Page 165 - Engineering Electromagnetics, 8th Edition

CHAPTER 6 Capacitance 147

D6.1. Find the relative permittivity of the dielectric material present in a
2
parallel-plate capacitor if: (a) S = 0.12 m , d = 80 µm, V 0 = 12 V, and the
3
capacitor contains 1 µJof energy; (b) the stored energy density is 100 J/m ,
V 0 = 200 V, and d = 45 µm; (c) E = 200 kV/m and ρ S = 20 µC/m .
2
Ans. 1.05; 1.14; 11.3

6.3 SEVERAL CAPACITANCE EXAMPLES
As a first brief example, we choose a coaxial cable or coaxial capacitor of inner
radius a, outer radius b, and length L.No great derivational struggle is required,
because the potential difference is given as Eq. (11) in Section 4.3, and we find the
capacitance very simply by dividing this by the total charge ρ L L in the length L.
Thus,
2π L
C = (5)
ln(b/a)
Next we consider a spherical capacitor formed of two concentric spherical con-
ducting shells of radius a and b, b > a. The expression for the electric field was
obtained previously by Gauss’s law,
Q
E r =
4π r 2
where the region between the spheres is a dielectric with permittivity . The expression
for potential difference was found from this by the line integral [Section 4.3, Eq. (12)].
Thus,
Q 1 1
V ab = −
4π a b
Here Q represents the total charge on the inner sphere, and the capacitance becomes

Q 4π
C = = 1 1 (6)
V ab
−
a b
If we allow the outer sphere to become infinitely large, we obtain the capacitance
of an isolated spherical conductor,

C = 4π a (7)
Fora diameter of 1 cm, or a sphere about the size of a marble,
C = 0.556 pF

in free space.