Transcript Chapter 26
Last time…
Fields, forces, work, and potential
+
+
Electric forces and work
Potential energy stored
in electric field
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Physics 208 Lecture 10
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Work, KE, and potential energy
When particle is not isolated,
W external K U
Work done
on system
Change in
kinetic energy
Change in
electric potential energy
U qE r
Works
for
constant
electric
field
if
Only electric potential energy difference
point is chosen
Sometimes a reference
E.g. Ur 0 at r (0,0,0)
Then U r qE r
for uniform electric field
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Electric potential V
Electric potential difference V is the electric
potential energy / unit charge = U/q
For uniform electric field,
U r qE r
V r
E r
q
q
This is only valid for a uniform electric field
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Check for uniform E-field
Push particle against E-field, or across E-field
Which requires work?
+
Increasing electric
potential in this direction
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Constant electric potential
in this direction
+
Decreasing electric
potential in this direction
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Quick Quiz
Two points in space A and B have electric potential
VA=20 volts and VB=100 volts. How much work
does it take to move a +100µC charge from A to
B?
A. +2 mJ
B. -20 mJ
C. +8 mJ
D. +100 mJ
E. -100 mJ
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Potential from electric field
dV E d
V Vo
d
d
E
V Vo E d
d
V=Vo
V Vo E d
dV largest in direction of E-field.
dV smallest (zero)
perpendicular to E-field
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Electric potential: general
U
F
Coulomb
ds
qE ds q E ds
Electric potential energy difference U
proportional to charge q that work is done on
U /q V Electric potential difference
E ds
Depends only on charges that create E-fields
Electric field usually created
by some charge
distribution.
V(r) is electric potential of that charge distribution
V has units of Joules / Coulomb = Volts
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Electric potential of point charge
kQ
Electric field from point charge Q is E 2 rˆ
r
What is the electric potential difference?
V
end
r final
E ds
start
k
Define V r 0
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rinitial
k
Q
dx
2
r
r final
Q
Q
Q
k
k
r rinitial
rinital
rfinal
Then
Q
V r k
for point charge
r
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Equipotential lines
Lines of constant potential
In 3D, surfaces of constant potential
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Topographic map
Each lines is
constant elevation
Same as constant
gravitational potential
gh (energy = mgh)
Height interval between lines constant
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Electric field from potential
Said before that dW Fext ds FCoulomb ds
dV E ds
Spell out the vectors:
for
This works
dV Ex dx Ey dy Ez dz
dV
dV
dV
Ex
, Ey
, Ez
dx
dy
dz
Usually written
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dV dV dV
E V ,
,
dx dy dz
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Quick Quiz
Suppose the electric potential is constant
everywhere. What is the electric field?
A) Positive
B) Negative
C) Increasing
D) Decreasing
E) Zero
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Electric Potential - Uniform Field
dV E ds
VB VA
B
B
B
E ds Exˆ ds
A
B
A
Edx E dx E x B x A
A
A
B
A
Constant E-field corresponds to linearly
decreasing (in direction of E) potential
Here V depends only on x, not on y
x
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Check of basic cases
Previous quick quiz: uniform potential
corresponds to zero electric field
E V constant 0
Linear potential corresponds to constant
electric field
E V Ex Ex, Ex, Ex Exˆ
y
z
x
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Potential and charge
Have shown that a conductor has an electric
potential, and that potential depends on its charge
For a charged conducting sphere:
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Q k
V R V k Q
R R
Electric potential proportional
to total charge
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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
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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.
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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
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Definition of Capacitance
Same as for single conductor
1
V Q
C
but V = potential difference between plates
Q = charge transferred between plates
SI unit of capacitance is farad (F) = 1 Coulomb / Volt
This is a very large unit: typically use
mF = 10-6 F, nF = 10-9 F, pF = 10-12 F
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How was charge 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
QCV
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Work done to charge a capacitor
Requires work to transfer charge dq from one plate:
q
dW Vdq dq
C
Total work = sum of incremental work
Q
W
0
q
Q2
dq
C
2C
Work done stored as potential energy in capacitor
Q2 1
1
2
U
QV CV
2C 2
2
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Example: Parallel plate capacitor
+Q
Charge Q moved from right outer
conductor to left conductor
-Q
inner
Charge only on inner surfaces
Plate surfaces are charge sheets,
each producing E-field
E left E right /2o /2o /o
Uniform field between plates
d
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Quick Quiz
Electric field between plates of infinite parallel-plate
capacitor has a constant value /o. What is the
field outside of the plates?
A. /o
B. /2o
C. - /2o
D. /4o
E. 0
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What is potential difference?
Potential difference = V+-V= - (work to move charge q
from + plate to plate) / q
qEd /q
d
V Ed d /o Q
o A
d
V V V Q
o A
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What is the capacitance?
d
V V V Q
o A
V Q /C
C
o A
d
-Q
This is a geometrical factor
+Q
Energy stored in parallel-plate capacitor
1
1 o A
1
2
2
U CV
Ed Ado E 2
2
2 d
Energy density
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U /Ad o E 2
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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
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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
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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
8.8510
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F /m4 5010 m
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8 109 m
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3.5 1011 F 35pF
~0.1µF/cm2
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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 )
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- +
- +
- +
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Cytoplasm
How many ions flow through each
channel?
Na+
Cl-
Charge xfer required Q=CV=(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
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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.
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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
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E=0
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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/ kand V = V0/ k
The capacitance increases
C = k C0
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E0
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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
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