ECE 329
Homework 2 - Solution
Due: Feb. 2, 2010
Spring 2010
1. a) According to Gauss' Law,
E dS =
S
1
0
dV =
V
1
0
(3) dV =
V
3
0
L3 (V m) .
b)
E dS =
S
1
0
dV =
V
1
0
1/2
1/2
1/2
x2 + y 2 + z 2 dxdydz
1/2 1/2 1/2
By symmetry,
E dS =
S
3
0
1/2
1/2
ECE 329
Homework 4 - Solution
Due: Feb. 16, 2010
Spring 2010
1. For E = 2xx + 2sin (z ) y 2z z , we have
x E=
x
y
y
z
z
= 2cos (z ) x.
2x 2sin (z ) 2z
Since 2.
E = 0, eld E is not electrostatic.
a) The bottom plate is holding the positive charge d
ECE 329 Hour Exam 1 July 1, 2009
a) (10 pts) In a certain region of space electrostatic potential is specied as
V(x, y, z) = 2333; + 22 V.
Find the corresponding electrostatic eld E.
b) (10 pts) In a certain region of space where e = 60 a static electr
ECE 329
Homework 9 - Solution
Due: Mar. 30, 2010
Spring 2010
1. a) For plane waves propagating in free space, E and H are related by
E = 0 H , where is the unit vector in the wave propagation direction. Therefore, H1 = 1 1 (y ) E1 = [x 2cos ( t + y )] (A
ECE 329
Homework 8 - Solution
Due: Mar. 16, 2010
Spring 2010
1. a) From the problem, we know that in the free space region z > 0, the electric ux density is
D1 = 3x + 2z C/m2 . a a
Therefore, the electric eld is
E1 = D1
1
=
D1
0
=
3
0
ax +
2
0
az (V/m) .
ECE 329
Homework 7 - Solution
Due: Mar. 9, 2010
Spring 2010
1. Verifying vector identity
H EE H= x H EE H = = 3y ez 3y e
z x 4ez
(E H) z
z
for E =
4ez x
and H =
3ez . y
The left-hand side of the identity gives
y
y
x 4xez
z x
0
z 2z
0 4xe 3xe
0
z
y
ECE 329
Homework 6 - Solution
Due: Mar. 2, 2010
Spring 2010
1. The magnetic eld at the origin is a superposition of those generated by the two sheets. a) Since the currents are owing in z direction and they extend to innity in both x and z directions, the
ECE 329
Homework 5 - Solution
Due: Feb. 23, 2010
Spring 2010
1. a) According to Page 2 in Lecture 11, the mobility is dened as
mobility = q q = . m m
Although m is not explicitly given in the problem, it is related to Ne and by (see Page 3 in Lecture 11)
ECE 329
Homework 3 - Solution
Due: Feb. 9, 2010
Spring 2010
1.
E=V = x +y +z x y z = z 2 x 2xz z , = = = D ( 0 E)
0(
xz 2
E) +y +z = 0x x y z = 2 0 x.
x z 2 z 2xz
2. According to the relationship between electric eld and potential, we have
V (0, 0, 0)
ECE 329 Introduction to Electromagnetic Fields Spring 10
University of Illinois Roach, Kim, Goddard, Kim
Exam 1
Thursday, Feb 18, 2010 7:00-8:15 PM
l 1
Section: 9 AM 12 Noon 1 PM 2 PM I
Please clearly PRINT your name in CAPITAL LETTERS