Version 005 – newfinalchiu – chiu – (56565)
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001
10.0points
An electron and a neutral carbon atom of
polarizability
α
are at a distance
r
apart (
r
is much greater than the diameter
d
of the
atom).
Due to polarization of the atom by
the electron, there is a force
F
between the
electron and the carbon atom. If we change
r
to 3
r
, what will be the new force
F
′
between
the two?
Hint: To find the
r
dependence of
the force, first find the induced polarization
(dipole moment) of the atom as a function of
r
. Then find the force exerted by this induced
dipole on the electron.
1.
F
′
=
F
243
correct
2.
F
′
= 27
3.
F
′
=
F
4.
F
′
=
F
3
5.
F
′
=
F
81
6.
F
′
=
F
9
7.
F
′
= 9
F
8.
F
′
= 3
F
9.
F
′
=
F
27
Explanation:
The force will be proportional to
r
−
5
and
hence the correct answer will be
F
′
=
F
243
.
The calculation is as follows.
The magnitude of electric field due to the
electron at the location of the carbon atom
will be given by
E
e
=
1
4
πǫ
0
e
r
2
From this,
the magnitude of the induced
dipole moment of the carbon atom can be
written down as
p
=
αE
e
.
This dipole mo
ment will be pointing towards the electron,
and hence the electron will lie on the axis of
this dipole.
The magnitude of the electric
field due to this dipole at the location of the
electron will be given by
E
p
=
1
4
πǫ
0
2
p
r
3
Here, we have used the fact that
r >> d
. The
magnitude of force on the electron due to the
dipole (and vice versa) can be obtained as
F
=
E
p
e
.
Combining all the equations, we
get
F
=
E
p
e
=
1
4
πǫ
0
2
αE
e
e
r
3
=
parenleftbigg
1
4
πǫ
0
parenrightbigg
2
2
αe
2
r
5
This implies that when
r
to 3
r
, the electric
field field, and consequently the force, is de
creased by a factor of 3
5
= 243, and hence the
answer is
F
′
=
F
243
.
002
10.0points
An electron oscillating along the
y
axis with
angular frequency
ω
and amplitude
A
passes
through the origin at time
t
= 0 and is moving
in the +
y
direction. Which formula desribes
the time dependence of
E
rad
at a point
vectorp
=
(
x, x/
√
3
,
0
)
?
(Note: ˆ
n
is the unit vector along the direc
tion of
vectora
⊥
an instant after
t
= 0.)
1.
−
√
3
8
πǫ
0
eω
2
A
c
2
x
sin(
ωt
)ˆ
n
2.
1
4
πǫ
0
eω
2
A
c
2
x
sin(
ωt
)ˆ
n
3.
√
3
8
πǫ
0
eω
2
A
c
2
x
sin(
ωt
)ˆ
n
4.
√
3
16
πǫ
0
eω
2
A
c
2
x
sin(
ωt
)ˆ
n
5.
−
√
3
16
πǫ
0
eω
2
A
c
2
x
sin(
ωt
)ˆ
n
6.
−
3
16
πǫ
0
eω
2
A
c
2
x
sin(
ωt
)ˆ
n
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2
7.
−
1
4
πǫ
0
eω
2
A
c
2
x
sin(
ωt
)ˆ
n
8.
3
16
πǫ
0
eω
2
A
c
2
x
sin(
ωt
)ˆ
n
correct
Explanation:
First we note that the point
vectorp
makes a 30
◦
angle with the
xz
plane.
Calling the angle
between
vectorp
and the
xz
plane
θ
, we can see this
from
tan
θ
=
parenleftbigg
x
√
3
parenrightbigg
x
=
1
√
3
and the fact that tan(30
◦
) = 1
/
√
3.
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 Spring '08
 Turner
 Magnetic Field, Electric charge, newfinalchiu chiu

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