Transcript Capacitors - UCF Physics

CAPACITORS
September 29, 2008
How did you do?
A. Great
B. OK
C. Poor
D. Really bad
E. I absolutely flunked!
Calendar of the Day
 Exams 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.
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. Zero
Kq/d
C. Kq/d
D. 2Kq/d
E. 4kq/d
B.
Capacitors
A simple Capacitor
TWO PLATES
WIRES
WIRES
 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
Battery
INSIDE THE DEVICE
Two Charged Plates
(Neglect Fringing Fields)
d
Air or Vacuum
E
-Q
Area A
V=Potential Difference
+Q
Symbol
ADDED CHARGE
Where is the charge?
- Q-
d
Air or Vacuum
E
Area A
V=Potential Difference
+
+
+
+
+
+ +Q
AREA=A
s=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 Capacitors
Gauss
d
q
 E  dA  
Air or Vacuum
-Q
E
+Q
Area A
V=Potential Difference
Gaussian
Surface
0
 EA  0 A   EA  
Q
0
Q   0 EA
Q
(Q / A) s
E


0 A
0
0
Same result from other plate!
DEFINITION - 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.
q
C
V
or
q  CV
UNITS
 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
q
C
V
or
q  CV
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
Continuing…
 The capacitance of a
q
C
V
q  sA   0 EA 
so
C
0 A
d
 0 AV
d
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
Coulomb 2
 0  
2
Coulomb 2

m  Joule
Nm
Coulomb 2

m  Coulomb  Volt
Coulomb Farad


m  Volt
m
and
 0  8.85 10
12
F / m  8.85 pF / m
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)
 Probably less these days!
 Question: How many electrons are stored on
one of these capacitors in the +1 state?
Small is better in the IC world!
q CV (55 1015 F )(5.3V )
6
n 


1
.
8

10
electrons
19
e
e
1.6 10 C
Cap-II
October 1, 2008
Note:
 I do not have the grades yet. Probably by Friday.
 Quiz on Friday … Potential or Capacitors.
 Watch WebAssign for new stuff.
Last Time
 We defined capacitance:
 C=q/V
 Q=CV
 We showed that
 C=0A/d
 And
 E=V/d
TWO Types of Connections
SERIES
PARALLEL
Parallel Connection
q1  C1V1  C1V
q2  C2V
q3  C3V
QE  q1  q2  q3
V
CEquivalent=CE
QE  V (C1  C2  C3 )
therefore
C E  C1  C2  C3
Series Connection
q
V
-q
C1
q
-q
C2
The charge on each
capacitor is the same !
Series Connection Continued
V  V1  V2
q
V
C1
-q
q
-q
C2
q
q
q


C C1 C 2
or
1
1
1


C C1 C 2
More General
Series
1
1

C
i Ci
Parallel
C   Ci
i
Example
C1
C2
(12+5.3)pf
V
C3
C1=12.0 uf
C2= 5.3 uf
C3= 4.5 ud
More on the Big C
 We move a charge dq from
E=0A/d
the (-) plate to the (+) one.
 The (-) plate becomes more
(-)
 The (+) plate becomes more
(+).
 dW=Fd=dq x E x d
+dq
+q
-q
dW  dq  Ed
Gauss
s
q 1
E

0 A 0
So….
q 1
dW 
d  dq
A 0
Q
W U  
0
d
q2 d
q2 1
qdq 

A 0
2 A 0
2 ( A 0 )
d
or
Q 2 C 2V 2 1
U

 CV 2
2C
2C
2
Not All Capacitors are Created Equal
 Parallel Plate
 Cylindrical
 Spherical
Spherical Capacitor
Gauss
q
 E  dA  
4r E 
2
0
q
0
q
E (r ) 
2
4r  0
surprise ???
Calculate Potential Difference V
positive. plate
Eds

V
negative. plate
q 1
V  
 2 dr
40  r 
b
a
(-) sign because E and ds are in OPPOSITE directions.
Continuing…
q
b
dr
q
1
V

( )
2

40 a r
40 r
q 1 1
q ba
V
  


40  a b  40  ab 
q
ab
C   40
V
ba
Lost (-) sign due to switch of limits.
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 potential
See what happens
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
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?
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 V
What are the new charges on each capacitor?
(b)46.2µC (on C1)
(c)162µC (on C2)
Anudder Thunker
Find the equivalent capacitance between points a
and b in the combination of capacitors shown in the
figure.
V(ab) same across each
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 is
REDUCED
We can measure the C of a capacitor (later)
C0 = Vacuum or air Value
C = With dielectric in place
C=kC0
(we show this later)
How to Check This
C0
V
V0
Charge to V0 and then disconnect from
The battery.
Connect the two together
C0 will lose some charge to the capacitor with the dielectric.
We can measure V with a voltmeter (later).
Checking the idea..
q0  C0V0
V
q1  C0V
q2  CV
q0  q1  q2
C0V0  C0V  CV
 V0 
C  C0   1  kC0
V

Note: When two Capacitors are the same (No dielectric), then V=V0/2.
Messing with
Capacitors
+
The battery means that the
potential difference across
the capacitor remains constant.
+
V
-
-
For this case, we insert the
dielectric but hold the voltage
constant,
+
+
q=CV
V
Remember – We hold V
constant with the battery.
-
since C  kC0
qk kC0V
THE EXTRA CHARGE
COMES FROM 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
q0 =C0Vo
V0
-
qk  kC0V
When
is inserted, no charge
q  Cthe
V  qdielectric
 kC V
or
is added
so the charge must be the same.
0
V
+
qk
V
-
0 0
V0
k
k
0
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..
q
Consider this virgin capacitor.
No dielectric experience.
Applied Voltage via a battery.
++++++++++++
V0
C0
-q
------------------
C0 
0 A
d
q  C0V0 
0 A
d
V0
Remove the Battery
q
++++++++++++
V0
The Voltage across the
capacitor remains V0
q remains the same as
well.
-q
-----------------The capacitor is fat (charged),
dumb and happy.
Slip in a Dielectric
Almost, but not quite, filling the space
Gaussian Surface
q
++++++++++++
- - - - - - - -
V0
-q’
in..small..gap
q
 E  dA 
+ + + + + +
-q
+q’

-----------------E 
0
0
q  s 
 
 0 A   0 
E0
E
E’ from induced
charges
A little sheet from the past..
-q’
+q’
- -q
-
+
q+
+
Esheet 
s
q'

2 0 2 0 A
Esheet / dialectric  2 
0
2xEsheet
0
q'
2 0 A

q'
0 A
Some more sheet…
Edielectricch arg e
q
E 0
0 A
so
q  q'
E
0 A
 q'

0 A
A Few slides back
No Battery
+
V0
-
+
q=C0Vo
q0
When the dielectric is inserted, no charge
is added so the charge must be the same.
qk  kC0V
qk
V
-
q0  C0V0  qk  kC0V
or
V
V0
k
From this last equation
V
V0
k
and
V  Ed
V0  E0 d
thus
V 1 E
 
V0 k E0
E
E0
k
Another look
Vo
+
-
Parallel  Plate
0 A
C0 
d
 0 AV0
Q0  C0V0 
d
Electric  Field
V0
E0 
d
Q0  0V0
s0 

A
d
Add Dielectric to Capacitor
Vo
 Original Structure
+
-
+
V0
 Disconnect Battery
+
 Slip in Dielectric
Note: Charge on plate does not change!
What happens?
so +
so -
si 
si 
E0
V0 1
E

k
d k
and
V0
V  Ed 
k
Potential Difference is REDUCED
by insertion of dielectric.
Q
Q
C 
 kC0
V V0 / k 
Charge on plate is Unchanged!
Capacitance increases by a factor of k
as we showed previously
SUMMARY OF RESULTS
V
V0
E
E0
k
C  kC0
k
APPLICATION OF GAUSS’ LAW
E0 
q
0 A
q  q ' E0
E

0 A
k
E
q
k 0 A
and
q  q' 
q
k
New Gauss for Dielectrics
k
E

d
A


sometimes
  k 0
q free
0