Modern Engineering Thermodynamics

4.7 Nonmechanical Work Modes of Energy Transport 119
Table 4.3 The Electric Susceptibility of Various Materials
Material Temperature (°C/°F) χ e (dimensionless)
Air (14.7 psia) 20/68 5.36 × 10 −4
Plexiglass 27/81 2.40
Neoprene rubber 24/75 5.7
Glycerine 25/77 41.5
Water 25/77 77.5
Source: Reprinted by permission of the publisher from Zemansky, M. W., Abbott, M. M., Van Ness, H. C., 1975. Basic Engineering
Thermodynamics, second ed. McGraw-Hill, New York.
Since the effect of the electric field is to orient the dipoles coincident with the field, then E ! and P ! are always
parallel and point in the same direction. Therefore, if we let the magnitude of E ! be E and the magnitude of P !
be P, then Eqs. (4.51) and (4.52) reduce to
ðdWÞ electrical
= − EdP (4.53)
polarization
and
ð 1 W 2 Þ electrical
= −
polarization
Z 2
1
EdP (4.54)
Many substances (particularly gases) correlate well with the following dielectric equation of state:
P = ε 0 χ e VE (4.55)
where V is the volume of the dielectric substance, ε 0 is the electric permittivity of vacuum (8.85419 × 10 −12 N/V 2 ),
and χ e is the electric susceptibility (a dimensionless number) of the material. Table 4.3 gives values of χ e for various
materials.
EXAMPLE 4.9
The parallel plate capacitor shown in Figure 4.16 is charged to a potential
difference of 120. V at 25.0°C. The plates are square with a side length of
0.100 m and are separated by 0.0100 m. If the gap between the plates is filled
with water, determine the polarization work required in the charging of the
capacitor.
Solution
Here, we can use the dielectric equation of state, Eq. (4.55). Then, Eq. (4.54)
becomes
Water at
25.0°C
120. volts
0.100 m
square
ð1W 2 Þ electric
= −
po1arization
Z 2
1
EdP = −
From the problem statement, we have
Z 2
1
ðε 0 χ e VEÞ dE = − ε 0 χ e V E 2 2 − E2 1
/2
V = AL = ð0:100 mÞ 2 ð0:0100 mÞ = 1:00 × 10 − 4 m 3
FIGURE 4.16
Example 4.9.
0.0100 m
If we assume that the electrical potential ϕ varies linearly between the plates, then
we can write
and
E = j−∇ðϕÞj = ðvoltage differenceÞ/ðplate gapÞ with E 1 = 0 ðuncharged platesÞ
E 2 =
120: V
0:0100 m = 1:20 × 104 V/m ðcharged platesÞ
(Continued )