Download Notes 26 3318 Electric Stored Energy

ECE 3318
Applied Electricity and Magnetism
Spring 2017
Prof. David R. Jackson
ECE Dept.
Notes 26
1
Stored Energy
Goal:
Calculate the amount of stored electric energy in
space, due to the presence of an electric field.
(Recall that the electric field is produced by a
charge density.)
2
Stored Energy (cont.)
Charge formulas:
1
U E   v  dV
2V
1
U E   s  dS
2S
(volume charge density)
(surface charge density)
Note:
We don’t have a linecharge version of the
charge formulas, since a
line charge would have an
infinite stored energy (as
would a point charge).
The charge formulas require      0
Electric-field formula:
1
U E   D  E dV
2V
Note:
The charge and electric field formulas will
give the same answer, provided that the
region V includes all places where there is a
charge or a field.
Please see the textbook for a derivation of these formulas.
3
Example
Find the stored energy.
A  m 2 
V0
+
-
r
h
x
We assume an ideal parallel-plate capacitor (no fields outside).
 V0 
ˆ
E  x 
h
Method #1
V 
D   0 r xˆ  0 
h
Use the electric field formula:
UE 
1
D  E dV

2V
so
2
1
V 
U E    0 r  0  dV
2V
h
4
Example (cont.)
Hence
1
 V0 
U E   0 r  
2
h
2
 Ah   J 
We can also write this as
1
A
U E    0 r  V02
2
h
Recall that
A
C   0 r
h
Therefore, we have
1
U E  CV02
2
5
Example (cont.)
Method #2
A  m 2 
SA
Use charge formula:
+++++++++++++
1
U E    s dS
2S
------------------
r
h
x
SB
(We use the surface-charge form of the formula.)
We then have
1
1
U E   A   s dS   B   s dS
2
2
SA
SB
6
Example (cont.)
1
1
U E   AQA   B QB
2
2
1
1
  A Q A   B QA
2
2
1
 Q  A  B 
2
so
1
U E  QV0
2
A  m 2 
SA
+++++++++++++
r
h
------------------
x
SB
Also, we can write this as
1
U E  QV0
2
1
  CV0  V0
2
or
1
U E  CV02
2
Note: This same derivation (method 2) actually holds for any shape capacitor.
7
Alternative Capacitance Formula
Using the result from the previous example, we can write
2U E
C 2
V
where
UE = stored energy in capacitor
V = voltage on capacitor
This gives us an alternative method for calculating capacitance.
8
Alternative Capacitance Formula (cont.)
Find C (using the alternative formula).
A  m 2 
A
r
Ex 0
h
x
B
Ideal parallel plate capacitor
Assume:
E  xˆ Ex 0
B
h
h
A
0
0
V  VAB   E  dr   Ex dx   Ex 0 dx
 Ex 0 h
9
Alternative Capacitance Formula (cont.)
A  m 2 
A
r
h
Ex 0
x
B
UE 
1
1
1
1 2
2
D

E
dV


E

E
dV


E
dV

 Ex 0  Ah 





2V
2V
2V
2
Therefore
1

2   Ex20  Ah  
2U E
2

C 2  
2
V
E
h
 x0 
so
C
A
h
10
Example
Find the stored energy.
a
1
U E   D  E dV
2V
0
E  E  r   rˆ Er  r 
 v 0  C/m 3 
Solid sphere of uniform
volume charge density
(not a capacitor!)
Gauss’s Law:
r<a

4 3

 v0  3  r  


E  rˆ 
2
 4 0 r





r>a
4 3


 v0 3  a 
E  rˆ 
2 
 4 0 r 


11
Example (cont.)
UE 
0
2 
E dV 
2
V


0
2
0
2
0
2
2
E
 r  r  dV
a
V
 v 0  C/m 3 
2  

2
2
E
r
r
sin  dr d d


 r
0
0 0 0

 2  2   Er2  r  r 2 dr
0
Hence we have
2
2
4 3
4 3


a v0
 v0
r 
a 


2
2
3
3
U E  2 0  
r
dr

2

r
dr
0
2 
2 
4 0 r 
4 0 r 
0
a




12
Example (cont.)
Result:
v20  4  5
UE 

a
 0  15 
Denote
This is the energy that it takes to
assemble (force together) the charges
inside the cloud from infinity.
4 3
Q  v 0   a 
3

so that
 3 
v 0  Q 
3 
 4 a 
We then have
1  Q2   3 
UE  


a   0   20 
J
13
Example (cont.)
Final result:
1Q
UE  
a  0
2
 3 


20



a
J
Q  C
0
Note:
UE  
as
a0
It takes an infinite amount of energy to make
an ideal point charge!
14
Example
Find the “radius” of an electron.
1  Q2   3 
UE    

a   0   20 
Q2  3 
 a


U E  20 0 
Q  qe  1.6022 1019 [C]
Assume that all of the stored energy is in the form of electrostatic energy:
U E  mc 2
We have:
m  9.1094 1031 [kg]
c  2.99792458 10 [m/s]
8
Hence
a  1.69 1015 [m]
U E  mc 2  8.1871  1014 [J]
This is one form of the “classical electron radius.”
15