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Slide Presentations for ECE 329,
Introduction to Electromagnetic Fields,
to supplement “Elements of Engineering
Electromagnetics, Sixth Edition”
by
Nannapaneni Narayana Rao
Edward C. Jordan Professor of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign, Urbana, Illinois, USA
Distinguished Amrita Professor of Engineering
Amrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, India
1.5
The Electric Field
1.5-2
The Electric Field
is a force field acting on charges by virtue of the
property of charge.
Q1Q2
Coulomb’s Law
F1
a 21
F2
2
a12
4
R
0
Q
2
R
F1
a21
Q1
Q2 Q1
F2
a12
2
4 0 R
0 permittivity of free space
10 –9
F/m
36
1.5-3
D1.13(b)
Q
2a
a
Q
Q2/40(2a2)
Q2/40(4a2)
2/4 (2a2)
Q
0
Q
Q 4 0
Q
From the construction, it is evident that the resultant
force is directed away from the center of the square.
The magnitude of this resultant force is given by
1.5-4
2
Q
2
4 0 2a2
1
1
2
2
2a 4a
0.957
2 N
a
cos 45
Q
2
4 0 4a2
1.5-5
Electric Field Intensity, E
is defined as the force per unit charge experienced by
a small test charge when placed in the region of the
field.
F
E
E Lim
qE
q0 q
–q
q
–qE
Thus
Fe qE
Units:
N N–m V
C C•m m
Sources: Charges;
Time-varying
magnetic field
1.5-6
Electric Field of a Point Charge
Qq
F
a
2 R
4 0 R
(Coulomb’s Law)
q
Q
q
a
2 R
4 0 R
q E due to Q
E due to Q
R
Q
Q
4 0 R
2
aR
aR
1.5-7
Constant magnitude surfaces
are spheres centered at Q.
Direction lines are radial lines
emanating from Q.
E
aR
R
Q
E due to charge distributions
(a) Collection of point charges
Q1
E
n
Qj
2
4
R
0 j
j 1
a Rj
aRn
R1
R2
R3
Q2
Q3
Qn
Rn
aR3
aR2
aR1
1.5-8
Ex.
z
Q (> 0)
d
d2 + x 2
a
e
d
Q (> 0)
y
d2 + x2
x
Electron (charge e and mass m) is displaced from the
origin by D (<< d) in the +x-direction and released
from rest at t = 0. We wish to obtain differential
equation for the motion of the electron and its
solution.
1.5-9
For any displacement x,
F 2
Qe
4 0 d x
2
2
Qex
2 0 d 2 x
2 32
is directed toward the origin,
and x D d.
Qe x
F–
ax
3
2 0 d
cos a ax
ax
1.5-10
The differential equation for the motion of the
electron is
d2 x
Qe x
m 2 –
dt
2 0 d 3
d2 x
Qe
x
0
dt 2 2m 0 d 3
Solution is given by
Qe
x A cos
tB
3
2m 0 d
1.5-11
dx
Using initial conditions x D and
0 at t = 0,
dt
we obtain
x D cos
Qe
t
2m 0 d 3
which represents simple harmonic motion about the
origin with period
2
Qe
.
2m 0 d 3
1.5-12
(b) Line Charges
Line charge
density, rL (C/m)
dl
P
dS
(c) Surface Charges
Surface charge
density, rS (C/m2)
(d) Volume Charges
Volume charge
density, r (C/m3)
dv
1.5-13
Ex. Finitely-Long Line Charge
z
a
rL rL0 4 0 C m
dz
z
f
x
–a
r2 z2
y
r
a
E
ar
dE 2
E
1.5-14
rL dz
4 0 r z
2
2
4 0
a
z 0
2
cos a ar
rL 0r dz
r
2
z
2 32
ar
rL 0r
2 0
z
2 2 2 ar
r r z z 0
rL 0a
a
ar
2a
2 0 r r 2 a2
r r 2 a2
r L0
For a , E
ar
2 0 r
ar
1.5-15
Infinite Plane Sheet of Charge
of Uniform Surface Charge Density
z
a
y2 z 2
z
y
rS0
x
dy
y
1.5-16
dEz 2
rS0 dy
2 0 y 2 z 2
cos a
rS0 z dy
0 y 2 z 2
rS0 z
Ez
0
rS0 z
0
rS0
2 0
dy
y0 y 2 z 2
1 2
da
a
0
z
1.5-17
rS0
E
a z for z 0
2 0
rS0
an
2 0
z<0
rS0
–
az
2 0
rS0
+
z>0
+
rS0
a
2 0 z
+
+
+
z=0
z
1.5-18
D1.16
Given
E(3,5,1) 0 V m
E(1, – 2,3) 6a z V m
E(3, 4,5) 4a z V m
rS1
rS2
rS3
z=0
z=2
z=4
1.5-19
1
rS 1 rS 2 rS 3 0
2 0
1
rS 1 rS 2 rS 3 6
2 0
1
rS 1 rS 2 rS 3 4
2 0
Solving, we obtain
(a) rS 1 4 0 C m
2
r
6
C
m
(b) S 2
0
2
(c) rS 3 2 0 C m
2
(d) E 2,1, 6 4az V m