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
Physics 208 Lecture 12 2
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
Physics 208 Lecture 12 3
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
Physics 208 Lecture 12 4
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
Physics 208 Lecture 12 5
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
Physics 208 Lecture 12 6
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
Physics 208 Lecture 12 7
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
Physics 208 Lecture 12 8
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
Physics 208 Lecture 12 9
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
Physics 208 Lecture 12 10
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
Physics 208 Lecture 12 11
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
Physics 208 Lecture 12 12
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
Physics 208 Lecture 12 13
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
Physics 208 Lecture 12 14
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
Physics 208 Lecture 12 15
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
Physics 208 Lecture 12 16
Current in a wire
Battery produces E-field in wire
Current flows in response to E-field
Tues. Oct. 11, 2007
Physics 208 Lecture 12 17
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
Physics 208 Lecture 12 18
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
Physics 208 Lecture 12 19
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
Physics 208 Lecture 12 20
Resistors
Schematic layout
Circuits
Physical layout
Tues. Oct. 11, 2007
Physics 208 Lecture 12 21
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
Physics 208 Lecture 12 22
Current conservation
Iin
Iout
Iout = Iin
I1
I2
I3I1=I2+I3
I2
I3
I1
I1+I2=I3
Tues. Oct. 11, 2007
Physics 208 Lecture 12 23
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
Physics 208 Lecture 12 24
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
Physics 208 Lecture 12 25
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
Physics 208 Lecture 12 26
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
Physics 208 Lecture 12 27
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
Physics 208 Lecture 12 28
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
Physics 208 Lecture 12 29
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
Physics 208 Lecture 12 30
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
Physics 208 Lecture 12 31
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
Physics 208 Lecture 12 32
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
Physics 208 Lecture 12 33
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
Physics 208 Lecture 12 34
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
Physics 208 Lecture 12 35
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
Physics 208 Lecture 12 36
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
Physics 208 Lecture 12 37
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
Physics 208 Lecture 12 38
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
Physics 208 Lecture 12 39
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
Physics 208 Lecture 12 40
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
Physics 208 Lecture 12 41
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
Physics 208 Lecture 12 42
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
Physics 208 Lecture 12 43
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
Physics 208 Lecture 12 44
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
Physics 208 Lecture 12 45
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
Physics 208 Lecture 12 46
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
Physics 208 Lecture 12 47
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
Physics 208 Lecture 12 48
Tues. Oct. 11, 2007
Physics 208 Lecture 12 49
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
Physics 208 Lecture 12 50
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
Physics 208 Lecture 12 51
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
Physics 208 Lecture 12 52
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
Physics 208 Lecture 12 53
Spherical capacitor
Capacitance ofSpherical Capacitor:
Capacitance ofCylindrical Capacitor:
Tues. Oct. 11, 2007
Physics 208 Lecture 12 54
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
Physics 208 Lecture 12 55
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
Physics 208 Lecture 12 56
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
Physics 208 Lecture 12 57
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
Physics 208 Lecture 12 58
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
Physics 208 Lecture 12 59
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
Physics 208 Lecture 12 61
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
Physics 208 Lecture 12 62
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
Physics 208 Lecture 12 63
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
Physics 208 Lecture 12 64
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
Physics 208 Lecture 12 65
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
Physics 208 Lecture 12 66
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
Physics 208 Lecture 12 67
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
Physics 208 Lecture 12 68
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
Physics 208 Lecture 12 69
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
Physics 208 Lecture 12 70
Equipotential lines
Lines of constant potential In 3D, surfaces of constant potential
Tues. Oct. 11, 2007
Physics 208 Lecture 12 71
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
Physics 208 Lecture 12 72
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
Physics 208 Lecture 12 73
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
Physics 208 Lecture 12 74
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
Physics 208 Lecture 12 75
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
Physics 208 Lecture 12 76
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
Physics 208 Lecture 12 77
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
Physics 208 Lecture 12 78
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
Physics 208 Lecture 12 79
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
Physics 208 Lecture 12 80
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
Physics 208 Lecture 12 81
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
Physics 208 Lecture 12 82
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
Physics 208 Lecture 12 83
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
Physics 208 Lecture 12 84
Resistivity
Resistivity
SI units Ω-m
€
=RA
LIndependent of sample geometry