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CAPACITORS. February, 2008. Capacitors Part I. 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. TWO PLATES. WIRES. WIRES. Battery. - PowerPoint PPT Presentation
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CAPACITORS
February, 2008
Capacitors Part I
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
INSIDE THE DEVICE
Two Charged Plates(Neglect Fringing Fields)
d
Air or Vacuum
Area A
- Q +QE
V=Potential Difference
Symbol
ADDED CHARGE
Where is the charge?
d
Air or Vacuum
Area A
- Q +QE
V=Potential Difference
------
++++++
AREA=A
=Q/A
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.
Capacitor
More on Capacitorsd
Air or Vacuum
Area A
- Q +QE
V=Potential Difference
GaussianSurface
000
0
0
0
)/(
0
AQ
A
QE
EAQ
QEAAEA
qd
Gauss
AE
Same result from other plate!
DEFINITIONDEFINITION - Capacity The 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
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
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
NOTE
Work 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=qEd E=V/d As before ( we omitted the
pesky negative sign but you know it is there, right?)
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!
Units of 0
mpFmF
andm
Farad
Voltm
CoulombVoltCoulombm
Coulomb
Joulem
Coulomb
Nm
Coulomb
/85.8/1085.8 120
2
2
2
2
0
pico
Simple Capacitor Circuits Batteries
Apply potential differences
Capacitors Wires
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.
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) Question: How many electrons are stored
on one of these capacitors in the +1 state?
Small is better in the IC world!
electronsC
VF
e
CV
e
qn 6
19
15
108.1106.1
)3.5)(1055(
TWO Types of Connections
SERIES
PARALLEL
Parallel Connection
VCEquivalent=CE
321
321
321
33
22
1111
)(
CCCC
therefore
CCCVQ
qqqQ
VCq
VCq
VCVCq
E
E
E
Series Connection
V C1 C2
q -q q -q
The charge on eachcapacitor is the same !
Series Connection Continued
21
21
21
111
CCC
or
C
q
C
q
C
q
VVV
V C1 C2
q -q q -q
More General
ii
i i
CC
Parallel
CC
Series
11
Example
C1 C2
V
C3
C1=12.0 fC2= 5.3 fC3= 4.5 d
(12+5.3)pf
series
(12+5.3)pf
More on the Big C We 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
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
Not All Capacitors are Created Equal
Parallel Plate
Cylindrical Spherical
Spherical Capacitor
???
4)(
4
02
0
2
0
surprise
r
qrE
qEr
qd
Gauss
AE
Calculate Potential Difference V
drr
qV
EdsV
a
b
platepositive
platenegative
20
.
.
1
4
(-) sign because E and ds are in OPPOSITE directions.
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.
Capacitor Circuits
ii
i i
CC
Parallel
CC
Series
11
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
Anudder Thunker
Find the equivalent capacitance between points a and b in the combination of capacitors shown in the figure.
V(ab) same across each
Thunk some more …
C1 C2
V
C3
C1=12.0 fC2= 5.3 fC3= 4.5 d
(12+5.3)pf
More on the Big C We 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
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
Sorta like (1/2)mv2
DIELECTRIC
Polar Materials (Water)
Apply an Electric Field
Some LOCAL ordering Larger Scale Ordering
Adding things up..
- +Net effect REDUCES the field
Non-Polar Material
Non-Polar Material
Effective Charge isREDUCED
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)
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).
Checking the idea..
V
00
0
000
210
2
01
000
1 CV
VCC
CVVCVC
qqq
CVq
VCq
VCq
Note: When two Capacitors are the same (No dielectric), then V=V0/2.
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.
Another Case
We 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.
No Battery
+
-
+
-
q0
q
q0 =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
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.
A Closer Look at this stuff..Consider this virgin capacitor.No dielectric experience.Applied Voltage via a battery.
C0
00
00
00
Vd
AVCq
d
AC
++++++++++++
------------------
V0
q
-q
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.
Slip in a DielectricAlmost, but not quite, filling the space
++++++++++++
------------------
V0
q
-q
- - - - - - - -
+ + + + + +
-q’
+q’
E0
E
E’ from inducedcharges
Gaussian Surface
000
0
....
A
qE
qd
gapsmallin
AE
A little sheet from the past..
+++
---q-q
-q’ +q’
A
q
A
qE
A
qE
dialectricsheet
sheet
00/
00
'
2
'2
2
'
2
0 2xEsheet 0
Some more sheet…
A
qqE
so
A
qE
A
qE echdielectric
0
00
0arg
'
'
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
From this last equation
0
00
00
0
1
EE
E
E
V
V
thus
dEV
EdV
and
VV
Another look
+
-
Vo
d
V
A
Qd
VE
FieldElectricd
AVVCQ
d
AC
PlateParallel
0000
00
00000
00
Add Dielectric to Capacitor Original Structure
Disconnect Battery
Slip in Dielectric
+
-
Vo
+
-
+
-
V0
Note: Charge on plate does not change!
What happens?
0
00 1
VEdV
andd
VEE
+
-
ii
oo
Potential Difference is REDUCEDby insertion of dielectric.
00 /
CV
Q
V
QC
Charge on plate is Unchanged!
Capacitance increases by a factor of as we showed previously
SUMMARY OF RESULTS
0
0
0
EE
CC
VV
APPLICATION OF GAUSS’ LAW
qqq
and
A
qE
E
A
qqE
A
qE
'
'
0
0
0
00
New Gauss for Dielectrics
0
0
sometimes
qd freeAE