Download Chapter 26

Last time…
Fields, forces, work, and potential
+
+
Electric forces and work
Electric potential energy
and electric potential
Tues. Oct. 9, 2007
Physics 208 Lecture 11
1
30
Exam 1 percentage
MEAN=79%
25
# SCORES
Exam 1
20
15
10
5
0
20


30
40
50
60
70
80
90 100
PERCENTAGE SCORE
Average = 79%
Letter grades indicate how you should
interpret this percentage:

Average is at the B/BC border.
Tues. Oct. 9, 2007
Physics 208 Lecture 11
2
Last time…
Fields, forces, work, and potential
+
+
Electric forces and work
Electric potential energy
and electric potential
Tues. Oct. 9, 2007
Physics 208 Lecture 11
3
Electric field from potential

Said before that dW  Fext  ds  FCoulomb  ds 
dV  E  ds


Spell out the vectors:
 for
This works
dV  E x dx  E y dy  E z dz
dV
dV
dV
Ex  
, Ey  
, Ez  
dx
dy
dz

Usually written

Tues. Oct. 9, 2007
dV dV dV 
E  V   ,
,

 dx dy dz 
Physics 208 Lecture 11
4
Quick Quiz
Suppose the electric potential is constant
everywhere. What is the electric field?
A) Positive
B) Negative
C) Increasing
D) Decreasing
E) Zero
Tues. Oct. 9, 2007
Physics 208 Lecture 11
5
Potential from electric field
dV  E  d



Electric field can be
V  Vo  E d
d
used to find changes V  V d
o
in potential
Potential changes
E
largest in direction of



E-field.

V=Vo
d
Smallest (zero)
perpendicular to
 V  Vo  E d
E-field
Tues. Oct. 9, 2007

Physics 208 Lecture 11
6
Equipotential lines


Lines of constant potential
In 3D, surfaces of constant potential
Tues. Oct. 9, 2007
Physics 208 Lecture 11
7
Quick Quiz
How does the electric potential outside a uniform
infinite sheet of positive charge vary with distance
from the sheet?
A. Is constant
B. Increasing as (distance)1
C. Decreasing as (distance)1
D. Increasing as (distance)2
E. Decreasing as (distance)2
Tues. Oct. 9, 2007
Physics 208 Lecture 11
8
E 
V
V VA
 B
d
d
Electric Potential - Uniform Field
V  E ds  VB  VA 
B
   E ds  E
A

E || ds
B
B
 E  ds
A
 ds  E x
B
 xA 
A
E cnst

B
A

x
Tues. Oct. 9, 2007
Constant E-field corresponds to linearly
decreasing (in direction of E) potential
Particle gains kinetic energy equal to the
potential energy lost
Physics 208 Lecture 11
9
Check of simple cases

Previous quick quiz: uniform potential
corresponds to zero electric field
E  V  constant   0

Constant electric field corresponds to
linear potential
 



E  V  Ex    Ex, Ex, Ex  Exˆ
y
z 
x
Tues. Oct. 9, 2007
Physics 208 Lecture 11
10
Complicated check: point charge
E  V


E points opposite
to direction of
steepest slope
Magnitude
proportional to
local slope
Tues. Oct. 9, 2007
Physics 208 Lecture 11
y
+Q
x
11
Potential of spherical conductor



Charge resides on surface,
so this is like the spherical charge shell.
Found E = keQ / R2 in the radial direction.
What is the electric potential of the conductor?
V R  V  
R
 E  ds


difficult
path

 E  ds
easy
path
R
Integral along
some path,
from point on
surface to inf.
Tues. Oct. 9, 2007
Easy because is same
direction as E,
E  ds  E dr  Edr
Physics 208 Lecture 11
12
Quick quiz
V R  V  


R

Q
E dr   k 2 dr
r
R

Q
Q
 k
k
rR
R
So conducting sphere of
radius R carrying charge Q
is at a potential kQ/ R
Two conducting spheres of diff radii connected by long
conducting wire. What is approximatelytrue of Q1, Q2?
R1
Q1
Q2
R2
A) Q2>Q1
B) Q2<Q1
C) Q2=Q1
Tues. Oct. 9, 2007
Physics 208 Lecture 11
13
Connected spheres

Since both must be at the same potential,
kQ1 kQ2
Q1 R1



R1
R2
Q2 R2
Surface charge densities?

Q
1 R2



2
4R
 2 R1
Electric field?


Since E 
,
2o
Tues. Oct. 9, 2007
Charge proportional
to radius
Surface charge density
proportional to 1/R
Local E-field proportional to 1/R
(1/radius of curvature)
Physics 208 Lecture 11
14
Varying E-fields on conductor



Expect larger electric fields near the small end. Electric field
approximately proportional to 1/(local radius of curvature).
Large electric fields at sharp points,
just like square (done numerically previously)
Fields can be so strong that air ionized and ions accelerated.
Tues. Oct. 9, 2007
Physics 208 Lecture 11
15
Potential and charge


Have shown that a conductor has an electric
potential, and that potential depends on its charge
For a charged conducting sphere:
+ + +
+
+
+
+
+
+
+ +
Tues. Oct. 9, 2007
Q k
V R V   k  Q
R R
Electric potential proportional
to total charge
Physics 208 Lecture 11
16
Quick Quiz
Consider this conducting object. When it has total
charge Qo, its electric potential is Vo. When it has
charge 2Qo, its electric potential
A. is Vo
B. is 2Vo
C. is 4Vo
D. depends on shape
Tues. Oct. 9, 2007
Physics 208 Lecture 11
17
Capacitance

Electric potential of any conducting object
proportional to its total charge.
1
V Q
C
C = capacitance
  Large capacitance: need lots of charge to change potential


Small capacitance: small charge can change potential.
Tues. Oct. 9, 2007
Physics 208 Lecture 11
18
Capacitors

Where did the charge come from?


Usually transferred from another conducting
object, leaving opposite charge behind
A capacitor consists of two conductors




Conductors generically called ‘plates’
Charge transferred between plates
Plates carry equal and opposite charges
Potential difference between plates
proportional to charge transferred Q
Tues. Oct. 9, 2007
Physics 208 Lecture 11
19
Definition of Capacitance

Same as for single conductor
1
V  Q
C





but V = potential difference between plates
Q = charge transferred between plates
The SIunit of capacitance is the farad (F)
1 Farad = 1 Coulomb / Volt
This is a very large unit: typically use
mF = 10-6 F, nF = 10-9 F, pF = 10-12 F
Tues. Oct. 9, 2007
Physics 208 Lecture 11
20
Parallel plate capacitor +Q
outer



-Q
inner
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
d
Tues. Oct. 9, 2007
Physics 208 Lecture 11
21
How did the charge get transferred?


Battery has fixed electric potential difference
across its terminals
Conducting plates connected to battery
terminals by conducting wires.

Vplates = Vbattery across plates

Electrons move

V

from negative battery terminal to -Q plate

from +Q plate to positive battery terminal
This charge motion requires work

The battery supplies the work
1
V  Q
C
Tues. Oct. 9, 2007
Physics 208 Lecture 11
22
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. 9, 2007
Physics 208 Lecture 11
23
What is the potential difference?

Electric field between plates
E left  E right   /2o   /2o   /o

Uniform electric field
Potential difference = V+-V= (1/q)x(- work to move + charge
from + to minus plate)
 1/q  qEd
 d 
V  Ed  d /o  Q 
o A 
Tues. Oct. 9, 2007
Physics 208 Lecture 11
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+Q
Etotal
d
-
-Q
24
What is the capacitance?
-Q
 d 
V  V  V  Q 
o A 
+Q
V  Q /C
C
o A
d
This is a geometrical factor
d
Tues. Oct. 9, 2007
Physics 208 Lecture 11
25
Human capacitors

Cell membrane:

100 µm


‘Empty space’ separating
charged fluids (conductors)
~ 7 - 8 nm thick
In combination w/fluids, acts as
parallel-plate capacitor
Extracellular fluid
Plasma membrane
Cytoplasm
Tues. Oct. 9, 2007
Physics 208 Lecture 11
26
Modeling a cell membrane
A-
K+

Extracellular fluid
+ + + + + +
7-8
nm

V~0.1 V
Plasma membrane
- - - - - -

Cytoplasm
Na+

Cl-
Ionic charge at surfaces of
conducting fluids
Capacitance:
o A
d
Tues. Oct. 9, 2007
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
100 µm sphere ~ 3x10-4 cm2
surface area
8.85 10


12
F /m4  50 10 m
6
8 109 m
Physics 208 Lecture 11
2
 3.5 1011 F  35 pF
~0.1µF/cm2
27
Cell membrane depolarization
K+
A-

Cell membrane can reverse potential
by opening ion channels.

Potential change ~ 0.12 V
Extracellular fluid
- +
- +
- +
- +
- +
+
7-8
nm

Ions flow through ion channels
V~0.1
V
V~-0.02
V
Plasma membrane
Channel spacing ~ 10xmembrane
thickness (~ 100 channels / µm2 )
+
- +
- +
- +
- +
- +
Cytoplasm

How many ions flow through each
channel?
Na+
Cl-
Charge xfer required Q=CV=(35 pF)(0.12V) =(35x10-12 C/V)(0.12V)
= 4.2x10-12 Coulombs
1.6x10-19 C/ion -> 2.6x107 ions flow
(100 channels/µm2)x4(50 µm)2=3.14x106 ion channels
Ion flow / channel =(2.6x107 ions) / 3.14x106 channels ~ 7 ions/channel
Tues. Oct. 9, 2007
Physics 208 Lecture 11
28
Cell membrane as dielectric
K+
A-
Extracellular fluid
+ + + + + +
7-8
nm
Plasma membrane

Membrane is not really
empty

It has molecules inside that
respond to electric field.

The molecules in the
membrane can be polarized
- - - - - Cytoplasm
Na+
Cl-
Dielectric: insulating materials can respond to an electric
field by generating an opposing field.
Tues. Oct. 9, 2007
Physics 208 Lecture 11
29
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=0
Tues. Oct. 9, 2007
+
E
Physics 208 Lecture 11
30
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/ kand V = V0/ k

The capacitance increases
C = k C0
Tues. Oct. 9, 2007
Physics 208 Lecture 11
Eind
E0
31
Cell membrane as dielectric
K+
A-
Extracellular fluid
+ + + + + +
7-8
nm
Plasma membrane
- - - - - -

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
Na+
A. 70
Cl-
B. 7
C. 0.7
C increases by factor of 10
10 times as much charged needed to reach potential
Tues. Oct. 9, 2007
Physics 208 Lecture 11
32
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:



-
-Q
-
+
+
+Q
+
+
+
-
-
-
E-field zero inside outer cond.
E-field zero outside outer cond.
No flux ->
no charge on outer surface!
Tues. Oct. 9, 2007
Physics 208 Lecture 11
33
Spherical capacitor
Charge Q moved from outer to inner sphere
Gauss’ law says E=kQ/r2
until second sphere
Potential difference V 
b
 E  ds
a
Along path shown
Gaussian surface
to find E
b
1 1 
kQ
1
V   2  kQ  kQ  
a b 
ra
a r
b

1 1 1
Q
C
 k  
a b 
V
Path to find V
Tues. Oct. 9, 2007
Physics 208 Lecture 11
+ + +
+
+
+
+ + +
34
Work done and energy stored

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:
Tues. Oct. 9, 2007
U = 1/2 oAdE2
Physics 208 Lecture 11
35
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. 9, 2007
Physics 208 Lecture 11
36
-q
Quick Quiz 1
pull
-
+
-
+
-
A parallel plate capacitor given a charge q.
The plates are then pulled a small distance
further apart. Which of the following apply to
the situation after the plates have been
moved?
-
d
+
+q
pull
+
1)The charge decreases
2)The capacitance increases
3)The electric field increases
C = 0A/d  C decreases!
E= Q/(0A)  E constant
4)The voltage between the
plates increases
V= Ed  V increases
5)The energy stored in the
capacitor increases
U= QV / 2
Q constant, V increased
 U increases
Tues. Oct. 9, 2007
Physics 208 Lecture 11
37
Capacitors in Parallel
 0A
C
d
Both ends connected together by wire
• Same voltage: V1 = V2 = Veq
 Add Areas:
Ceq = C1+C2
 Share Charge: Qeq = Q1+Q2

15 V
15 V
C1
C2
10 V
10 V
Tues. Oct. 9, 2007
Physics 208 Lecture 11
15 V
Ceq
10 V
38
Capacitors in Series

Same Charge: Q1 = Q2 = Qeq

Share Voltage:V1+V2=Veq

Add d:
+Q
+
+
+
++
-Q
+Q
-Q
Tues. Oct. 9, 2007
--++
+
+
++
- +
-+
+
+
+
 0A
C
d
1
1
1
 
Ceq C1 C2
C1
+Q
+
Ceq
C2
-Q
Physics 208 Lecture 11
-
39
Tues. Oct. 9, 2007
Physics 208 Lecture 11
40
Electric Dipole alignment
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: t = p x E
of magnitude:t= Fa sin  pE sin 
 The potential energy is = work done by the torque to rotate
dipole: dW = t d
U = - pE cos  = - p · E
When the dipole is aligned to
the field it is minimum
U = -pE equilibrium!

Tues. Oct. 9, 2007
Physics 208 Lecture 11
41
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. 9, 2007
Physics 208 Lecture 11
42
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. 9, 2007
Physics 208 Lecture 11
43
Capacitance of Parallel Plate
Capacitor
The electric field from a charged plane of charge per unit area
 = Q/A is E = /20
For 2 planes of opposite charge
V
E= /0 = Q/(0A)
+E
A
+
Tues. Oct. 9, 2007
-A
-
E+
E+
E+
E-
E-
E-
d
44 2
0=1/(4ke)=8.85x10-12 C2/Nm
Physics 208 Lecture 11
Spherical capacitor
Capacitance of
Spherical Capacitor:
Capacitance of
Cylindrical Capacitor:
Tues. Oct. 9, 2007
Physics 208 Lecture 11
45
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
Charge Q on plates
V =VA – VB = +E0 d
+
+
+
+
+
E=E0
d
Charge 2Q on plates
VA – VB = +2E0 d
+
+
+
+
+
+
+
+
+
-
E=2E0
d
-
Potential difference is proportional to charge:
Double Q  Double V
EQ, VE, QV
Tues. Oct. 9, 2007
Physics 208 Lecture 11
46
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
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.


Due to the ion distribution between the inside and outside of
living cells there is a potential difference called resting
potential
Tues.
Oct. 9, 2007
Physics 208 Lecture 11
47
http://www.cytochemistry.net/Cell-biology/membrane.htm
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=%2
2membrane%20channels%22
Tues. Oct. 9, 2007
Physics 208 Lecture 11
48
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 mm2 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. 9, 2007
Physics 208 Lecture 11
49
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 mC/cm2

surface density of channels = C = 10 channels/mm2 = 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 = /t = (10-7 C/cm2)/(10-3 s) = 100 mA/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. 9, 2007
Physics 208 Lecture 11
50

Honors lecture this Friday



Superconductivity, by yours truly.
12:05 pm, 2241 Chamberlin
Everyone welcome!
HW 2 due Thursday midnite
Lab 2 this week - bring question sheet
available on course web site.
Tues. Oct. 9, 2007
Physics 208 Lecture 11
51
Capacitance and Dielectrics






1.
2.
This 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:
Gauss’ Law and applications
Electric field calculations from Potential
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. 9, 2007
Physics 208 Lecture 11
53
Capacitors with Dielectrics

Placing a dielectric between
the plates increases the
capacitance:
Dielectric
constant (k > 1)
C = k C0
Capacitance with
dielectric

Capacitance
without dielectric
The dielectric reduces the
potential difference
V = V0/ k
Tues. Oct. 9, 2007
Physics 208 Lecture 11
54
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. 9, 2007
Physics 208 Lecture 11
55
The Electric Field




qo dV  FCoulomb  d 
F

dV   Coulomb  d
 qo 
E is the Electric Field
 E  d
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,
E = Coulomb force per unit charge on a test charge due to
interaction with the other charges.
We’ll see later that E-fields in
electromagnetic waves exist w/o charges!
Tues. Oct. 9, 2007
Physics 208 Lecture 11
56
Electric field and potential
Said before that

dV  E  d
Electric field strength/direction shows how the potential
changesin different directions
For example,



Potential decreases in direction of local E field at rate  E
Potential increases in direction opposite to local E-field at rate 
potential constant in direction perpendicular to local E-field

Tues. Oct. 9, 2007
Physics 208 Lecture 11
E  d

0
E

57
Potential from electric field
dV  E  d



Electric field can be
V  Vo  E d
d
used to find changes V  V d
o
in potential
Potential changes
E
largest in direction of



E-field.

V=Vo
d
Smallest (zero)
perpendicular to
 V  Vo  E d
E-field
Tues. Oct. 9, 2007

Physics 208 Lecture 11
58
Quick Quiz 3

Suppose the electric potential is
constant everywhere. What is the
electric field?
A) Positive
B) Negative
C) Zero
Tues. Oct. 9, 2007
Physics 208 Lecture 11
59
E 
V
V VA
 B
d
d
Electric Potential - Uniform Field
+ V  E ds  V  V 
B
A
B
B
A
A
   E ds  E

E || ds
B
 E  ds
A
 ds  Ex
E cnst

B
A



x
Tues. Oct. 9, 2007
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
Physics 208 Lecture 11
60
Electric field from potential

Said before that dV  E  d
Spell out the vectors:

 for
This works

dV  E x dx  E y dy  E z dz
dV
dV
dV
Ex  
, Ey  
, Ez  
dx
dy
dz

Usually written

Tues. Oct. 9, 2007
dV dV dV 
E  V   ,
,

 dx dy dz 
Physics 208 Lecture 11
61
Equipotential lines


Lines of constant potential
In 3D, surfaces of constant potential
Tues. Oct. 9, 2007
Physics 208 Lecture 11
62
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
Tues. Oct. 9, 2007
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
Physics
208 Lecture
11
63
The Electric Field




qo dV  FCoulomb  d 
F

dV   Coulomb  d
 qo 
E is the Electric Field
 E  d
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,
E = Coulomb force per unit charge on a test charge due to
interaction with the other charges.
We’ll see later that E-fields in
electromagnetic waves exist w/o charges!
Tues. Oct. 9, 2007
Physics 208 Lecture 11
64
Electric field and potential
Said before that

dV  E  d
Electric field strength/direction shows how the potential
changesin different directions
For example,



Potential decreases in direction of local E field at rate  E
Potential increases in direction opposite to local E-field at rate 
potential constant in direction perpendicular to local E-field

Tues. Oct. 9, 2007
Physics 208 Lecture 11
E  d

0
E

65
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
Tues. Oct. 9, 2007
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
Physics
208 Lecture
11
66
Qinner
Qouter


Q = Qinner+Qouter
Total E-field = Eleft plate+Eright plate
In between plates,


Q
Eleft plate  inner  outer 
2o
2o
2Ao
 inner  outer
Q
Er ight plate 


2o
2o
2Ao
+
+
+
+
+

Q
Etotal  Eleft plate  Er ight plate 
o A
r ight plate
Q
Qd
V  E  ds 
dx 
+
left plate
+


o A
o A
C
Tues. Oct. 9, 2007
Q o A

V
d
Physics 208 Lecture 11
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+Q
-
Eleft plate +
Eright plate =
Etotal
-
-
-
d
-Q
67