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PHY 251 Fall 2009: homework problem set 4, due in the PHY 251 drop box in room A129 by
noon on Friday, Oct. 9.
1. Do Serway problem 4.2 on p. 146. This problem illustrates how electrochemistry experi
ments already gave some idea about the discrete charge of the electron, once you believe in
discrete atoms and know Avogadro’s number!
Answer:
Copper sulfate involves Cu
+2
and SO
−
2
4
, so there are two valence electrons per
atom. We then find that the fraction of a mole is
(1
C/sec
)
·
(3600
sec
)
(96
,
500
C/mol
)
·
2
= 0
.
0187
mol
.
The number of coppper atoms deposited is then
(0
.
0187
mol
)
·
(6
.
02
×
10
23
atoms/mol
) = 1
.
13
×
10
22
,
the weight of a copper atom is
1
.
185
×
10
−
3
kg
1
.
13
×
10
22
atoms
= 1
.
05
×
10
−
25
kg
,
and the molar mass of copper is
1
.
185
g
0
.
0187
mol
= 63
.
4
g/mol
which is about right (of course they gave us numbers that would work out!).
2. Do Serway problem 4.3.
Answer:
In the apparatus, we have
F
y
=
q
V
d
=
m
Δ
v
y
Δ
t
and we also have
v
x
=
ℓ/
Δ
t
or
Δ
t
=
ℓ/v
x
. Therefore we can write
Δ
v
y
=
qV
Δ
t
md
=
qV
md
ℓ
v
x
.
We can then figure out the angle
θ
:
tan
θ
=
v
y
v
x
=
qV ℓ
mdv
2
x
so
q
m
=
dv
2
x
tan
θ
V ℓ
.
Now from the magnetic field cancelling the electric field we can say
qV
d
=
qv
x
B
so
v
x
=
V
Bd
=
2000
4
.
57
×
10
−
2
·
0
.
02
= 2
.
19
×
10
6
m/s
which means relativistic effects are small (
γ
= 1
.
000 027
). We can also substitute this result
for velocity into our expression for
q/m
:
q
m
=
d
V ℓ
V
2
B
2
d
2
tan
θ
=
V
ldB
2
tan
θ
=
2000
0
.
10
·
0
.
02
·
(4
.
57
×
10
−
2
)
2
tan(0
.
2) = 9
.
7
×
10
7
C/kg
.
A proton has
q/m
= 1
.
6
×
10
−
19
/
1
.
67
×
10
−
27
= 9
.
7
×
10
7
so it seems like we have a good
candidate
. . .
1
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View Full Document3. Do Serway problem 4.4.
Answer:
The electric field the electron experiences when it travels between the two plates is
E
y
=
V/d
, and the acceleration it experiences is
a
=
F/m
=
qE/m
=
qV/
(
md
)
. Since we
expect
v
y
≪
v
x
, we can say that the time that the electron experiences this acceleration is
Δ
t
=
ℓ/v
x
. Consequently the electron receives a velocity kick in the
ˆ
y
direction of
v
y
=
a
y
·
Δ
t
=
qV
md
·
ℓ
v
x
so that its angle
θ
leaving the field region is
θ
=
v
y
/v
x
=
y
2
/D
(the latter by geometry in
the small angle limit) or
θ
=
y
2
D
=
v
y
v
x
=
qV
md
ℓ
v
2
x
which can be solved for
y
2
to give
y
2
=
qV Dℓ
mdv
2
x
.
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 Fall '01
 Rijssenbeek
 Physics, Charge, Work

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