Prof. David R. Jackson ECE Dept. Fall 2014 Notes 7 ECE 2317 Applied Electricity and Magnetism Notes prepared by the EM Group University of Houston 1.
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Prof. David R. JacksonECE Dept.
Fall 2014
Notes 7
ECE 2317 Applied Electricity and Magnetism
Notes prepared by the EM Group University of Houston
1
Coulomb’s Law
1 22 2
0
120
N
F/m
ˆ4
8.854187818 10
q qF r
r
Experimental law:
c = speed of light = 2.99792458108 [m/s] (defined)
0 0
1c
7
0 H/m4 10 ( ) exact
Charles-Augustin de Coulomb
2
Here is how we can calculate 0 accurately:
(from ECE 3317)
0 20
1
c
x
y
z
r
q1
q2
r
Coulomb’s Law (cont.)
Hence:
But
1 22 2
0
ˆ4
q qF r
r
E1 = E due to q1
2 2 1F q E r
Note: There is no self-force on charge 2 due to its own electric field.
3
r = location of observer
11 2
0
ˆ4
qE r r
r
x
y
z
r
q1
q2
r
r1= (0, 0, 0)
r2= r
Generalization (q1 not at the origin):
Coulomb’s Law (cont.)
r1= (x1, y1, z1)
r2= (x2, y2, z2)
2 1
12 2 2 2
2 1 2 1 2 1
R R r r
x x y y z z
121 2
0
ˆ4
qE r R
R ˆ R
RR
2 1R r r q2 (x2, y2, z2)
x
y
zR R
r2
r1
q1 (x1, y1, z1)
4
2 1 2 1 2 1ˆ ˆ ˆR x x x y y y z z z
Example
q1 = 0.7 [mC] located at (3,5,7) [m]
q2 = 4.9 [C] located at (1,2,1) [m]
E1 (r2) = electric field due to charge q1, evaluated at point r2
22 2 1F q E r
Find: F1, F2
For F2:
F1 = force on charge q1
F2 = force on charge q2
Rq2
q1
5
Example (cont.)
121 2
0
2 1
2 2 2
m
m
ˆ4
ˆ ˆ ˆ1 3 2 5 1 7
ˆ ˆ ˆ2 3 6 [ ]
2 3 6 7 [ ]
2 3 6ˆ ˆ ˆ ˆ7 7 7
qE r R
R
R r r x y z
x y z
R R
RR x y z
R
q1 = 0.7 [mC] located at (3,5,7) [m]
q2 = 4.9 [C] located at (1,2,1) [m]
22 2 1F q E r
6
Rq2
q1
Example (cont.)
3
21 212
4
62 22 2 1 1
V/m
N
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 [ ]
E r x y z
x y z
F q E r E r
x y z
(Newton’s Law)
2
1
N
N
ˆ ˆ ˆ0.180 0.270 0.539 [ ]
ˆ ˆ ˆ0.180 0.270 0.539 [ ]
F x y z
F x y z
1 2F F
7
General Case: Multiple Charges
q1 : r1 = (x1, y1, z1)
R1 = r - r1
q2 : r2 = (x2, y2, z2)
...
qN : rN = (xN, yN, zN)
R2 = r - r2
...
RN = r - rN
E=E1+E2+…+EN (superposition)
1 21 22 2 2
0 1 0 2 0
ˆ ˆ ˆ...4 4 4
NN
N
qq qE r R R R
R R R
x
y
z
R1
q2
q1 R2 RNR3
q3 qN
r = (x, y, z)
8
Field from Volume Charge
1/22 2 2
ˆ ˆ ˆ
ˆ
R x x x y y y z z z
R R x x y y z z
RR
R
9
r = (x, y, z)
x
y
z
R , ,r x y z
R r r
dV´
, ,v vr x y z ˆ
ˆ
r
r
No
The
The bl
red d
ue dot is as
ot is associ
soc
ate
iated wit
d wit
h
h
te :
Field from Volume Charge (cont.)
20
20
ˆ4
, , ˆ4
v
dQdE R
R
x y z dVR
R
2
0
ˆ4
v
V
rE r R dV
R
10
x
y
zr = (x, y, z)
R
dV´
, ,r x y z
vdQ r dV
R r r R R r r
Exampleq1 = +20 [nC] located at (1,0,0) [m]
q2 = -20 [nC] located at (0,1,0) [m]
Find E (0,0,1)
R1 = (0,0,1) - (1,0,0)
R2 = (0,0,1) - (0,1,0)
1
1
1
1,0,1
2
1ˆ 1,0,12
R
R
R
2
2
2
0, 1,1
2
1ˆ 0, 1,12
R
R
R
r = (0,0,1)
x
y
z
R1
q2 = -20 [nC]
R2
q1 = +20 [nC]
Solution:
13
1 21 22 2
0 1 0 2
ˆ ˆ4 4
q qE R R
R R
Example (cont.)
1 21 22 2
0 1 0 2
9 9
2 212 12
ˆ ˆ4 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 qE R R
R R
V/mˆ ˆ63.55E x y
14
Example
2
ˆ ˆ ˆ
ˆ
ˆ ˆ
R x x x y y y z z z
R z h z
R h z h z h z
R z
Semi-infinite uniform line charge
dl dz
15
2
0
ˆ4
l
C
rE R dl
R
x
y
z
r = (0, 0, h)
R
l = l0 [C/m]
Find E (0, 0, h)
0,0,r z
Example (cont.)
00
20
00
20
0
0
0
ˆ
4
ˆ 1
4
ˆ 1
4
l
l
l
zE dz
h z
zdz
h z
z
h z
0
0
V/mˆ [ ]4
lE zh
Note: The upper limit must be greater than the lower limit to keep dl positive.
16
Example
Find E (0, 0, z)
2
0C
d a d
17
l = l0 [C/m] (uniform)
x
y
z
a
R
r = (0, 0, z)(rectangular coordinates)
, ,0 ( )r a cylindrical
Note: The upper limit must be greater than the lower limit to keep dl positive.
0d so
xa
y
d a d
d
2
0
ˆ4
l
C
rE R dl
R
Example (cont.)
ˆ ˆ ˆ0 cos 0 sin 0
ˆ ˆ ˆcos sin
R x a y a z z
a x y z z
ˆ ˆ ˆcos sinx y Note:
ˆ ˆ ˆR x x x y y y z z z
ˆ ˆR a z z
Hence
y
x
z
a
R
r = (0, 0, z)
18
x
y
Example (cont.)
2 2
2 2
ˆ ˆ
ˆ ˆˆ
R a z z
R a z
a zzR
a z
We can also get these results geometrically, by simply looking at the picture.19
y
x
z
a
R
r = (0, 0, z)
ˆa
ˆz z
2
0
ˆ4
l
C
rE R dl
R
Example (cont.)
20
20
12 22 200 2
2 20
32 20 0 02
ˆ4
ˆ ˆ1
4
ˆ ˆ4
l
C
l
l
rE R dl
R
a zza d
a za z
aa d z z d
a z
20
Reminder: The upper limit must be greater than the lower limit to keep dl positive.
Example (cont.)
0
32 2 2
0
ˆ 2
4
l aE z z
a z
2
0
ˆ 0d
ˆ ˆ ˆcos sinx y
0
32 20 2
V/mˆ [ ]2
l a zE z
a z
2
0
2d
Also,
Hence
21
a
x
y
Example (cont.)
Limiting case: a 0 (while the total charge remains constant)
0 0 03 23
20 0 02
0
20
20
1ˆ ˆ ˆ
2 2 2
2 1ˆ 1 0, 0
4
ˆ [V/m]4
l l l
l
a a az zE z z z
zzz
az z z
z
Qz
z
when when
(point-charge result)
0 2lQ a (total charge on ring)
Note: If we wish Q to remain fixed, than l0 must increase as a gets smaller.
22
where
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