september 29, 2008 capacitors how did you do? a. great b. ok c. poor d. really bad e. i absolutely...
TRANSCRIPT
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September 29, 2008
CAPACITORS
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How did you do?
A. GreatB. OKC. PoorD. Really badE. I absolutely
flunked!
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Calendar of the DayCalendar of the DayExams will be returned within a week.
If you did badly in the exam you need to have a plan to succeed. Let me know if you want any help with this.
Quiz on Friday – Potential or Capacitance.WebAssign will appear shortly if it hasn’t
done so already.There is a WA on board for potential.Quizzes are in the bin on the third floor
through the double doors.
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Two +q charges are separated by a distance d. What is the potential at a point midway between the charges on the line connecting them
A. ZeroB. Kq/dC. Kq/dD. 2Kq/dE. 4kq/d
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Capacitors
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A simple Capacitor
Remove the battery Charge Remains on the
plates. The battery did WORK to
charge the plates That work can be
recovered in the form of electrical energy – Potential Difference
WIRES
TWO PLATES
Battery
WIRES
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INSIDE THE DEVICE
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Two Charged Plates(Neglect Fringing Fields)
d
Air or Vacuum
Area A
- Q +QE
V=Potential Difference
Symbol
ADDED CHARGE
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Where is the charge?
d
Air or Vacuum
Area A
- Q +QE
V=Potential Difference
------
++++++
AREA=A
=Q/A
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One Way to Charge:
Start with two isolated uncharged plates.Take electrons and move them from the + to
the – plate through the region between.As the charge builds up, an electric field
forms between the plates.You therefore have to do work against the
field as you continue to move charge from one plate to another.
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Capacitor
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More on Capacitors
000
0
0
0
)/(
0
AQ
A
QE
EAQ
QEAAEA
qd
Gauss
AE
d
Air or Vacuum
Area A
- Q +QE
V=Potential Difference
GaussianSurface
Same result from other plate!
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DEFINITION - CapacityThe Potential Difference is
APPLIED by a battery or a circuit.
The charge q on the capacitor is found to be proportional to the applied voltage.
The proportionality constant is C and is referred to as the CAPACITANCE of the device.
CVq
orV
qC
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UNITSUNITS
A capacitor which acquires a charge of 1 coulomb on each plate with the application of one volt is defined to have a capacitance of 1 FARAD
One Farad is one Coulomb/Volt
CVq
orV
qC
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The two metal objects in the figure have net charges of +79 pC and -79 pC, which result in a 10 V potential difference between them.
(a) What is the capacitance of the system? [7.9] pF(b) If the charges are changed to +222 pC and -222 pC, what does the capacitance become? [7.9] pF(c) What does the potential difference become?[28.1] V
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NOTEWork to move a charge from one side of a capacitor to the other is qEd.
Work to move a charge from one side of a capacitor to the other is qV
Thus qV=qEdE=V/d As before
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Continuing…
d
AC
sod
AVEAAq
V
qC
0
00
The capacitance of a
parallel plate capacitor depends only on the Area and separation between the plates.
C is dependent only on the geometry of the device!
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Units of Units of 00
mpFmF
andm
Farad
Voltm
CoulombVoltCoulombm
Coulomb
Joulem
Coulomb
Nm
Coulomb
/85.8/1085.8 120
2
2
2
2
0
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Simple Capacitor CircuitsBatteries
Apply potential differencesCapacitorsWires
Wires are METALS.Continuous strands of wire are all at the same
potential.Separate strands of wire connected to circuit
elements may be at DIFFERENT potentials.
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Size Matters!A Random Access Memory stores
information on small capacitors which are either charged (bit=1) or uncharged (bit=0).
Voltage across one of these capacitors ie either zero or the power source voltage (5.3 volts in this example).
Typical capacitance is 55 fF (femto=10-15)Probably less these days!
Question: How many electrons are stored on one of these capacitors in the +1 state?
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Small is better in the IC world!
electronsC
VF
e
CV
e
qn 6
19
15
108.1106.1
)3.5)(1055(
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October 1, 2008
Cap-II
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Note:I do not have the grades yet. Probably by
Friday.Quiz on Friday … Potential or Capacitors.Watch WebAssign for new stuff.
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Last Time
We defined capacitance:C=q/VQ=CV
We showed thatC=0A/d
AndE=V/d
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TWO Types of Connections
SERIES
PARALLEL
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Parallel Connection
VCEquivalent=CE
321
321
321
33
22
1111
)(
CCCC
therefore
CCCVQ
qqqQ
VCq
VCq
VCVCq
E
E
E
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Series Connection
V C1 C2
q -q q -q
The charge on eachcapacitor is the same !
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Series Connection Continued
21
21
21
111
CCC
or
C
q
C
q
C
q
VVV
V C1 C2
q -q q -q
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More GeneralMore General
ii
i i
CC
Parallel
CC
Series
11
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Example
C1 C2
V
C3
C1=12.0 fC2= 5.3 fC3= 4.5 d
(12+5.3)pf
series
(12+5.3)pf
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More on the Big CWe move a charge
dq from the (-) plate to the (+) one.
The (-) plate becomes more (-)
The (+) plate becomes more (+).
dW=Fd=dq x E x d+q -q
E=0A/d
+dq
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So….
2222
0
2
0
2
0 0
0
00
2
1
22
)(
1
22
1
1
CVC
VC
C
QU
ord
Aq
A
dqqdq
A
dUW
dqdA
qdW
A
qE
Gauss
EddqdW
Q
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Not All Capacitors are Created Equal
Parallel Plate
CylindricalSpherical
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Spherical Capacitor
???
4)(
4
02
0
2
0
surprise
r
qrE
qEr
qd
Gauss
AE
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Calculate Potential Difference V
drr
qV
EdsV
a
b
platepositive
platenegative
20
.
.
1
4
(-) sign because E and ds are in OPPOSITE directions.
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Continuing…
ab
ab
V
qC
ab
abq
ba
qV
r
q
r
drqV
b
a
0
00
02
0
4
4
11
4
)1
(44
Lost (-) sign due to switch of limits.
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A Thunker
If a drop of liquid has capacitance 1.00 pF, what is its radius?
STEPS
Assume a charge on the drop.Calculate the potentialSee what happens
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In the drawing below, find the equivalent capacitance of the combination. Assume that C1 = 8 µF, C2 = 4 µF, and C3 = 3 µF.
5.67µF
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In the diagram, the battery has a potential difference of 10 V and the five capacitors each have a capacitance of 20 µF.
What is the charge on( a) capacitor C1 and (b) capacitor C2?
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In the figure, capacitors C1 = 0.8 µF and C2 = 2.8 µF are each charged to a potential difference of V = 104 V, but with opposite polarity as shown. Switches S1 and S2 are then closed.
(a) What is the new potential difference between points a and b? 57.8 VWhat are the new charges on each capacitor?(b)46.2µC (on C1)(c)162µC (on C2)
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Anudder ThunkerAnudder ThunkerFind the equivalent capacitance between points a and b in the combination of capacitors shown in the figure.
V(ab) same across each
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DIELECTRIC
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Polar Materials (Water)
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Apply an Electric Field
Some LOCAL ordering Larger Scale Ordering
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Adding things up..
- +Net effect REDUCES the field
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Non-Polar Material
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Non-Polar Material
Effective Charge isREDUCED
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We can measure the C of a capacitor (later)
C0 = Vacuum or air Value
C = With dielectric in place
C=C0
(we show this later)
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How to Check This
Charge to V0 and then disconnect fromThe battery.C0 V0
Connect the two togetherV
C0 will lose some charge to the capacitor with the dielectric.We can measure V with a voltmeter (later).
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Checking the idea..
00
0
000
210
2
01
000
1 CV
VCC
CVVCVC
qqq
CVq
VCq
VCq
V
Note: When two Capacitors are the same (No dielectric), then V=V0/2.
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Messing with Capacitors
+
V-
+
V-
+
-
+
-
The battery means that thepotential difference acrossthe capacitor remains constant.
For this case, we insert the dielectric but hold the voltage constant,
q=CV
since C C0
qC0V
THE EXTRA CHARGE COMES FROM THE BATTERY!
Remember – We hold V constant with the battery.
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Another CaseWe charge the capacitor to a voltage V0.
We disconnect the battery.We slip a dielectric in between the two
plates.We look at the voltage across the capacitor
to see what happens.
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No Battery
0
0000
0
VV
or
VCqVCq
VCq
+
-
+
-
q0
q
q0 =C0Vo
When the dielectric is inserted, no chargeis added so the charge must be the same.
V0
V
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Another Way to Think About This
There is an original charge q on the capacitor.
If you slide the dielectric into the capacitor, you are adding no additional STORED charge. Just moving some charge around in the dielectric material.
If you short the capacitors with your fingers, only the original charge on the capacitor can burn your fingers to a crisp!
The charge in q=CV must therefore be the free charge on the metal plates of the capacitor.
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A Closer Look at this stuff..
00
00
00
Vd
AVCq
d
AC
Consider this virgin capacitor.No dielectric experience.Applied Voltage via a battery.
C0
++++++++++++
------------------
V0
q
-q
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Remove the Battery
++++++++++++
------------------
V0
q
-q
The Voltage across thecapacitor remains V0
q remains the same aswell.
The capacitor is fat (charged),dumb and happy.
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Slip in a DielectricAlmost, but not quite, filling the space
000
0
....
A
qE
qd
gapsmallin
AE
++++++++++++
------------------
V0
q
-q
- - - - - - - -
+ + + + + +
-q’
+q’
E0
E
E’ from inducedcharges
Gaussian Surface
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A little sheet from the past..
A
q
A
qE
A
qE
dialectricsheet
sheet
00/
00
'
2
'2
2
'
2
+++
---q-q
-q’ +q’
0 2xEsheet 0
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Some more sheet…
A
qqE
so
A
qE
A
qE echdielectric
0
00
0arg
'
'
![Page 61: September 29, 2008 CAPACITORS How did you do? A. Great B. OK C. Poor D. Really bad E. I absolutely flunked!](https://reader036.vdocuments.mx/reader036/viewer/2022062308/56649e4f5503460f94b466ef/html5/thumbnails/61.jpg)
A Few slides backNo Battery
+
-
+
-
q0
q
q=C0Vo
When the dielectric is inserted, no chargeis added so the charge must be the same.
0
0000
0
VV
or
VCqVCq
VCq
V0
V
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From this last equation
0
00
00
0
1
EE
E
E
V
V
thus
dEV
EdV
and
VV
![Page 63: September 29, 2008 CAPACITORS How did you do? A. Great B. OK C. Poor D. Really bad E. I absolutely flunked!](https://reader036.vdocuments.mx/reader036/viewer/2022062308/56649e4f5503460f94b466ef/html5/thumbnails/63.jpg)
Another look
d
V
A
Qd
VE
FieldElectricd
AVVCQ
d
AC
PlateParallel
0000
00
00000
00
+
-
Vo
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Add Dielectric to Capacitor
Original Structure
Disconnect Battery
Slip in Dielectric
+
-
Vo
+
-
+
-
V0
Note: Charge on plate does not change!
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What happens?
0
00 1
VEdV
andd
VEE
00 /
CV
Q
V
QC
+
-
ii
oo
Potential Difference is REDUCEDby insertion of dielectric.
Charge on plate is Unchanged!
Capacitance increases by a factor of as we showed previously
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SUMMARY OF RESULTS
0
0
0
EE
CC
VV
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APPLICATION OF GAUSS’ LAW
qqq
and
A
qE
E
A
qqE
A
qE
'
'
0
0
0
00
![Page 68: September 29, 2008 CAPACITORS How did you do? A. Great B. OK C. Poor D. Really bad E. I absolutely flunked!](https://reader036.vdocuments.mx/reader036/viewer/2022062308/56649e4f5503460f94b466ef/html5/thumbnails/68.jpg)
New Gauss for Dielectrics
0
0
sometimes
qd freeAE