Equation Sheet for Chapter 10
Temperature conversions:
TC = (5/9)(TF 32)
TF = (9/5)TC + 32
TK = TC + 273.2
Q = mcT, Q is heat, m is mass, c is specific heat, and T is change in temperature.
Q = mL, where Q is heat, m is mass, and L is the latent heat
1st
Equation Sh eet for Chapter 2
distance traveled
d
s=
time
t
change in velocity
"v
acceleration =
a=
time
t
average speed =
for uniform acceleration : v = vo + at,
!
1
d = vo t + at 2
2
Equation Sh eet for Chapter 3
(Note that this is identical to the Chap
Equation Sh eet for Chapter 6
Work = Force " distance W = Fd
work
W
Power =
P=
time
t
1
KE = mv 2 ,
PE = mgh
2
1
Hooke' s Law F = #kx,
PE = kx 2
2
E total mechanical = KE + PE
!
Equation Sheet for Chapter 10
Temperature conversions:
TC = (5/9)(TF 32)
TF = (9/5)TC + 32
TK = TC + 273.2
Q = mcT, Q is heat, m is mass, c is specific heat, and T is change in temperature.
Q = mL, where Q is heat, m is mass, and L is the latent heat
1st
Equation Sh eet for Chapter 2
distance traveled
d
s=
time
t
change in velocity
"v
acceleration =
a=
time
t
average speed =
for uniform acceleration : v = vo + at,
!
1
d = vo t + at 2
2
Equation Sh eet for Chapter 3
(Note that this is identical to the Chap
Equation Sheet for Chapter 12
F=
kq1q2
where F is the force between two static charges, of charges q1 and q2, which
r2
are a distance r apart.
!
E=
Fe
,
q
"V =
!
Fe = qE Where E is the electric field and q is the charge
"PE
where V is change in electric p
PHYSICS 142
Spring 2008
First Test
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containing physical principles, expressions, physical constants and o - . er material that may
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PHYSICS 142 L 4
Spring 2010 "7 a 3 V ,
Firstth 7 c; (A
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containing physical principles, expressions, physical constants and ther material that may
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PHYSICS 142
Spring 2011
First Test
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119
Avg 2/ =
\50
PHYSICS 142
Fall 2014
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containing physical principles, expressions, physical constants and other material that may
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PHYSICS 142
Fall 2009
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containing physical principles, expressions, physical constants and ot er material that may
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PHYSICS 142
Spring 2007
First Test
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PHYSICS 142
2009
Spring
First Test
nu."-
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containing physical principles, expressions, physical constants and other material that may
housed:
3045:
a EN 5.2 _:_< I 33.5
NAME:
2. A pair of parallel plates are separated by a distance of 0.050 m . There is vacuum between the
plates. The plates are connected to a voltage source. The voltage difference between the two
plates is unknown. A free electron at the center of the pl
PHYSICS 142
Fall 2012
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PHYSICS 1-42
Fall 2008
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containing physical principlu, expressions, physical constants and other material that may
he used in anyofthepmhlemswillheglvnsepmtely. Donottm in
PHYSICS 142
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First Test
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containing physical principles, expressions, physical constants and other material that may
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PHYSICS 142
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First Test
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Chapter 24
26. (You might need to read page 683 to understand more about this kind of problem). Given
the intensity S = 1.0 103 W/m2 . Find the electromagnetic energy (or total energy) contained
in 5.5 m3 .
Answer:
Total energy = (total energy density) Vo
Homework #1: Solution
1. Using the Dirac-delta function find the expression for charge density
(a) for a charge Q uniformly distributed over a spherical shell of radius R.
(b) for a charge Q uniformly distributed over a circular ring of radius R lying in
(Problem 1) Starting with the geometric series
X
zn =
n=0
1
1z
Show that
a.
X
n cos n =
cos 2
1 + 2 2 cos
n sin n =
sin
1 + 2 2 cos
n=1
b.
X
n=1
c.
X
n cos n
1
= log(1 + 2 2 cos )
n
2
n=1
Answer: Consider
X
(ei )n =
n=0
=
1+
X
(ei )n =
n=1
=
=
=
X
n=
NAME :
1. Two point-charges of charges +q and4q are held at the corners of an isosceles rectangle
triangle, as showri in gure below. The absolute value of q is 1.414 uC. The distance d as shown
in the gure is 0.5 m. The angle at A is 90. The gravitational
Nam Nguyen
98224593
Physics 927
8.2-Density of Levels
(a) In the free electron case the density of levels at Fermi energy can be written in the form (Eq.
(2.64) g(F ) = mkF /~2 2 . Show that the general form (8.63) reduces to this when F = ~2 k 2 /2m and
Chapter 18
13. Two point charges are fixed on the y axis: a negative point charge
q1 = 25 C at y1 = +0.22 m and a positive point charge q2 at y2 = +0.34
m. A third point charge q = +0.84 C is fixed at the origin. The net
electrostatic force exerted on the
Nam Nguyen
98224593
Physics 911
Homework 2
Problem 1
Z
Z
Z p
Z p
Z
p
ds
1
c
1
n0
T = dt =
=
(dx)2 + (dy)2 =
n 1 + y 02 dx =
ey 1 + y 02 dx
v
c
v
c
c
Z
p
n0
= T =
A(y, y 0 )dx
where A(y, y 0 ) = ey 1 + y 02
c
Now consider
Z
S[x] =
A(y, y 0 )dx
(1)
Since A(
Nam Nguyen
98224593
Physics 911
Homework 3
1. For a particle of charge q subject to an electromagnetic field with potentials and A. Show
that the Jacobi energy is
1
h = mv 2 + q
2
Using the equations of motion, show that h is constant provided and A have