ions/s !!

Tues. Oct. 11, 2007


Honors lecture this Friday Superconductivity, by yours truly.

12:05 pm, 2241 ChamberlinEveryone welcome!

HW 2 due Thursday midnite Lab 2 this week - bring question sheet

available on course web site.

Capacitance and DielectricsThis lecture:

Definition of capacitance Capacitors Combinations of capacitors in circuits Energy stored in capacitors Dielectrics in capacitors and their polarization Cell Membranes

From previous lecture:

1. Gauss’ Law and applications

2. Electric field calculations from Potential

Tues. Oct. 11, 2007


Charge distribution on conductors

Rectangular conductor (40 electrons)

Edges are four lines Charge concentrates at

corners Equipotential lines closest

together at corners Potential changes faster near

corners. Electric field larger at corners.

Tues. Oct. 11, 2007


Placing a dielectric between the plates increases the capacitance:

C = C0

The dielectric reduces the potential difference

V = V0/

Capacitance with dielectric

Dielectric constant ( > 1)

Capacitance without dielectric

Capacitors with Dielectrics

Tues. Oct. 11, 2007


Dielectrics – An Atomic View Molecules in a dielectric can be

modeled as dipoles The molecules are randomly oriented

in the absence of an electric field When an external electric field is

applied, this produces a torque on the molecules

The molecules partially align with the electric field (equilibrium)

The degree of alignment depends on the magnitude of the field and on the temperature

Tues. Oct. 11, 2007


The Electric Field

is the Electric Field It is independent of the test charge,

just like the electric potential It is a vector, with a magnitude and direction, When potential arises from other charges,

= Coulomb force per unit charge on a test charge due to interaction with the other charges.

€

qodV = −r F Coulomb • d

r l ⇒

dV =−

r F Coulomb

qo

⎛

⎝ ⎜

⎞

⎠ ⎟• d

r l

€

=− r

E • dr l

€

rE

€

rE

We’ll see later that E-fields in electromagnetic waves exist w/o charges!

Tues. Oct. 11, 2007


Electric field and potential

Electric field strength/direction shows how the potential changes in different directions

For example, Potential decreases in direction of local E field at rate Potential increases in direction opposite to local E-field at rate potential constant in direction perpendicular to local E-field

€

dV = −r E • d

r l Said before that

€

rE • d

r l = 0( )

€

∝ r

E

€

∝ r

E

Tues. Oct. 11, 2007


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

Tues. Oct. 11, 2007


Quick Quiz 3

Suppose the electric potential is constant everywhere. What is the electric field?

A) Positive

B) Negative

C) Zero

Tues. Oct. 11, 2007


Electric Potential - Uniform Field

Constant E-field corresponds to linearly increasing electric potential

The particle gains kinetic energy equal to the potential energy lost by the charge-field system

€

ΔV = −r E • d

r s ⇒ VB −VA = −

r E

A

B

∫ • dr s

€

=− r

E A

B

∫ dr s = −E d

r s

A

B

∫ = −Ex

€

rE || d

r s

€

rE || d

r s E cnstE cnst

A B

x

€

E = −ΔV

d= −

VB −VA

d

+

Tues. Oct. 11, 2007


Electric field from potential

Said before that Spell out the vectors:

This works for

€

dV = −r E • d

r l

€

dV = − Exdx + Eydy + E zdz( )

€

Ex = −dV

dx, Ey = −

dV

dy, E z = −

dV

dz

Usually written

€

E = −

r ∇V = −

dV

dx,dV

dy,dV

dz

⎛

⎝ ⎜

⎞

⎠ ⎟

€

E = −

r ∇V = −

dV

dx,dV

dy,dV

dz

⎛

⎝ ⎜

⎞

⎠ ⎟

Tues. Oct. 11, 2007


Equipotential lines

Lines of constant potential In 3D, surfaces of constant potential

Tues. Oct. 11, 2007


Electric Field and equipotential lines for + and - point charges

The E lines are directed away from the source charge A positive test charge would be

repelled away from the positive source charge

The E lines are directed toward the source charge

A positive test charge would be attracted toward the negative source charge

Blue dashed lines are equipotential

Tues. Oct. 11, 2007


The Electric Field

is the Electric Field It is independent of the test charge,

just like the electric potential It is a vector, with a magnitude and direction, When potential arises from other charges,

= Coulomb force per unit charge on a test charge due to interaction with the other charges.

€

qodV = −r F Coulomb • d

r l ⇒

dV =−

r F Coulomb

qo

⎛

⎝ ⎜

⎞

⎠ ⎟• d

r l

€

=− r

E • dr l

€

rE

€

rE

We’ll see later that E-fields in electromagnetic waves exist w/o charges!

Tues. Oct. 11, 2007


Electric field and potential

Electric field strength/direction shows how the potential changes in different directions

For example, Potential decreases in direction of local E field at rate Potential increases in direction opposite to local E-field at rate potential constant in direction perpendicular to local E-field

€

dV = −r E • d

r l Said before that

€

rE • d

r l = 0( )

€

∝ r

E

€

∝ r

E

Tues. Oct. 11, 2007


Electric Field and equipotential lines for + and - point charges

The E lines are directed away from the source charge A positive test charge would be

repelled away from the positive source charge

The E lines are directed toward the source charge

A positive test charge would be attracted toward the negative source charge

Blue dashed lines are equipotential

Tues. Oct. 11, 2007


Q = Qinner+Qouter

Total E-field = Eleft plate+Eright plate

-Q

QinnerQouter

€

Eleft plate =σ inner

2εo

+σ outer

2εo

=Q

2Aεo

+++

+

+

+++

+

++

+++

+

+

+

+

+

+

+

+

+

---

-

-

---

-

--

---

-

-

-

-

-

-

-

-

-

+Q

€

Eright plate =σ inner

2εo

+σ outer

2εo

=Q

2Aεo

In between plates,

d

€

Etotal = Eleft plate + Eright plate =Q

εoA

€

ΔV = E∫ • ds =Q

εo Aleft plate

right plate

∫ dx =Qd

εo A

€

C =Q

ΔV=

εoA

d

€

C =Q

ΔV=

εoA

d

Eleft plate

Eright plate

Etotal

+

=

Tues. Oct. 11, 2007


Drift Speed

Conductor of cross-sectional area A and length Δx n = # of charge carriers per unit volume nA Δx = # of charge carriers in this elementary volume total charge = number of carriers x charge of carrier q

ΔQ = (nA Δx)q Average current:

Iav = ΔQ/ Δt = nqvdAwhere drift speed at whichcarriers move is

vd = Δx / Δt Current density

Magnitude J = I/A (A/m2) J = nqvd in the direction of positive charge

carriers

Tues. Oct. 11, 2007


How small is the drift velocity?

Copper wire: A = 3.31 x 10-6 m2 and = 8.95 g/cm3

Copper molar mass m = 63.5 g/mol

1 mol contains NA = 6.02 x 1023 atoms

Hence for I = 10 A

Even if the drift velocity is so low the effect of flipping a switch is

instantaneous since electrons do not have to travel from the light

switch to the light bulb in order for the light to operate but they are

already in the light filament

The volume occupied bya mol V=m/ = 7.09 cm3

Density of electronsn = NA/V = 8.49 x 1028 m-3

€

vd =I

nqA= 2.22 ×10−4 m / s

Tues. Oct. 11, 2007


During the charging of a capacitor, when a charge q is on the plates, the work needed to transfer further dq from one plate to the other is:

The total work required to charge the capacitor is

The energy stored in any capacitor is: For a parallel capacitor:

Work done and energy stored

U = 1/2 oAdE2

Tues. Oct. 11, 2007


Electric field produces accel a=qE/m velocity v = vo + qEt / m

Lots of particles: averageover all particle velocities vave = (vo)ave + qEtave / m

vave = qE / m vd= Drift velocity= qE / m

Drift Velocity

vo= velocity after last collision

t = time since last collision

x

x

x

x

xx

=ave. time since last collision

=0

Tues. Oct. 11, 2007


Drift Speed

n = # of charge carriers per unit volume nA Δx = # of charge carriers in test volume (length Δx charge in volume = (# carriers) x (charge of carrier q)

ΔQ = (nA Δx)q

Average current:

Iav = ΔQ/ Δt = nqvdA

Δx / Δt = vd

Cross-sectional area A

vd= Drift velocity= qE / m

Tues. Oct. 11, 2007


Resistance

I = current = nqvdA vd= Drift velocity= qE / m

€

I = nqqEτ

m

⎛

⎝ ⎜

⎞

⎠ ⎟A =

nq2τ

mAE

Constant E-field -> E=V/L (L=sample length)

€

I = nqqEτ

m

⎛

⎝ ⎜

⎞

⎠ ⎟A =

nq2τ

m

A

L

⎛

⎝ ⎜

⎞

⎠ ⎟V ⇒ V = IR

Voltage proportional to current! This is Ohm’s law

€

R =nq2τ

m

A

L

⎛

⎝ ⎜

⎞

⎠ ⎟

−1

€

R =nq2τ

m

A

L

⎛

⎝ ⎜

⎞

⎠ ⎟

−1

Tues. Oct. 11, 2007


Producing an electric current

To make an electric current, must get charges to move.

To more charges requires a force Force on a charge can be produced

by an electric field.

Tues. Oct. 11, 2007


Energy density

The energy stored per unit volume is

U/(Ad) = 1/2 oE2

This is a fundamental relationship for the local energy stored in an electric field

Not restricted to capacitors: E-field can be from any source

Interpretation: energy is stored in the field

Tues. Oct. 11, 2007


Resistivity

Resistivity

SI units Ω-m

€

=RA

LIndependent of sample geometry

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Tues. Oct. 11, 2007Physics 208 Lecture 121 Last time… Potential and electric field Capacitors...

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