76
2 – Methods and techniques
d. The capacitance C of the sample is
C
R
1
(6.19 10 )(5.69 10 )
1
0
4
5
ω
=
=
#
#
.
C
F
2 84 10
11
=
−
#
e. The decentering angle β is
(1 p) 2
(1 0.879) 2
β
π
π
= −
= −
.
10 9 °
β =
f. The capacitance C of the sample depends on the dielectric constant ε r of
the sample and its geometric factor k:
C
S
k
0 r
0 r
,
ε ε
ε ε
=
=
#
where ε 0 is the vacuum permittivity with
10
4 c
0
7
2
ε
π
=
^
h and c is the
speed of light.
4 (3 10 )
10
0
8 2
7
ε
π
=
#
#
8.85 10 F m
0
12
1
ε =
−
−
#
From this we deduce
C k
8.85 10
2.84 10
14
r
0
12
11
ε
ε
=
=
−
−
#
#
#
44.9
r
ε =
g. The permittivity is thus
ε = 44.9 # 8.85 # 10
−12
3.97 10 F m
10
1
#
ε =
−
−
h. The conductivity of the sample is given by
R
1 S
R
k
,
σ =
=
#
or
6.19 10
0.14
4
σ =
#
2.26 10 S cm
6
1
σ =
−
−
#
i. The expression relating resistivity ρ to the conductivity σ is
1
ρ
σ
=
4.42 10
cm
5
ρ
Ω
=
#