ECE 3318
Applied Electricity and Magnetism
Spring 2016
Prof. David R. Jackson
ECE Dept.
Notes 7
Notes prepared by the EM Group
University of Houston
1
Coulomb’s Law
z
r̂
r
Experimental law:
q1q2
F2 
rˆ
2
4 0 r
q2
 N
y
 0  8.854187818 10 12
q1
 F/m 
(permittivity of free space)
x
Here is how we can calculate 0 accurately:
Charles-Augustin de Coulomb
c = speed of light = 2.99792458108 [m/s] (defined)
c
1
 0 0
(from ECE 3317)
0  4 107  H/m (exact)
1
0 
0c 2
2
Coulomb’s Law (cont.)
A “test” charge q2 is placed at r to measure the electric field there from charge q1.
z
r̂
r2 = r
r
F2 
q2
But
q1q2
rˆ
2
4 0 r
F2  q2 E1  r 
y
q1
x
r1= (0, 0, 0)
E1 = E due to q1
Note: There is no self-force on
charge 2 due to its own electric field.
r = location of charge 2
Hence:
E1  r  
q1
4 0 r
2
rˆ
3
Coulomb’s Law (cont.)
r1= (x1, y1, z1)
Generalization (q1 not at the origin):
z
q1 (x1, y1, z1)
r1
r2= (x2, y2, z2)
R̂
R
q2 (x2, y2, z2)
R  r 2  r1
R  xˆ  x2  x1   yˆ  y2  y1   zˆ  z2  z1 
r2
y
R  R  r 2  r1
  x2  x1    y2  y1    z2  z1  


2
2
2
1
2
x
E1  r 2  
q1
4 0 R 2
Rˆ
R
Rˆ 
R
4
Example
q1 = 0.7 [mC] located at (3,5,7) [m]
q2 = 4.9 [C] located at (1,2,1) [m]
q1
Find: F1, F2
R
q2
F2
F1
F1 = force on charge q1
F2 = force on charge q2
For F2:
F2  q2 E1  r 2 
E1 (r2) = electric field due to charge q1, evaluated at point r2
5
Example (cont.)
q1 = 0.7 [mC] located at (3,5,7) [m]
q2 = 4.9 [C] located at (1,2,1) [m]
q1
E1  r 2  
q1
4 0 R
2
Rˆ
F1
R
q2
F2
F2  q2 E1  r 2 
R  r 2  r1  xˆ 1  3  yˆ  2  5  zˆ 1  7 
 xˆ  2   yˆ  3  zˆ  6 
R R 
 2 
2
[ m]
  3     6   7
R
 2 
Rˆ 
 xˆ   
R
 7 
2
2
[ m]
 3 
 6 
yˆ    zˆ  
 7 
 7 
6
Example (cont.)
E1  r 2  
0.7 103
4  8.854 1012   7 
2
  2 
 3 
 6  
ˆ
ˆ
ˆ
x

y

z
 
 
  7 
 7 
 7 
  
 1.834 104  xˆ  2   yˆ  3  zˆ  6  
[ V/m]
F2  q2 E1  r 2   4.9 106 E1  r 2 
 0.08988  xˆ  2   yˆ  3  zˆ  6  
F1   F2
[ N]
(Newton’s Law)
F2  xˆ  0.180   yˆ  0.270   zˆ  0.539 
[ N]
F1  xˆ  0.180   yˆ  0.270   zˆ  0.539 
[ N]
7
General Case: Multiple Charges
z
r = (x, y, z)
q2 : r2 = (x2, y2, z2)
R1
q1
q2
q1 : r1 = (x1, y1, z1)
...
R2 R
3 RN
q3
qN : rN = (xN, yN, zN)
qN
R1 = r - r1
y
R2 = r - r2
...
x
RN = r - rN
E=E1+E2+…+EN (superposition)
E r  
q1
4 0 R12
Rˆ 1 
qN
q2
ˆ
ˆ
R

...

R
2
N
4 0 R2 2
4 0 RN 2
8
Field from Volume Charge
z
r = (x, y, z)
R  r  r
R
r    x, y, z 
dV´
Note :
v  r   v  x, y, z
y
x
r
The red dot is associated with r 
The blue dot is associated with
R  xˆ  x  x   yˆ  y  y   zˆ  z  z  
1/2
2
2
2




R  R  x  x   y  y  z  z  


R
ˆ
R
R
9
Field from Volume Charge (cont.)
z
r = (x, y, z)
R
r    x, y, z
dQ ˆ
dE 
R
2
4 0 R
dV´
dQ  v  r  dV 
y
x
v  x, y, z   dV  ˆ

R
2
4 0 R
v  r  ˆ
E r   
R dV 
2
4 0 R
V
R  r  r
R  R  r  r
10
Field from Surface Charge
z
r = (x, y, z)
R
dSdS´
y
r    x, y, z 
x
dQ  s  r  dS 
s  r  ˆ
E r   
R dS 
2
4 0 R
S
R  r  r
R  R  r  r
11
Field from Line Charge
z
r = (x, y, z)
R
l  r   ˆ
E
R dl 
2
4 0 R
C
dl
y
r    x, y, z 
x
dQ  l  r  dl 
R  r  r
R  R  r  r
Note : dl   dr
12
Example
q1 = +20 [nC] located at (1,0,0) [m]
z
q2 = -20 [nC] located at (0,1,0) [m]
r = (0,0,1)
Find E (0,0,1)
Solution:
R2
R1 = (0,0,1) - (1,0,0)
R1
y
q2 = -20 [nC]
q1 = +20 [nC]
x
E
q1
4 0 R12
Rˆ 1 
q2
Rˆ 2
2
4 0 R2
R2 = (0,0,1) - (0,1,0)
R1   1, 0,1
R 2   0, 1,1
R1  2
R2  2
1
Rˆ 1 
 1, 0,1
2
1
Rˆ 2 
 0, 1,1
2
13
Example (cont.)
E

q1
4 0 R12
Rˆ 1 
q2
Rˆ 2
2
4 0 R2
20 109
4  8.854 10
12
 2 
2
20 10 9
 1

 1, 0,1 

 2
 4  8.854 10 12 
 2
2
 1

0,

1,1




 2

 63.55  1, 0,1   0, 1,1 
 63.55  1, 1, 0  
E  63.55   xˆ  yˆ 
 V/m
14
Example
l  r   ˆ
E
R dl 
2
4 0 R
C
z
Find E (0, 0, h)
r = (0, 0, h)
R  xˆ  x  x   yˆ  y  y   zˆ  z  z  
R  zˆ  h  z  
R
y
x
r    0,0, z
R
 h  z
2
 h  z  h  z
Rˆ  zˆ
l = l0 [C/m]
dl  dz
Semi-infinite uniform line charge
15
Example (cont.)
l 0
E
4 0
zˆ l 0

4 0
zˆ l 0

4 0
0
zˆ
  h  z
2
dz 
Note: The upper limit must be
greater than the lower limit to
keep dl positive.
dz 
(This is different from voltage
drop calculations, where the
upper limit can be smaller.)

0
1
  h  z
2

0
 1 


  h  z    
 l 0 
E  zˆ 
 [V/m]
 4 0 h 
16
Example
z
r = (0, 0, z)
Find E (0, 0, z)
(rectangular coordinates)
R
l = l0 [C/m] (uniform)
a
y
l  r   ˆ
E
R dl 
2
4 0 R
C
r   a,  ,0 (cylindrical)
x
y
d 

a
d   a d 
x
d   0 so
Note: The upper limit must
be greater than the lower
limit to keep dl positive.
 d
C

2
   a d 
0
17
Example (cont.)
R  xˆ  x  x  yˆ  y  y  zˆ  z  z
z
r = (0, 0, z)
R  xˆ  0  a cos     yˆ  0  a sin     zˆ  z  0 
R


 a xˆ cos    yˆ sin    zˆ z
a
y
Note:
y
x
̂ 
ˆ   xˆ cos    yˆ sin  

x
Hence
R   a ˆ   z zˆ
18
Example (cont.)
l  r   ˆ
E
R dl 
2
4 0 R
C
z
r = (0, 0, z)
R
R  a ˆ   z zˆ
z zˆ
a
y
 a ˆ 
̂ 
x
R  a2  z 2
ˆ   zzˆ

a

Rˆ 
a2  z 2
We can also get these results geometrically, by simply looking at the picture.
19
Example (cont.)
Continuing with the calculation…
l  r   ˆ
E
R dl 
2
4 0 R
C
  a ˆ   zzˆ 
 l 0  
1

 a d 

 
  2
2
2
2 1/2 


4

a

z


0 
0 
 a  z  
2



a

  l0
3/2
 4 0  a 2  z 2  


2
 2

 a  ˆ  d   zˆ z  d 
0
 0

Reminder: The upper limit must be greater than the lower limit to keep dl positive.
20
Example (cont.)
y
̂ 
2
a
x
 ˆ  d   0
 ˆ   xˆ cos  yˆ sin 
0
2
Also,
 d   2
0



a
l0
  zˆ z  2 
Hence E  
3/2

 4  a 2  z 2   
0


or

 l 0 a  
z
E  zˆ 
  2 2 3/2 
 20   a  z  
[V/m]
21
Example (cont.)
Limiting case: a  0 (while the total charge remains constant)


  a  z 
E  zˆ  l 0  

3 
 2 0    z 2  2 


l 0  2 a  1
ˆ
z
 1
4  0 z 2
  zˆ
Q
4  0 z
2
[V/m]
 l 0 a   z 
zˆ 
  3  
 2 0   z 
  a  1 
zˆ  l 0    2 
 2 0   z 
  when z  0,
 when z  0 
(point-charge result)
where Q  l 0  2 a  (total charge on ring)
Note: If we wish for Q to remain fixed, than l0 must increase as a gets smaller.
22