Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston Fall 2004...
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Transcript of Prof. Jeffery T. Williams Dept. of Electrical & Computer Engineering University of Houston Fall 2004...
Prof. Jeffery T. WilliamsDept. of Electrical & Computer Engineering
University of Houston
Fall 2004
Coulomb’s Law
ECE 2317: ECE 2317: Applied Electricity and MagnetismApplied Electricity and Magnetism
Charles Coulomb
Coulomb’s Force Law
• proportional to the product of the charges,
• inversely proportional to the square of the distance between them,
• directed along the line joining them, and
• repulsive for like charges and attractive for unlike charges.
Charles A. Coulomb, a French military engineer, measured the forces between electric charges, reporting his results in 1785. He found that the electric force between two charged particles was
y
z
x
1q
2q
r
21F
r̂
Coulomb’s LawCoulomb’s Law
1 22 2
0
120
ˆ N4
8.854187818 10 F/m
q q
r
F r
Experimental law:
Note: c = speed of light = 2.99792458108 [m/s] (exactly)
0 0
1c
7
0 4 10 H/m (exactly)
x
y
z
q1
q2
r̂
r
Electric FieldElectric Field
Hence:
But
1 22 2
0
ˆ4
q q
rF r
12
0
ˆ4
q
r1E r
2 2 1 2qF E r
Note: no self-force on charge 2 due to its own electric field
x
y
z
q1
q2
r̂
r
( : electric field due to q1 )1E
Generalization (q1 not at the origin):
Generalizing Coulomb’s LawGeneralizing Coulomb’s Law
r1= (x1, y1, z1)
q2 (x2,y2,z2)
x
y
z R̂q1 (x1,y1,z1)
r2= (x2, y2, z2)
2 1
12 2 2 2
2 1 2 1 2 1
R
x x y y z z
R r r
12
0
ˆ4
q
R1E R
R
RR
2 1 R r r
1r2r
R
ExampleExample
q1=0.7 [mC] located at (3,5,7) [m]
q2=4.9 [C] located at (1,2,1) [m]q2
q1
E1 = electric field due to charge q1, evaluated at point r2
2q2 1F E
Find: F1, F2
For F2:
F1 = force on charge q1
F2 = force on charge q2
R
Example (cont.)Example (cont.)q1=0.7 [mC] located at (3,5,7) [m]
q2=4.9 [C] located at (1,2,1) [m]
Rq2
q1
11 2
0
2 1
2 2 2
4
ˆ ˆ ˆ1 3 2 5 1 7
ˆ ˆ ˆ2 3 6
2 3 6 7 [ ]
2 3 6ˆ ˆ ˆ
7 7 7
q
R
R m
E R
R r r x y z
R x y z
R
RR x y z
R
2 2q 1F E
Example (cont.)Example (cont.)
3
1 212
41
62 2 1 1
2
2
1
ˆ ˆ ˆ0.180 0.
0.7 10 2 3 6ˆ ˆ ˆ
7 7 74 8.854 10 7
ˆ ˆ ˆ1.834 10 2 3 6
4.9 10
ˆ ˆ ˆ0.08988 2 3 6 [ ]
270 0.539 [ ]
ˆ ˆ ˆ0.180 0.270
q
N
N
E x y z
E x y z
F
F x y z
F
F x
x
E
y
E
y z
0.539 [ ]Nz(Newton’s Law)
x
y
z
General Case: Multiple ChargesGeneral Case: Multiple Chargesq1 : r1 = (x1,y1,z1)
q2
q1
q3 qN
q2 : r2 = (x2,y2,z2)
...
qN : rN = (xN,yN,zN)
(superposition)
1 21 22 2 2
0 1 0 2 0
...4 4 4
NN
N
qq q
R R R E R R R
r
1 1 R r r
2 2 R r r
N N R r r1 2 N E E E E
1R
2R3R
NR
x
y
z
Field from Volume ChargeField from Volume Charge
r= (x,y,z)
V (r´) = V (x´,y´,z´)
1/ 22 2 2
ˆ ˆ ˆ' ' '
' ' '
x x y y z z
R x x y y z z
R r r x y z
R
RR
R
' ', ', 'x y zrR
x
y
z
Field from Volume Charge (cont.)Field from Volume Charge (cont.)
r = (x,y,z)
R
dQ = V (r´) dV´
dV´
20
20
4
', ', ' '
4V
dQd
R
x y z dV
R
E R
R
2
0
''
4V
V
r dVR
r
E R
' ', ', 'x y zr
r = (x,y,z)
dQ = S (r´) dS´
2
0
''
4S
S
dSR
r
E R
Field from Surface ChargeField from Surface Charge
x
y
z
R
dSdS´
' ', ', 'x y zr
dQ = l (r´) dl´
2
0
''
4l
C
rdl
RE R
Field from Line ChargeField from Line Charge
x
y
z r = (x,y,z)
R
dl´
' ', ', 'x y zr
ExampleExample
r = (0,0,1)
x
y
z
R1
q2 = -20 [nC]
R2
q1 = +20 [nC]
q1 = +20 [nC] located at (1,0,0)
q2 = -20 [nC] located at (0,1,0)
Find E (0,0,1)
Solution:
R1 = (0,0,1) - (1,0,0)
R2 = (0,0,1) - (0,1,0)
1
1
1
1,0,1
2
11,0,1
2
R
R
R
2
2
2
0, 1,1
2
10, 1,1
2
R
R
R
Example (cont.)Example (cont.)
1 21 22 2
0 1 0 2
9 9
2 212 12
4 4
20 10 1 20 10 11,0,1 0, 1,1
2 24 8.854 10 2 4 8.854 10 2
63.55 1,0,1 0, 1,1
63.55 1, 1,0
q q
R R
E R R
ˆ ˆ63.55 V/m E x y
x
y
z
ExampleExample
r = (0,0,h)
R
l = lo [C/m]
r´= (0, 0, z´ )
20
2
''
4
ˆ '
' ' '
ˆ
l
C
rdl
R
h z
R h z h z h z
E R
R r r
z
R z
Find E(0,0,h)
h>0
semi-infinite uniform line charge
Example (cont.)Example (cont.)
0
20
0
20
0
0
ˆ'
4 '
ˆ 1'
4 '
ˆ 1
4 '
lo
lo
lo
dzh z
dzh z
h z
zE
z
z
0
ˆ [V/m]4
lo
h
E z
ExampleExample
x
a
´
d´
dl´= a d´
l = lo [C/m] (uniform)
Find E(0,0,z)
y
x
y
z
a
R
r = (0,0,z)
r´= (a, ´, 0 )
d´ > 0 so 2
0
' 'C
d a d
2
0
''
4l
C
dR
r
E R
Example (cont.)Example (cont.)
ˆ ˆ ˆ0 cos ' 0 sin ' 0
ˆ ˆ ˆcos ' sin '
a a z
a z
R x y z
x y z
ˆ ˆ ˆ' cos ' sin ' ρ x y
x
´
ˆ 'ρ
Note:
R r r
y
x
y
z
a
R
r = (0,0,z)
r´= (a, ´, 0 )
Example (cont.)Example (cont.)
2 2
2 2
ˆ ˆ'
ˆ ˆ'
a z
R a z
a z
a z
R ρ z
ρ zR
R
a
z
x
Example (cont.)Example (cont.)
20
2
12 22 200 2
2 2
32 20 0 02
''
4
ˆ ˆ1 ''
4
ˆ ˆ' ' '4
l
C
lo
lo
rdl
R
a za d
a za z
aa d z d
a z
E R
ρ z
ρ z
Example (cont.)Example (cont.)
3
2 2 20
ˆ 2
4
lo az
a z
E z
2
0
ˆ ' ' 0d ρ
ˆ ˆ ˆ' cos ' sin ' ρ x y
3
2 20 2
ˆ [V/m]2
lo a z
a z
E z
2
0
' 2d
ˆρ
Example (cont.)Example (cont.)
Limiting case: a 0 (while the total charge remains constant)
30
0
20
20
ˆ2
2 1ˆ 1 when 0, when 0
4
ˆ4
lo
l
a z
z
az z
z
Q
z
E z
z
z
(point-charge result)