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Oct. 3, 2014 /
PHYS 381: Electromagnetic Theory 1
Hegmann; Fall 2014
Assiment #3
DUB: Friday, Oct. 10, drop box for PHYS 381, by 11 am.
1. A point charge of +3 pC is situated 2 cm away from a point charge of -3 pC.
(1 pC = 10'12 C) Using innit
Oct. 29, 2014
PHYS 381: Electromagnetic Theory I
Hegmann; Fall 2014
Assignment #5
DUE: Friday, Nov. 7, drop box for PHYS 381, by 11 a.m.
1.
Problem 3.9
(force between a point charge and a neutral conducting sphere at V0)
2.
Problem 3.41
(negatively charge
Sept. 24, 2014 3 0L U T WIVS
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PHYS 381: Electromagnetic Theory I
Hegmann; Fall 2014
Assign_rr_1ent #2
DUB: Thursday, Oct. 2, drop box for PHYS 381, by 4 pm.
1. What is the divergence of the electric eld 2 m away from a point charge of +3 pC?
2. Problem 2
Canadian Light Source (CLS) synchrotron
5
Lorentz force used
for bending magnets
for electron beam
Cathode ray tubes (CRTs)
Lorentz force used to
steer electron beam in
CRTs
Magnetron in a microwave oven
f = 2.45 GHz
Transverse electric and magnetic
field
Solutions to assignment 8
1. Si, with electronic structure 1s2 2s2 2p6 3s2 3p2 , has
4 electrons (3s2 3p2 ) outside its closed n = 2 shell.
(a) Al has 3 valence electrons (3s2 3p1 ). Thus, it
will act as an acceptor for one electron in the Si
structure; a
Solutions to assignment 7
1. (a) The current density is j = I/A = I/(d/2)2
and has the value
4(103 A)
= 479 A m2 .
j=
(1.63 103 m)2
(b) The drift velocity is vd = I/Ane = j/ne. Using
the j value from part (a), we nd
vd =
(8.47
1028
479 A m2
m3 )(1.602 10
Solutions to assignment 6
1. The dipole moment for a fully ionic bond involves
a transfer of one complete unit of charge over a
distance r0 :
or 44% ionic.
2. For N2 , the fundamental rotational energy is
pionic = er0
E0r = 2.48 104 eV =
= (1.60 1030 C)(0
Solutions to assignment 2
= 1 mv 2 is given in
2
1. The average value of the energy
the usual way by
=
Written explicitly, this becomes
U=
d3 v e
.
d3 v e
=
=
dv 4v 2 .
dvx dvy dvz =
(2)
0
=
dv (4v 2 ) e
0
dv (4v 2 ) e
0
1
mv 2
m 0 dv v 4 e 2
1
2
2
dv v
Solutions to assignment 3
1. Surface waves on liquid 4 He have a dispersion relation
(k) = (/)1/2 k 3/2 . Here, the wavevector k = (kx , ky )
is restricted to two dimensions, so its sum has the form
(Area)
hc = 1.2398 106 eV m,
kB = 8.6173 105 eV K1 .
Si
Solutions to assignment 1
2
2
2
1(a). E = 1 m(vx + vy + vz ) + mgz
2
2(b). In one dimension, there are exactly two states with
2
speed v = vx = |vx |, namely vx = v and vx = v.
Thus,
1(b). The average height of each particle is
z =
=
d3 v d3 r zeE
d3 v d3
Study Guide
Major concepts
What is thermodynamic equilibrium? How is temperature dened?
How is thermodynamic equilibrium established? What role do interactions play?
Why does a macroscopic number of degrees of freedom require a probabilistic theory?
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