prof. d. wilton ece dept. notes 24 ece 2317 applied electricity and magnetism notes prepared by the...
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Prof. D. WiltonECE Dept.
Notes 24
ECE 2317 ECE 2317 Applied Electricity and MagnetismApplied Electricity and Magnetism
Notes prepared by the EM group,
University of Houston.
Stored EnergyStored Energy
1
2E v
V
U dV
1
2E
V
U D E dV
0
Note: Please see the text book or supplementary notes for a derivation.
1
2E s
S
U dS
charge formula:
electric-field formula:
(volume charge density)
(surface charge density)
v D D D
Charge form to electric field form uses
The divergence theorem
ExampleExample
+ -
r hxV0
A [m2]
Method #1
0
0 00
2
00
1
2
1
2
E
V
r
E r
V
U D E dV
VE x
h
V VD x x
h h
VU dV
h
Find the stored energy
Example (cont.)Example (cont.)
2
00
1J
2E r
VU Ah
h
20 0
1
2E r
AU V
h
Or,
20
1
2EU CV
Recall that 0 r
A AC
h h
Hence
Example (cont.)Example (cont.)
Method #2
1
2
1
2
1 1
2 2
constant od
nan A B
A B
A B
E v
V
S S
s
S S S
A s B s
S S
U dV
dS
dS dS
(since we have surface charge in this problem)
r h x
A [m2]A
B
+ + + + + + + + + + + + +
- - - - - - - - - - - - - - - - - -
Example (cont.)Example (cont.)
1 1
2 21 1
2 21
2
E A A B B
A A A B
A B
U Q Q
Q Q
Q
0
0 0
1
21
2
EU QV
CV V
0
1
2EU QV 20
1
2EU CV
Capacitance from Stored EnergyCapacitance from Stored Energy
2
2
1
2
2
E
E
Q Q
V
U CV
UC
V
The previous derivation assumes only that ...
there is a charge and on the two
capacitor plates
the potential difference between them is
Since the derivati
on is
it applies in general and provides another way to compute capacitance!
independent of capacitor geometry,
ExampleExample
v0 [C/m3]0 a
Method #1:1
2E
V
U D E dV
Gauss’s Law:
r < a
30
20
43
4
v aE r
r
3
0
20
43
4
v rE r
r
r > a
Find the stored energy
Solid sphere of uniform volume charge density
Example (cont.)Example (cont.)
2 20 0
22 20
0 0 0
2 20
0
2 23 3
0 02 2
0 02 20 00
2 2
sin2
2 22
4 43 32 2
4 4
E r
V V
r
r
a v v
E
a
U E dV E dV
E r r dr d d
E r r dr
r aU r dr r dr
r r
v0 [C/m3]0 a
Example (cont.)Example (cont.)
Result:2
50
0
4
15v
EU a
Let:3
0
0 3
4
3
3
4
v
v
Q a
Qa
as 0EU
a
2
0
1 3J
20E
QU
a
Note:
KCL Law (cont.)KCL Law (cont.)
A
C
+Q
totin
dvi C
dt
ground
C is the stray capacitance between the “node” and ground
+
-
v
KCL Law (cont.)KCL Law (cont.)
1) In “steady state” (no time change)
2) As A (area of node) 0, C 0
0
0totin
dv
dt
i
0totini
totin
dvi C
dt
Two cases for which the KCL law is valid:
KCL Law (Differential Form)KCL Law (Differential Form)
1
0 or 0N
totn in
n
i i
Differential form:
0J
0
1limV
S
J J n dSV
0
out
S
in
J n dS i
i
J
V
(valid for D.C. currents)
For D.C. : (circuit form)
Integral form:
DC Current FormulasDC Current Formulas
Ohm’s Law
J E
Charge-Current Formula
vJ v
(an experimental law)
(this was derived earlier in the semester)
Joule’s LawJoule’s Law
AB
B
v
A
v
v v
W Q V
V E dr
V E r
rV r E V E t
t
W = work done on a small volume of charge as it moves a small distance inside the conductor from point A to point B.
This goes to heat!
V
A B
v
E
+ ++ +
+ ++ +
v
conducting bodyJ E
Joule’s Law (cont.)Joule’s Law (cont.)
Power v
WV v E
tV J E
3[W/m ]
W
d
Vd
V
P J E
P J E dV
Since, we also have
22 3[W/m ]d
JP E
v v
rW V E t V v E t
t
Power/unit volume d
WP J E
V t
J E
Power Dissipation by ResistorPower Dissipation by Resistor
Wvolume
Vd x xP J E LA
I
A
V
L
L
A
IV2
Vd
Vd
P VI
P RI
Hence,
Note: passive sign convention appears in the final result.
Resistor -V
I
A
+
L
x
RC AnalogyRC Analogy
Goal: Assuming we know how to solve the C problem (find C), can we solve the R problem (find R)?
“C Problem”
r F r
A
+ -VAB
B EC
“R Problem”
r F r
A
+ -VAB
B ER
I
RC Analogy (cont.)RC Analogy (cont.)
Theorem: EC = ER (same field)
“C Problem”
r F r
A
+ -VAB
B EC
“R Problem”
r F r
A
+ -VAB
B ER
IA
Hence: EC = ER
“C Problem” “R Problem”
0 0
0
C
C
D E
E
0 0
0
R
R
J E
E
• Same D. E. since (r) = (r)• Same B. C. since the same voltage is applied
RC Analogy (cont.)RC Analogy (cont.)
A
A
AB
s
S
B
A
s
QQC
V V
ds
E dr
D n E n
“C Problem”
r F r
A
+ -VAB
B EC
RC Analogy (cont.)RC Analogy (cont.)
AS
B
A
E n ds
C
E dr
AS
B
A
E n ds
G
E dr
“R Problem”
r F r
A
+ -VAB
B ER
IA
RC Analogy (cont.)RC Analogy (cont.)
1
A
A
AB
A
B
A
S
B
A
S
VVR
I I
E dr
J n ds
J E
E dr
RRE n ds
C G
RC AnalogyRC Analogy
Recipe for calculating resistance:
1. Calculate the capacitance of the C problem.
2. Replace everywhere with to obtain G.
3. Take the reciprocal to obtain R.
In equation form:
Special case: A single homogeneous medium of conductivity surrounds the two conductors.
RC FormulaRC Formula
AS
B
A
E n ds
C
E dr
AS
B
A
E n ds
G
E dr
Hence,
RC
orC G
geometry(GF)
factor
ExampleExampleFind resistance
h1
h2 2
1
Note: We cannot use the RC formula, since there is more than one region.
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