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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/2016/s29/3/a/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
/
2016
/
S29
/
3
/
a
/
Solution
by
Codex
0
2026-10-25
For
a
thin uniform
spherical shell
,
d
C
=
(
R
sin
θ
)
2
d
m
and angular averaging gives
⟨
sin
2
θ
⟩
=
2/3
. Thus
C
=
2
M
R
2
/3
. The printed
formula
requires
a
negligibly thin outer shell. For
a
finite-thickness
hollow
sphere
of
uniform density
between
r
0
and
R
,
C
=
5
2
M
R
3
−
r
0
3
R
5
−
r
0
5
.
(1)
For example
r
0
=
R
/2
gives
C
/
(
M
R
2
)
=
31/70
, not
2/3
. The thin-shell limit
r
0
→
R
restores the printed result.
A
uniform outer layer of unspecified thickness does not by itself establish it.
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