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Fall 2008
Physics 231 Lecture 4-1
Capacitance and Dielectrics
Fall 2008
Physics 231 Lecture 4-2
Capacitors
Device for storing electrical energy which can then be
released in a controlled manner
Consists of two conductors, carrying charges of q and –q,
that are separated, usually by a nonconducting material - an
insulator
Symbol in circuits is
It takes work, which is then stored as potential energy in the
electric field that is set up between the two plates, to place
charges on the conducting plates of the capacitor
Since there is an electric field between the plates there is also a
potential difference between the plates
Fall 2008
Physics 231 Lecture 4-3
Capacitors
We usually talk about
capacitors in terms of
parallel conducting
plates
They in fact can be
any two conducting
objects
Fall 2008
Physics 231 Lecture 4-4
Capacitance
ab
V
Q
C 
Units are
Volt
Coulomb
farad
1
1
1 
The capacitance is defined to be the ratio of the
amount of charge that is on the capacitor to the
potential difference between the plates at this point
Fall 2008
Physics 231 Lecture 4-5
Calculating the Capacitance
We start with the simplest form – two parallel conducting
plates separated by vacuum
Let the conducting plates have area A and be
separated by a distance d
The magnitude of the electric field
between the two plates is given by A
Q
E
0
0 




We treat the field as being uniform
allowing us to write A
d
Q
Ed
Vab
0



Fall 2008
Physics 231 Lecture 4-6
Calculating the Capacitance
Putting this all together, we have for the
capacitance
d
A
V
Q
C
ab
0



The capacitance is only dependent
upon the geometry of the capacitor
Fall 2008
Physics 231 Lecture 4-7
1 farad Capacitor
Given a 1 farad parallel plate capacitor having a
plate separation of 1mm. What is the area of the
plates?
We start with
d
A
C 0


And rearrange to
solve for A, giving
  
2
8
12
3
0
10
1
.
1
/
10
85
.
8
10
0
.
1
0
.
1
m
m
F
m
F
d
C
A





 


This corresponds to a square about 10km on a side!
Fall 2008
Physics 231 Lecture 4-8
Series or Parallel Capacitors
Sometimes in order to obtain needed values
of capacitance, capacitors are combined
in either
or
Series
Parallel
Fall 2008
Physics 231 Lecture 4-9
Capacitors in Series
Capacitors are often combined in series and the question
then becomes what is the equivalent capacitance?
Given what is
We start by putting a voltage,
Vab, across the capacitors
Fall 2008
Physics 231 Lecture 4-10
Capacitors in Series
Capacitors become charged because of Vab
If upper plate of C1 gets a charge of +Q,
Then the lower plate of C1 gets a
charge of -Q
What happens with C2?
Since there is no source of charge at point c, and we
have effectively put a charge of –Q on the lower plate
of C1, the upper plate of C2 gets a charge of +Q
This then means that lower plate of C2 has a charge of -Q
Charge Conservation
Fall 2008
Physics 231 Lecture 4-11
We also have to have that the potential across C1 plus the
potential across C2 should equal the potential drop across
the two capacitors
2
1 V
V
V
V
V cb
ac
ab 



We have
2
2
1
1
C
Q
V
C
Q
V 
 and
Then
2
1 C
Q
C
Q
Vab 

Dividing through by Q, we have
2
1
1
1
C
C
Q
Vab 

Capacitors in Series
Fall 2008
Physics 231 Lecture 4-12
2
1
1
1
C
C
Q
Vab 

The left hand side is the inverse of
the definition of capacitance Q
V
C

1
So we then have for the equivalent capacitance
2
1
1
1
1
C
C
Ceq


If there are more than two capacitors in series,
the resultant capacitance is given by


i i
eq C
C
1
1
The equivalent capacitor will also have the
same voltage across it
Capacitors in Series
Fall 2008
Physics 231 Lecture 4-13
Capacitors in Parallel
Capacitors can also be connected in parallel
Given what is
Again we start by putting
a voltage across a and b
Fall 2008
Physics 231 Lecture 4-14
The upper plates of both capacitors
are at the same potential
Likewise for the bottom plates
We have that ab
V
V
V 
 2
1
Now
V
C
Q
V
C
Q
or
C
Q
V
C
Q
V
2
2
1
1
2
2
2
1
1
1




and
and
Capacitors in Parallel
Fall 2008
Physics 231 Lecture 4-15
The equivalent capacitor will have the same voltage across
it, as do the capacitors in parallel
But what about the charge on the equivalent capacitor?
The equivalent capacitor will have the same total charge
2
1 Q
Q
Q 

Using this we then have
2
1
2
1
2
1
C
C
C
or
V
C
V
C
V
C
Q
Q
Q
eq
eq






Capacitors in Parallel
Fall 2008
Physics 231 Lecture 4-16
The equivalent capacitance is just the sum of the
two capacitors
If we have more than two, the resultant capacitance is just
the sum of the individual capacitances


i
i
eq C
C
Capacitors in Parallel
Fall 2008
Physics 231 Lecture 4-17
Example 1
C
C1 C2
C3
a
b
a b

2
1
3
1
1
1
C
C
C
C 

 
3
2
1
2
1
3 )
(
C
C
C
C
C
C
C




Where do we start?
Recognize that C1 and C2 are parallel with each other and
combine these to get C12
This C12 is then in series with with C3
The resultant capacitance is then given by
Fall 2008
Physics 231 Lecture 4-18
Which of these configurations has the lowest overall capacitance?
Three configurations are constructed using identical capacitors
a) Configuration A
b) Configuration B
c) Configuration C
C
C
C
C C
Configuration A
Configuration B Configuration C
Example 2
The net capacitance for A is just C
In B, the caps
are in series and
the resultant is
given by
2
2
1
1
1 C
C
C
C
C
C
net
net





In C, the caps are in parallel and
the resultant is given by
C
C
C
Cnet 2



Fall 2008
Physics 231 Lecture 4-19
A circuit consists of three unequal capacitors C1, C2, and C3
which are connected to a battery of emf E. The capacitors
obtain charges Q1 Q2, Q3, and have voltages across their
plates V1, V2, and V3. Ceq is the equivalent capacitance of the
circuit.
Check all of the following that
apply:
a) Q1= Q2 b) Q2= Q3 c) V2= V3
d) E = V1 e) V1 < V2 f) Ceq > C1
Example 3
A detailed worksheet is available detailing the answers
Fall 2008
Physics 231 Lecture 4-20
(a) Ceq = (3/2)C (b) Ceq = (2/3)C (c) Ceq = 3C
o
o
C C
C
Ceq
C
C
C
 C C1
C
C
C
1
1
1
1


2
1
C
C  C
C
C
Ceq
2
3
2



Example 4
What is the equivalent
capacitance, Ceq, of the
combination shown?
Fall 2008
Physics 231 Lecture 4-21
Energy Stored in a Capacitor
Electrical Potential energy is stored in a capacitor
The energy comes from the work that is done in charging
the capacitor
Let q and v be the intermediate charge and potential on the
capacitor
The incremental work done in bringing an incremental
charge, dq, to the capacitor is then given by
C
dq
q
dq
v
dW 

Fall 2008
Physics 231 Lecture 4-22
Energy Stored in a Capacitor
The total work done is just the integral of this
equation from 0 to Q
C
Q
dq
q
C
W
Q
2
1 2
0

 
Using the relationship between capacitance, voltage and
charge we also obtain
V
Q
V
C
C
Q
U
2
1
2
1
2
2
2



where U is the stored potential energy
Fall 2008
Physics 231 Lecture 4-23
Example 5
d
A
- - - - -
+ + + +
Now suppose you pull the plates further
apart so that the final separation is d1
Which of the quantities Q, C, V, U, E change?
How do these quantities change?
Answers: C
d
d
C
1
1  V
d
d
V 1
1 
Suppose the capacitor shown here is charged
to Q and then the battery is disconnected
U
d
d
U 1
1 
Charge on the capacitor does not change
Capacitance Decreases
Voltage Increases
Potential Energy Increases
Q:
C:
V:
U:
E: Electric Field does not change
Fall 2008
Physics 231 Lecture 4-24
Suppose the battery (V) is kept
attached to the capacitor
Again pull the plates apart from d to d1
Now which quantities, if any, change?
How much do these quantities change?
Answers:
Example 6
C
d
d
C
1
1  E
d
d
E
1
1 
U
d
d
U
1
1 
Q:
C:
V:
U:
E:
Voltage on capacitor does not change
Capacitance Decreases
Charge Decreases
Potential Energy Decreases
Electric Field Decreases
Q
d
d
Q
1
1 
Fall 2008
Physics 231 Lecture 4-25
Electric Field Energy Density
The potential energy that is stored in the capacitor can be
thought of as being stored in the electric field that is in the
region between the two plates of the capacitor
The quantity that is of interest is in fact the energy density
d
A
u
Density
Energy
V
C 2
2
1


where A and d are the area of the capacitor plates and their
separation, respectively
Fall 2008
Physics 231 Lecture 4-26
Electric Field Energy Density
Using
d
A
C 0

 and d
E
V  we then have
2
0
2
1
E
u 

Even though we used the relationship for a parallel capacitor,
this result holds for all capacitors regardless of configuration
This represents the energy density of the electric field in
general
Fall 2008
Physics 231 Lecture 4-27
Dielectrics
Most capacitors have a nonconducting material between
their plates
This nonconducting material, a dielectric, accomplishes
three things
1) Solves mechanical problem of keeping the plates
separated
2) Increases the maximum potential difference allowed
between the plates
3) Increases the capacitance of a given capacitor over
what it would be without the dielectric
Fall 2008
Physics 231 Lecture 4-28
Dielectrics
Suppose we have a capacitor of value C0 that is charged to
a potential difference of V0 and then removed from the
charging source
We would then find that it has a charge of 0
0V
C
Q 
We now insert the dielectric material into the capacitor
We find that the potential difference decreases by a factor K
K
V
V 0

Or equivalently the capacitance has increased by a factor of K
0
C
K
C 
This constant K is known as the dielectric constant and is
dependent upon the material used and is a number
greater than 1
Fall 2008
Physics 231 Lecture 4-29
Polarization
Without the dielectric in
the capacitor, we have
The electric field points undiminished from the positive to
the negative plate
Q
+++++++++++++++
- - - - - - - - - - - - - - -
E0
V0
With the dielectric
in place we have
The electric field between the plates of the capacitor is
reduced because some of the material within the dielectric
rearranges so that their negative charges are oriented
towards the positive plate
- - - - - - - - - - - - - - -
+++++++++++++++
Q
V E
+
- +
-
+
-
+
-
+
-
+
-
+
-

Fall 2008
Physics 231 Lecture 4-30
Polarization
These rearranged charges set up an
internal electric field that opposes
the electric field due to the charges
on the plates
The net electric field is given by
K
E
E 0

Fall 2008
Physics 231 Lecture 4-31
Redefinitions
We now redefine several quantities using the dielectric
constant
Capacitance:
d
A
d
A
K
KC
C 
 

 0
0
with the last two relationships holding for a parallel
plate capacitor
Energy Density 2
2
0
2
1
2
1
E
E
K
u 
 

We define the permittivity of the dielectric as being
0

 K

Fall 2008
Physics 231 Lecture 4-32
Two identical parallel plate capacitors are given the same charge
Q, after which they are disconnected from the battery. After C2
has been charged and disconnected it is filled with a dielectric.
Compare the voltages of the two capacitors.
a) V1 > V2 b) V1 = V2 c) V1 < V2
Example 7
We have that Q1 = Q2 and that C2 = KC1
We also have that C = Q/V or V= Q/C
Then
1
1
1
C
Q
V  and 1
1
1
2
2
2
1
V
K
KC
Q
C
Q
V 



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Sjd lecture24

  • 1. Fall 2008 Physics 231 Lecture 4-1 Capacitance and Dielectrics
  • 2. Fall 2008 Physics 231 Lecture 4-2 Capacitors Device for storing electrical energy which can then be released in a controlled manner Consists of two conductors, carrying charges of q and –q, that are separated, usually by a nonconducting material - an insulator Symbol in circuits is It takes work, which is then stored as potential energy in the electric field that is set up between the two plates, to place charges on the conducting plates of the capacitor Since there is an electric field between the plates there is also a potential difference between the plates
  • 3. Fall 2008 Physics 231 Lecture 4-3 Capacitors We usually talk about capacitors in terms of parallel conducting plates They in fact can be any two conducting objects
  • 4. Fall 2008 Physics 231 Lecture 4-4 Capacitance ab V Q C  Units are Volt Coulomb farad 1 1 1  The capacitance is defined to be the ratio of the amount of charge that is on the capacitor to the potential difference between the plates at this point
  • 5. Fall 2008 Physics 231 Lecture 4-5 Calculating the Capacitance We start with the simplest form – two parallel conducting plates separated by vacuum Let the conducting plates have area A and be separated by a distance d The magnitude of the electric field between the two plates is given by A Q E 0 0      We treat the field as being uniform allowing us to write A d Q Ed Vab 0   
  • 6. Fall 2008 Physics 231 Lecture 4-6 Calculating the Capacitance Putting this all together, we have for the capacitance d A V Q C ab 0    The capacitance is only dependent upon the geometry of the capacitor
  • 7. Fall 2008 Physics 231 Lecture 4-7 1 farad Capacitor Given a 1 farad parallel plate capacitor having a plate separation of 1mm. What is the area of the plates? We start with d A C 0   And rearrange to solve for A, giving    2 8 12 3 0 10 1 . 1 / 10 85 . 8 10 0 . 1 0 . 1 m m F m F d C A          This corresponds to a square about 10km on a side!
  • 8. Fall 2008 Physics 231 Lecture 4-8 Series or Parallel Capacitors Sometimes in order to obtain needed values of capacitance, capacitors are combined in either or Series Parallel
  • 9. Fall 2008 Physics 231 Lecture 4-9 Capacitors in Series Capacitors are often combined in series and the question then becomes what is the equivalent capacitance? Given what is We start by putting a voltage, Vab, across the capacitors
  • 10. Fall 2008 Physics 231 Lecture 4-10 Capacitors in Series Capacitors become charged because of Vab If upper plate of C1 gets a charge of +Q, Then the lower plate of C1 gets a charge of -Q What happens with C2? Since there is no source of charge at point c, and we have effectively put a charge of –Q on the lower plate of C1, the upper plate of C2 gets a charge of +Q This then means that lower plate of C2 has a charge of -Q Charge Conservation
  • 11. Fall 2008 Physics 231 Lecture 4-11 We also have to have that the potential across C1 plus the potential across C2 should equal the potential drop across the two capacitors 2 1 V V V V V cb ac ab     We have 2 2 1 1 C Q V C Q V   and Then 2 1 C Q C Q Vab   Dividing through by Q, we have 2 1 1 1 C C Q Vab   Capacitors in Series
  • 12. Fall 2008 Physics 231 Lecture 4-12 2 1 1 1 C C Q Vab   The left hand side is the inverse of the definition of capacitance Q V C  1 So we then have for the equivalent capacitance 2 1 1 1 1 C C Ceq   If there are more than two capacitors in series, the resultant capacitance is given by   i i eq C C 1 1 The equivalent capacitor will also have the same voltage across it Capacitors in Series
  • 13. Fall 2008 Physics 231 Lecture 4-13 Capacitors in Parallel Capacitors can also be connected in parallel Given what is Again we start by putting a voltage across a and b
  • 14. Fall 2008 Physics 231 Lecture 4-14 The upper plates of both capacitors are at the same potential Likewise for the bottom plates We have that ab V V V   2 1 Now V C Q V C Q or C Q V C Q V 2 2 1 1 2 2 2 1 1 1     and and Capacitors in Parallel
  • 15. Fall 2008 Physics 231 Lecture 4-15 The equivalent capacitor will have the same voltage across it, as do the capacitors in parallel But what about the charge on the equivalent capacitor? The equivalent capacitor will have the same total charge 2 1 Q Q Q   Using this we then have 2 1 2 1 2 1 C C C or V C V C V C Q Q Q eq eq       Capacitors in Parallel
  • 16. Fall 2008 Physics 231 Lecture 4-16 The equivalent capacitance is just the sum of the two capacitors If we have more than two, the resultant capacitance is just the sum of the individual capacitances   i i eq C C Capacitors in Parallel
  • 17. Fall 2008 Physics 231 Lecture 4-17 Example 1 C C1 C2 C3 a b a b  2 1 3 1 1 1 C C C C     3 2 1 2 1 3 ) ( C C C C C C C     Where do we start? Recognize that C1 and C2 are parallel with each other and combine these to get C12 This C12 is then in series with with C3 The resultant capacitance is then given by
  • 18. Fall 2008 Physics 231 Lecture 4-18 Which of these configurations has the lowest overall capacitance? Three configurations are constructed using identical capacitors a) Configuration A b) Configuration B c) Configuration C C C C C C Configuration A Configuration B Configuration C Example 2 The net capacitance for A is just C In B, the caps are in series and the resultant is given by 2 2 1 1 1 C C C C C C net net      In C, the caps are in parallel and the resultant is given by C C C Cnet 2   
  • 19. Fall 2008 Physics 231 Lecture 4-19 A circuit consists of three unequal capacitors C1, C2, and C3 which are connected to a battery of emf E. The capacitors obtain charges Q1 Q2, Q3, and have voltages across their plates V1, V2, and V3. Ceq is the equivalent capacitance of the circuit. Check all of the following that apply: a) Q1= Q2 b) Q2= Q3 c) V2= V3 d) E = V1 e) V1 < V2 f) Ceq > C1 Example 3 A detailed worksheet is available detailing the answers
  • 20. Fall 2008 Physics 231 Lecture 4-20 (a) Ceq = (3/2)C (b) Ceq = (2/3)C (c) Ceq = 3C o o C C C Ceq C C C  C C1 C C C 1 1 1 1   2 1 C C  C C C Ceq 2 3 2    Example 4 What is the equivalent capacitance, Ceq, of the combination shown?
  • 21. Fall 2008 Physics 231 Lecture 4-21 Energy Stored in a Capacitor Electrical Potential energy is stored in a capacitor The energy comes from the work that is done in charging the capacitor Let q and v be the intermediate charge and potential on the capacitor The incremental work done in bringing an incremental charge, dq, to the capacitor is then given by C dq q dq v dW  
  • 22. Fall 2008 Physics 231 Lecture 4-22 Energy Stored in a Capacitor The total work done is just the integral of this equation from 0 to Q C Q dq q C W Q 2 1 2 0    Using the relationship between capacitance, voltage and charge we also obtain V Q V C C Q U 2 1 2 1 2 2 2    where U is the stored potential energy
  • 23. Fall 2008 Physics 231 Lecture 4-23 Example 5 d A - - - - - + + + + Now suppose you pull the plates further apart so that the final separation is d1 Which of the quantities Q, C, V, U, E change? How do these quantities change? Answers: C d d C 1 1  V d d V 1 1  Suppose the capacitor shown here is charged to Q and then the battery is disconnected U d d U 1 1  Charge on the capacitor does not change Capacitance Decreases Voltage Increases Potential Energy Increases Q: C: V: U: E: Electric Field does not change
  • 24. Fall 2008 Physics 231 Lecture 4-24 Suppose the battery (V) is kept attached to the capacitor Again pull the plates apart from d to d1 Now which quantities, if any, change? How much do these quantities change? Answers: Example 6 C d d C 1 1  E d d E 1 1  U d d U 1 1  Q: C: V: U: E: Voltage on capacitor does not change Capacitance Decreases Charge Decreases Potential Energy Decreases Electric Field Decreases Q d d Q 1 1 
  • 25. Fall 2008 Physics 231 Lecture 4-25 Electric Field Energy Density The potential energy that is stored in the capacitor can be thought of as being stored in the electric field that is in the region between the two plates of the capacitor The quantity that is of interest is in fact the energy density d A u Density Energy V C 2 2 1   where A and d are the area of the capacitor plates and their separation, respectively
  • 26. Fall 2008 Physics 231 Lecture 4-26 Electric Field Energy Density Using d A C 0   and d E V  we then have 2 0 2 1 E u   Even though we used the relationship for a parallel capacitor, this result holds for all capacitors regardless of configuration This represents the energy density of the electric field in general
  • 27. Fall 2008 Physics 231 Lecture 4-27 Dielectrics Most capacitors have a nonconducting material between their plates This nonconducting material, a dielectric, accomplishes three things 1) Solves mechanical problem of keeping the plates separated 2) Increases the maximum potential difference allowed between the plates 3) Increases the capacitance of a given capacitor over what it would be without the dielectric
  • 28. Fall 2008 Physics 231 Lecture 4-28 Dielectrics Suppose we have a capacitor of value C0 that is charged to a potential difference of V0 and then removed from the charging source We would then find that it has a charge of 0 0V C Q  We now insert the dielectric material into the capacitor We find that the potential difference decreases by a factor K K V V 0  Or equivalently the capacitance has increased by a factor of K 0 C K C  This constant K is known as the dielectric constant and is dependent upon the material used and is a number greater than 1
  • 29. Fall 2008 Physics 231 Lecture 4-29 Polarization Without the dielectric in the capacitor, we have The electric field points undiminished from the positive to the negative plate Q +++++++++++++++ - - - - - - - - - - - - - - - E0 V0 With the dielectric in place we have The electric field between the plates of the capacitor is reduced because some of the material within the dielectric rearranges so that their negative charges are oriented towards the positive plate - - - - - - - - - - - - - - - +++++++++++++++ Q V E + - + - + - + - + - + - + - 
  • 30. Fall 2008 Physics 231 Lecture 4-30 Polarization These rearranged charges set up an internal electric field that opposes the electric field due to the charges on the plates The net electric field is given by K E E 0 
  • 31. Fall 2008 Physics 231 Lecture 4-31 Redefinitions We now redefine several quantities using the dielectric constant Capacitance: d A d A K KC C      0 0 with the last two relationships holding for a parallel plate capacitor Energy Density 2 2 0 2 1 2 1 E E K u     We define the permittivity of the dielectric as being 0   K 
  • 32. Fall 2008 Physics 231 Lecture 4-32 Two identical parallel plate capacitors are given the same charge Q, after which they are disconnected from the battery. After C2 has been charged and disconnected it is filled with a dielectric. Compare the voltages of the two capacitors. a) V1 > V2 b) V1 = V2 c) V1 < V2 Example 7 We have that Q1 = Q2 and that C2 = KC1 We also have that C = Q/V or V= Q/C Then 1 1 1 C Q V  and 1 1 1 2 2 2 1 V K KC Q C Q V   