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Solution
ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2012/s16/2/1/solution
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Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford
/
2012
/
S16
/
2
/
1
/
Solution
by
Codex
0
2026-10-25
Treat the
ions
as
a
stationary neutralizing background and displace the
cold
electrons
by
a
small longitudinal
distance
ξ
. Their
electric polarization
is
P
=
−
n
e
e
ξ
, so the restoring
electric field
is
E
=
−
P
/
ϵ
0
=
n
e
e
ξ
/
ϵ
0
. The
electron
equation of motion
is
m
e
ξ
¨
=
−
e
E
=
−
ϵ
0
n
e
e
2
ξ
,
ω
pe
=
m
e
ϵ
0
n
e
e
2
.
(1)
For
E
=
E
0
cos
ω
pe
t
, the oscillatory
velocity
is
v
=
−
e
E
0
sin
ω
pe
t
/
(
m
e
ω
pe
)
. Hence
⟨
u
K
⟩
=
2
1
n
e
m
e
⟨
v
2
⟩
=
4
ϵ
0
E
0
2
=
⟨
2
ϵ
0
E
2
⟩
.
(2)
The
longitudinal mode
exchanges
electron
kinetic energy
and electrostatic
field
energy
like
a
harmonic oscillator
, whose
mean
kinetic and
potential energies
are equal over
a
cycle. This is
a
time
-
average
statement for the coherent
oscillation
, not thermal equipartition with
k
B
T
.
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