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Capacitance
Energy & Dielectrics
Q
+++++++++++++
q
dW  V (q) dq    dq
C
W
1
CV 2
2
E0
V0
-------------
Q
+++++++++++++
V

E
-------------
1
u  0E 2
2


 0  E  dS  q
Today…
• Review items from last lecture:
• Capacitors (series and parallel)
• Calculate Energy Stored in Capacitor
• Calculate Energy Density in Electric Field
• Define Dielectric Constant
• Modify Gauss’ Law to include Dielectrics
Summary of important capacitor geometries
• The definition of the capacitance relates Q to V via C:
C
Q
V
• The capacitance depends on the geometry:
-Q
A
++++
d
-----
r
+Q
-Q
C
o A
d
In SI unit system:
a
a
b
b
L
Cylindrical
Parallel Plates
+Q
C
2 o L
b
ln  
a
Spherical
ab
C  4 o
ba
C has units of “Farads” or F (1F = 1C/V)
 o has units of F/m
1
Lecture 8, ACT 1
• What is the relationship between V0 and V in the
systems shown below?
+Q
V0
(Area A)
+Q
d
-Q
(a) V = (2/3)V0
(Area A)
d/3
V
conductor
-Q
(b) V = V0
d/3
(c) V = (3/2)V0
Lecture 8, ACT 1
• What is the relationship between V0 and V in the
systems shown below?
+Q
V0
(Area A)
+Q
d
-Q
(a) V = (2/3)V0
(Area A)
d/3
V
conductor
-Q
(b) V = V0
d/3
(c) V = (3/2)V0
• The electric field in the conductor = 0.
• The electric field everywhere else is: E = Q/(A0)
• To find the potential difference, integrate the electric field:
 
V0   E  dl  Ed
 
d
d
V   E  dl  E  0  E
3
3
2
V  Ed
3
Lecture 8, ACT 1, Another Way
• What is the relationship between V0 and V in the
systems shown below?
+Q
+Q
(Area A)
(Area A)
d/3
V0
V
d
conductor
-Q
-Q
(b) V = V0
(a) V = (2/3)V0
•
d/3
(c) V = (3/2)V0
The arrangement on the right is equivalent to capacitors (each with
separation = d/3) in SERIES!!
+Q
+Q
d/3

d/3
V
d/3
-Q
Ceq
(Area A)
1
Ceq  C
2
1 A 0
3 A 0 3


 C0
2 d / 3 2 d
2
d/3
-Q
V
conductor
Q
Q
2

 V0
Ceq 3 / 2 C0 3
Energy of a Capacitor
• How much energy is stored in a charged capacitor?
– Calculate the work provided (usually by a battery) to charge a
capacitor to +/- Q:
Calculate incremental work dW needed to add charge dq to capacitor
at voltage V (there is a trick here!):
- +
q
dW  V (q)  dq     dq
C
• The total work W to charge to Q is then given by:
Q
1
1 Q2
W   qdq 
C0
2 C
• In terms of the voltage V:
Look at this!
Two ways to write W
W 
1
CV 2
2
Capacitor Variables
• The total work to charge to Q equals the energy U stored in
the capacitor:
Q
2
1
1Q
U   qdq 
C0
2 C
• In terms of the voltage V:
Look at this!
Two ways to write U
U 
1
CV 2
2
• You can do one of two things to a capacitor :
• hook it up to a battery
• put some charge on it
specify V and Q follows
Q  CV
specify Q and V follows
Q
V
C
Question!
A
++++
d
-----
• Suppose the capacitor shown here is
charged to Q and then the battery is
disconnected.
• Now suppose I pull the plates further apart so that the final
separation is d1.
• How do the quantities Q, C, E, V, U change?
•
•
•
•
•
Q:
C:
E:
V:
U:
remains the same.. no way for charge to leave.
decreases.. since capacitance depends on geometry
remains the same... depends only on charge density
increases.. since C , but Q remains same (or d  but E the same)
increases.. add energy to system by separating
• How much do these quantities change?.. exercise for student!!
Answers:
C1 
d
C
d1
V1 
d1
V
d
U1 
d1
U
d
Related Question
• Suppose the battery (V) is kept
attached to the capacitor.
A
++++
d
-----
V
• Again pull the plates apart from d
to d1.
• Now what changes?
•
•
•
•
•
C:
V:
Q:
E:
U:
decreases (capacitance depends only on geometry)
must stay the same - the battery forces it to be V
must decrease, Q=CV charge flows off the plate
2
V

must decrease ( E  , E  )
E0
D
must decrease (U  12 CV 2)
• How much do these quantities change?.. exercise for student!!
Answers:
d
C1 
C
d1
E1 
d
E
d1
U1 
d
U
d1
Preflight 8:
Two identical parallel plate capacitors are connected to a battery, as
shown in the figure. C1 is then disconnected from the battery, and the
separation between the plates of both capacitors is doubled.
2) What is the relation between the charges on the two capacitors ?
a) Q1 > Q2
b) Q1 = Q2
c) Q1 < Q2
3) How does the electric field between the plates of C2 change as
separation between the plates is increased ? The electric field:
a) increases
b) decreases
c) doesn’t change
Preflight 8:
Two identical parallel plate capacitors are connected to a battery,
as shown in the figure. C1 is then disconnected from the battery,
and the separation between the plates of both caps is doubled.
5) What is the relation between the voltages on the two capacitors?
a) V1 > V2
b) V1 = V2
c) V1 < V2
Lecture 8, ACT 2
• Two identical parallel plate capacitors are connected to a battery.
– C1 is then disconnected from the battery and the separation
between the plates of both capacitors is doubled.
d
V
2d
d
C1
V
C2
2d
– What is the relation between the U1, the energy stored in C1, and
the U2, energy stored in C2?
(a) U1 < U2
(b) U1 = U2
(c) U1 > U2
Lecture 8, ACT 2
• Two identical parallel plate capacitors are connected to a battery.
– C1 is then disconnected from the battery and the separation
between the plates of both capacitors is doubled.
d
V
2d
d
C1
V
C2
2d
– What is the relation between the U1, the energy stored in C1, and
the U2, energy stored in C2?
(a) U1 < U2
(b) U1 = U2
(c) U1 > U2
• What is the difference between the final states of the two capacitors?
• The charge on C1 has not changed.
• The voltage on C2 has not changed.
• The energy stored in C1 has definitely increased since work must be
done to separate the plates with fixed charge, they attract each other.
• The energy in C2 will actually decrease since charge must leave in order
to reduce the electric field so that the potential remains the same.
2
Q
1
1
1
2
0
Later: U 

2
U
U

C
V

U0
Initially: C1  C2
1
0
2
2 0
2 C1
2
2
Where is the Energy Stored?
• Claim: energy is stored in the electric field itself.
Think of the energy needed to charge the capacitor
as being the energy needed to create the field.
• To calculate the energy density in the field, first consider the
constant field generated by a parallel plate capacitor, where
-Q
-------- ------
++++++++ +++++++
+Q
1 Q2 1 Q2
U

2 C 2 ( A 0 / d )
• The electric field is given by:

Q
E 
0 0 A

This is the energy
density, u, of the
electric field….
1
U   0 E 2 Ad
2
• The energy density u in the field is given by:
U
U 1 2
u

 0 E
volume Ad 2
Units:
J
m3
Energy Density
Claim: the expression for the energy density of the
electrostatic field
1
u
2
0E 2
is general and is not restricted to the special case of the
constant field in a parallel plate capacitor.
• Example (and another exercise for the student!)
– Consider E- field between surfaces of cylindrical
capacitor:
– Calculate the energy in the field of the capacitor by
integrating the above energy density over the volume of
the space between cylinders.
1
1
U   0  E 2 dV   0   E 2 r dr dl  etc.
2
2
– Compare this value with what you expect from the
general expression:
W 
1
CV 2
2
3
Lecture 8, ACT 3
• Consider two cylindrical capacitors,
each of length L.
C1
1.1
1
– C1 has inner radius 1 cm and outer radius 1.1cm.
– C2 has inner radius 1 cm and outer radius 1.2cm.
If both capacitors are given the same amount of
charge, what is the relation between U1, the energy
stored in C1, and U2, the energy stored in C2?
(a) U2 < U1
(b) U2 = U1
C2
1.2
1
(c) U2 > U1
Lecture 8, ACT 3
• Consider two cylindrical capacitors,
each of length L.
C1
1.1
1
– C1 has inner radius 1 cm and outer radius 1.1cm.
– C2 has inner radius 1 cm and outer radius 1.2cm.
If both capacitors are given the same amount of
charge, what is the relation between U1, the energy
stored in C1, and U2, the energy stored in C2?
(a) U2 < U1
(b) U2 = U1
C2
1.2
1
(c) U2 > U1
The magnitude of the electric field from r = 1 to 1.1 cm is the same
for C1 and C2. But C2 also has electric energy density in the volume
1.1 to 1.2 cm. In formulas:
2o L
2
1
1
U
Q
/ 2C2 C1
1.2
C
C2 ~
2
C1 ~
 2

 ln( )
 1.2 
 1.1
 router 
U1 Q / 2C1 C2
1.1
ln 
ln 


ln 

 1 
 1 
r

inner

Dielectrics
• Empirical observation:
Inserting a non-conducting material between the plates of a
capacitor changes the VALUE of the capacitance.
• Definition:
The dielectric constant of a material is the ratio of the
capacitance when filled with the dielectric to that without it:
 
C
C0
–  values are always > 1 (e.g., glass = 5.6; water = 78)
– They INCREASE the capacitance of a capacitor (generally
good, since it is hard to make “big” capacitors)
– They permit more energy to be stored on a given capacitor
than otherwise with vacuum (i.e., air):
 C0V 2
CV 2
U 

 U
2
2
Parallel Plate Example
• Deposit a charge Q on parallel plates
filled with vacuum (air)—capacitance
C0.
• Disconnect from battery
Q
+++++++++++++++
V0
E0
---------------
• The potential difference is V0 = Q / C0.
• Now insert material with dielectric
constant  .
E
+
+
+
+
+
+
V 
-
– Voltage decreases from V0 to:
V 
-
– Capacitance increases C =  C0
+++++++++++++++
+
– Charge Q remains constant
Q
---------------
V
Q
Q

 0
C  C0

– Electric field decreases also:
V
E
V
E 
 0  0
d
d

Note: The field only
decreases when the
charge is held constant!
Parallel Plate Example, assuming Q constant
 modify Gauss Law, see Appendix
V0
E0
-------------
Q
+++++++++++++
V
+ + +
- -
• The molecules partially align with
the field so that their negative
charge tends toward the positive
plate.
• The field due to this rearrangement
inside the dielectric (a la dipole)
opposes the original field and is
therefore responsible for the
reduction in the effective field.
• Some of the original field lines now
end on charges in the dielectric
+++++++++++++
+
-
– How can field decrease if charge
remains the same?
– Answer: the dielectric becomes
polarized in the presence of the
field due to Q.
Q
+
+
- + +
+
- +
-
• HOW ABOUT GAUSS' LAW?

E
-------------
V0
V
κ
E0
E

4
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.
Preflight 8:
7) Compare the voltages of the two capacitors.
a) V1 > V2
b) V1 = V2
c) V1 < V2
8) Compare the electric fields between the plates of both
capacitors.
a) E1 > E2
b) E1 = E2
c) E1 < E2
Preflight 8:
10) When we insert the dielectric into the capacitor C2 we do:
a) positive work
b) negative work
c) no work
Recall the meaning of negative work  the energy of the system is
reduced. The dielectric is sucked into the capacitor. When the charge
is constant, the total energy of the capacitor decreases because the
presence of the dielectric increases the capacitance . It turns out that
the dielectric is pulled in even if the voltage is held constant, e.g., via
a battery. On a microscopic scale the force on the dielectric arises due
to “fringing fields” at the edges of the capacitor.
Lecture 8, ACT 4
• Two parallel plate capacitors are identical
(same A, same d) except that C1 has half of
the space between the plates filled with a
material of dielectric constant  as shown.
– If both capacitors are given the same
amount of charge Q, what is the relation
between E1, the electric field in the air of
C1, and E2, the electric field in the air of C2
(a) E1 < E2
(b) E1 = E2
+Q

E1=?
C1
-Q
+Q
E2=?
-Q
(c) E1 > E2
C2
Lecture 8, ACT 4
• Two parallel plate capacitors are identical
(same A, same d) except that C1 has half of
the space between the plates filled with a
material of dielectric constant  as shown.
– If both capacitors are given the same
amount of charge Q, what is the relation
between E1, the electric field in the air of
C1, and E2, the electric field in the air of C2
(a) E1 < E2
(b) E1 = E2
+Q

E1=?
C1
-Q
+Q
E2=?
C2
-Q
(c) E1 > E2
• The key here is to realize that the electric field in the air in C1 must be
equal to the electric field in the dielectric in C1!
• WHY?
• The top plane is a conductor; therefore it is an equipotential
surface.
• The bottom plane is a conductor; therefore it is also an
equipotential surface.
• For this to happen, the charge density on each plane must be nonuniform to create equal electric fields!!
• Since C1 > C2, for the same charge, V1 < V2.
• Consequently, E1 < E2.
NOTE: If the voltage were held constant, E would NOT have changed!!
Why Salt Dissolves
Normally NaCl is in a rigid
crystal structure, maintained
by the electrostatic attraction
between the Na+ and Cl- ions.
Water has a very high
dielectric constant (78). This
reduces the field between the
atoms, hence their attraction
to each other. The crystal
lattice comes apart and
dissolves.
Note: This should leave lots of ions in the water
Summary
• Capacitors in parallel and in series
• energy stored in a charged capacitor
– work done to assemble charges on plates
• energy stored in electric field
– capacitor example
– field energy density J/m3
1
1 Q2
2
W  CV 
2
2 C
1
1
2
u  0 E  0 E  E
2
2
• electric response of dielectric materials
– atomistic view: polarized atoms -> dielectric constant 
– modified Gauss’ Law
 
 o  dS  E  qenclosed
– dielectric constant

 o
Replace  o with
!
Modified Gauss’ Law
(in presence of dielectric)
In vacuum, Gauss' Law is:


 0  E0  dS  q
In a dielectric, the field is reduced:
Rewrite Gauss' Law:
E
E0


 E0  
 0      dS  q
 
Reinterpret if in the presence of dielectric:
 
 0  E  dS  q
 o
Replace  o with
This form of Gauss' Law can be used in vacuum or
dielectric alike where q represents only the "free" charge.
- free charge is the charge you can move around
!