Energy storage in CAPACITORs
Charge capacitor by transferring bits of
charge dq at a timefrom bottom to top plate.
Can use a battery to do this. Battery does
work which increase potential energy of
capacitor.
+q
-q
V
dq
q is magnitude of charge on plates
V= q/C
V across plates
dU = V dq
increase in potential energy

q
Q
CV 
1

 CV 2
U   dU   dq 
C
2C
2C
2
0
0
U
Q
2
2
two ways to write
Y&F, eqn. 24.9
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 E2 Ad
2
The energy density u in the field is given
by:
U
U 1 2
  0E
u
volume Ad 2
Units:
J
m3
Energy Density
Claim: the expression for the energy density of the
electrostatic field
1
u  0E 2
2
is general and is not restricted to the special case of the
constant field in a parallel plate capacitor.
• Example
– 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.
U
1
1
 0  E 2 dV   0   E 2 r dr dl  etc.
2
2
– Compare this value with what you expect from the
general expression:
1
W  CV 2
2
Capacitor Summary
•
•
A Capacitor is an object with two spatially separated conducting
surfaces.
The definition of the capacitance of such an object is:
C 
Q
V
• The capacitance depends on the geometry :
-Q
A
++++
d
----Parallel Plates
C
A 0
d
r
+Q
-Q
+Q
a
a
b
b
L
Cylindrical
Spherical
2  0 L
 b 
ln 

 a 
4  0 ab
C 
ba
C 
Example 1
• 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
Example 1
• 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
/ 2C2 C1
1.2
U
Q
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
Consider parallel plate
capacitor with vacuum
separating plates (left)
Suppose we place a material
called a dielectric in between
the plates (right)
The charge on the plates
remain the same, but a
Y&F
dielectric has a property of
having induced charges on its surface that REDUCE
the electric field in between and the voltage difference.
Since C = Q/V, the resulting capacitance will INCREASE.
Figure 24.13
DIELECTRICS
Suppose the charges on the plate and the
dielectric are, s and si. The electric
Fields before and after are

  i
E0

E0 
; E
; K

0
0
E   i
We define the ratio of the original field over
the new field as the dielectric constant, K.
Hence, the voltage difference changes by
1/K and the capacitance, Co=Q/V, changes
by C=KQ/V=K Co
For same Q:
But
C = KCo
C = KCo
E = Eo/K
General
V = Vo/K
DIELECTRICS Materials
Glass, mica, plastics are very good dielectrics
DIELECTRICS and permittivity
We introduce a convenient redefinition of ε0, called
permittivity, as
0
Consider a parallel plate capacitor with no dielectric
A
C0   0
d
A capacitor with a dielectric becomes simply,
A
A
C  KC 0 K 0  
d
d
The change in capacitance can be accounted for
by changing permittivity.
EXAMPLE of parallel plate capacitor problem
A parallel plate capacitor is made by placing polyethylene (K = 2.3)
between two sheets of aluminum foil. The area of each sheet is
400 cm2, and the thickness of the polyethylene is 0.3 mm. Find the
capacitance.
C =K εo A/d = (2.3) (8.85 x 10-12 C2/Nm2) (400 cm2)(1m2/104 cm2)
0.3 x 10-3 m
= 2.71 nF
Example 2:
Two identical parallel plate capacitors are connected to a battery.
Remaining connected, C2 is filled with a dielectric.
 Compare the voltages of the two capacitors.
a) V1 > V2
b) V1 = V2
c) V1 < V2
Example 2:
Two identical parallel plate capacitors are connected to a battery.
Remaining connected, C2 is filled with a dielectric.
 Compare the voltages of the two capacitors.
a) V1 > V2
b) V1 = V2
c) V1 < V2
Example 3:
Two identical parallel plate capacitors are connected to a battery.
Remaining connected, C2 is filled with a dielectric.
 Compare the charges on the plates of the capacitors.
a) Q1 > Q2
b) Q1 = Q2
c) Q1 < Q2
Example 3:
Two identical parallel plate capacitors are connected to a battery.
Remaining connected, C2 is filled with a dielectric.
 Compare the charges on the plates of the capacitors.
a) Q1 > Q2
b) Q1 = Q2
c) Q1 < Q2
Note: Unlike constant Q case, here V and E remain the same
but C = K Co still.
EXAMPLE
Two parallel plate capacitors, C1 = C2 = 2 μF, are connected across
a 12 V battery in parallel.
a.) What energy is stored?
1
U1  U 2  CV 2  144 J U T  288J
2
b.) A dielectric (K = 2.5) is inserted between the plates of C2. Energy?
C2'  KC2  2.5  2F  5F
1 ' 2
U  C2V  360J U T  504J
2
'
2
Note: a dielectric increases amount of energy stored in C2.
Y&F Problems 24.72 and 24.71
A parallel plate capacitor has two
dielectrics, side by side, show the
capacitance is,
A K1  K 2
C  0
2
d
A parallel plate capacitor has two
dielectrics, stacked, show the
capacitance is,
A 2 K1K 2
C  0
d K1  K 2
More weekend Fun
• HW #4  get cracking (Hints on Monday)
• Office Hours immediately after this class (9:30
– 10:00) in WAT214 [1-1:30pm today]
• 2nd Quiz Now