Prof. David R. Jackson ECE Dept. Fall 2014 Notes 4 ECE 2317 Applied Electricity and Magnetism Notes...
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Transcript of Prof. David R. Jackson ECE Dept. Fall 2014 Notes 4 ECE 2317 Applied Electricity and Magnetism Notes...
Prof. David R. JacksonECE Dept.
Fall 2014
Notes 4
ECE 2317Applied Electricity and Magnetism
Notes prepared by the EM Group University of Houston
1
so
Electric Field
F q E
1E F
q
The electric-field vector is the force vector for a unit charge.
Units of electric field: [V/m]
+ + + + + + + + + + + +
- - - - - - - - - - - -
+-
V0 [V] Eq
Note: The “test” charge is small enough so that it does not disturb the charge on the capacitor and hence the field from the capacitor.
2
Electric Field (cont.)
20
ˆ4
qE r
r
Note: The electric-field vector may be non-uniform.
E
qPoint charge q in free space:
3
Voltage Drop
The voltage drop between two points is now defined.
Volta
4
B
AB
A
V V A V B E dr
CA
B
E (x,y,z)
Comment: In statics, the line integral is independent of the shape of the path.
(This will be proven later after we talk about the curl.)
Voltage Drop (Cont.)
EF q E
We now explore the physical interpretation of voltage drop.
CA
B
E (x,y,z)
q test charge
F ext (x,y,z)
A test charge is moved from point A to point B at a constant speed (no increase in kinetic energy).
ext EF F q E 5
Voltage Drop (cont.)
W ext = work done by observer in moving the charge.
B B Bext ext
A A A
W F dr q E dr q E dr
6
CA
B
E (x,y,z)
q test charge
F ext (x,y,z)
extABW qVSo
Voltage Drop (cont.)
so
extABW qV
extW PE B PE A But we also have
ABqV PE B PE A
or
1ABV PE A PE B
q
7
Physical Interpretation of Voltage
8
The voltage drop is equal to the change in potential energy of a unit test charge.
Point A is at a higher voltage, and hence a higher potential energy, than point B.
+ + + + + + + + + + + +
- - - - - - - - - - - -
+-
V0 [V] Eq = 1A
B
1ABV PE A PE B q
Comments
9
The electric field vector points from positive charges to negative charges.
Positive charges are at a higher voltage, and negative charges are at a lower voltage.
+ + + + + + + + + + + +
- - - - - - - - - - - -
+-
V0 [V] E
Review of Doing Line Integrals
10
B
AB
A
V E dr In rectangular coordinates,
ˆ ˆ ˆx y zE x E y E z E
ˆ ˆ ˆ ˆˆ ˆr x x y y z z dr x dx y dy z dz
B
AB x y z
A
V E dx E dy E dz Hence
Note: The limits are vector points.
, , , , , ,B B B
A A A
x y z
AB x y z
x y z
V E x y z dx E x y z dy E x y z dz so
Each integrand must be parameterized in terms of the respective integration variable. This requires knowledge of the path C.
Example
Find: E
Assume: 0ˆ, ,E x y z x E
0 V, , [ ]B
AB
A
V E x y z dr V
+ + + + + + + + + + + +
- - - - - - - - - - - -
+-
V0 [V] E
x = 0
x = h
A
B
11
Ideal parallel-plate capacitor assumption.
Example (cont.)
Evaluate in rectangular coordinates:
0 V, , [ ]B
AB
A
V E x y z dr V
ˆ ˆ ˆx y zE x E y E z E
ˆ ˆ ˆdr x dx y dy z dz
x y zE dr E dx E dy E dz
0 V[ ]B
A
x
AB x
x
V E dx V 12
Example (cont.)
Hence, we have
0
B
A
x
AB x
x
V E dx V
0 0
0
h
E dx V
0 0E h V 0 0 /E V h
0ˆ, ,E x y z x E
0 V/mˆ, , [ ]V
E x y z xh
Recall that
13
Example
A proton is released at point A on the top plate with zero velocity. Find the velocity v(x) of the proton at distance x from the top plate (at point B).
0 0KE x KE PE PE x
210 0
2mv x PE PE x q V V x
14
Conservation of energy:
h = 0.1 [m]
V0 = 9 [V]
+ + + + + + + + + + + +
- - - - - - - - - - - -
+-V0 [V] E
x = 0
x = h
ProtonA
B x = x
x = 0
Example (cont.)
210
2mv x q V V x
0 0
0 0
0B x x
x
A
V VV V x E dr E dx dx x
h h
Hence: 2 01
2
Vmv x q x
h
02qVv x x
h m
15
Example (cont.)
Hence:
02qVv x x
h m
h = 0.1 [m]V0 = 9 [V]
q = 1.602 x 10-19 [C]
m = 1.673 x 10-27 [kg]
6 m/s5.627 10 [ ]v x x
(See Appendix B of the Hayt & Buck book or Appendix D of the Shen & Kong book.)
16
Reference Point A reference point R is a point where the voltage is assigned.
This makes the voltage unique at all points.
The voltage at a given point is often called the potential .
R 0R
C
r = (x,y,z)
Note: 0 is often chosen to be zero, but this is not necessary.
17
Integrating the electric field along C will determine the potential at point r.
Example
Find the potential on the left terminal (cathode) assuming R is on right terminal (anode) and = 0 at the reference point.
9BAV B A
V9 [ ]A
18
9 [V]
A B 0R
9 [V]
+-
9BAV
A B
Find the potential function at x, assuming that the reference point R is on the bottom plate, and the voltage at R is zero.
Example
19
+ + + + + + + + + + + +
- - - - - - - - - - - -
+-
V0 [V] E
x = 0
x = h
r (x, y, z)
0R
0 0
0 0
0x x
x
V Vx E dx dx x
h h
00 h V
00 V[ ]
Vx V x
h
Also,
Hence