Download 9/23/15 1 ANNOUNCEMENT Capacitors with a dielectric for Fixed

9/23/15
Capacitors with a dielectric for Fixed Amount of
Charge (Q is constant)
ANNOUNCEMENT
*Exam 1: Monday September 28, 2015, 8 PM - 10 PM
*Location: Elliot Hall of Music
*Covers all readings, lectures, homework from Chapters 21
through 23.
*The exam will be multiple choice (15-18 questions).
For Fixed Charge, the insertion of a dielectric
Increases the capacitance:
decreases the Electric Potential
*Practices exams can be found on the course website
and on CHIP.
decreases the Electric Field
*Be sure to bring your student ID card and your own
one-page (two-side) crib sheet.
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Ewithdielectric =
decreases the Electric Potential Energy
NOTE THAT FEW EQUATIONS WILL BE GIVEN – YOU ARE REMINDED
THAT IT IS YOUR RESPONSIBILITY TO CREATE WHATEVER TWO-SIDED
CRIB SHEET YOU WANT TO BRING TO THIS EXAM.
The equation sheet that will be given with the exam is
posted on the course homepage. Click on the link on the
left labeled “Equationsheet”
Vwithdielectric =
Cwithdielectric = KCo
Vo
K
Eo
K
U withdielectric =
Uo
K
The decrease in energy can be accounted for by the work done in moving
the dielectric slab between the plates of the capacitor.
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Capacitors with a dielectric connected to a
battery (V = Constant Potential).
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Chapter 25
For a constant applied potential, the insertion of a dielectric::
Cwithdielectric = KCo
increases the amount of charge on the plates
increases the Electric Potential Energy
• 
• 
• 
• 
• 
Qwithdielectric = Qo K
U withdielectric
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Qwithdielectric
∝
K
Current
Resistivity as a property of materials
Temperature dependence of resistivity
Emf
Power
The increase in energy occurs because work is done to maintain the
potential across the plates as the dielectric is inserted.
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Electrostatics versus Electric Current
Electric Current
Electrostatic Equilibrium
Steady-State
+
-
Battery
Note: Current arrow is drawn in direction in which positive charge carriers would
move, even if the actual charges are negative & move in the opposite direction.
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Electric Current
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Drift Velocity
•  Notice that flow of positive charge in one direction,
Becomes
more
negative
+
+
+
+
+
+
+
+
+
+
Becomes
more
positive
CURRENT I
•  It’s completely equivalent to negative charge in the
opposite direction.
Becomes
Becomes
more
more
- positive
negative
CURRENT I
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Microscopic View of an Electron in Motion
Drift Speed, Total Charge & Current
# charge carriers
volume
q = charge of each particle
vd = drift velocity
n=

with E 
without E
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Relationship between Current and Drift Speed
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Electric Current
Find vd for 14-gauge copper wire carrying a current of 1
A. Assume there is 1 free electron/atom.
Copper
n = natoms
ρN
=
M
10
one q in
 = 8.93 g/cm3
one q out
M = 63.55 g/mol
14 gauge wire
Area= 2.081 mm2"
vd =
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I
qnA
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Resistance
Resistance & Ohm’s Law
Resistance is a property
of the object, i.e . It
depends on the shape
and material
ohmic
Non-ohmic
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Resistivity
Resistivity is a property
of the material.
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Resistivity & Temperature Coefficients
area A
I
I
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Material
L
Ag
Cu
W
Si
Si, n-type
Si, p-type
glass
length
Resistivity "
(m)
1.6 x 10-8
1.7 x 10-8
5.5 x 10-8
640
8.7 x 10-4
2.8 x 10-3
1010–1014
Temp. coeff.
 (K-1)
3.8 x 10-3
3.9 x 10-3
4.5 x 10-3
–7.5 x 10-2
See TM Table 25-1 for more.
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Temperature Dependence of Resistances
ρ − ρo = ρoα (T − To )
R=ρ
L
A
Temperature Dependence
ρ = ρo (1 + α (T − To ))
R = Ro (1 + α (T − To ))
−3
Heated Tungsten Light Bulb filament at 3000 K: α = 4.5x10 / K
 for copper (Cu) as a function of temperature
Notice: Resisitivity increases as temperature increases.
This curve does not deviate greatly from a straight line.
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DEMO: Temperature Dependence
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DEMO: Temperature Dependence
lamp
liquid
nitrogen
~ 77oK
B
S
vacuum
bottle
Ge
Lower Cu initially at
room temperature (~ 300oK)
into liquid N2.
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Heat the Ge with a
candle.
semiconductor
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Temperature Stable Resistor
Power in Electric Circuits
A temperature-stable resistor is to be made by connecting a resistor
made of silicon in series with one made of nichrome. If the required
total resistance is 1300  in a wide temperature range around 20ºC,
what should be the resistances of the two resistors?
Rn
Rtotal = RN + RSi = 1300Ω
In general: R = Ro (1 + α (T − To ))
RoN (1 + α N (T − To )) + RoSi (1 + α Si (T − To )) = 1300Ω
dU = dqV = idtV
+
RSi
P=
For a Resistor: P=I2R = V2/R
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Power in Electric Circuits
Units of Power:
Volts Ampere=
Joule/second=Watt
Power associated with
dissipation of U into
thermal energy in the
resistor.
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Real Battery
Take a 100 W light bulb powered by 110 Volts (RMS AC).
What is the resistance of the (hot) filament?
We know P and V, don’t know I or R, are asked for R.
So we choose the last form below, and solve for R:
Examine potential as
we start from point b
and end at a:
internal
resistance
emf
ε − Ir − IR = 0
ε
I=
R+r
R = V2/P = 110 x 110/100 = 121Ω
In an ideal battery:
r=0
Va − Vb = ε
V
I=
R
2
V
P = IV = I 2 R =
R
Note that the cold filament will have ~13 times less resistance,
and therefore there will be a big surge in current as the bulb is
turned on. Often, used lightbulbs burn out at this instant.
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dU
= iV
dt
Power associated with transfer: P=IV
(RoN + RoSi ) + RoN α N (T − To ) + RoSiα Si (T − To )) = 1300Ω
(RoN α N + RoSiα Si )(T − To ) = 0
(RoN α N + RoSiα Si ) = 0
(1300Ω − RoSi )α N + RoSiα Si = 0
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-
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Note:

Over the battery:
Va − Vb = ε − Ir
arrows always points from negative to positive.
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Effect of Internal Resistance
r =0
(ideal battery)
real battery
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