Page 164 - Engineering Electromagnetics, 8th Edition

146 ENGINEERING ELECTROMAGNETICS

distribution are then almost uniform at all points not adjacent to the edges, and this
latter region contributes only a small percentage of the total capacitance, allowing us
to write the familiar result
Q = ρ S S
ρ S
V 0 = d

Q S
C = = (3)
V 0 d

More rigorously, we might consider Eq. (3) as the capacitance of a portion of the
infinite-plane arrangement having a surface area S. Methods of calculating the effect
of the unknown and nonuniform distribution near the edges must wait until we are
able to solve more complicated potential problems.

EXAMPLE 6.1
Calculate the capacitance of a parallel-plate capacitor having a mica dielectric, r = 6,
2
a plate area of 10 in. , and a separation of 0.01 in.
Solution. We may find that
2
S = 10 × 0.0254 = 6.45 × 10 −3 m 2
d = 0.01 × 0.0254 = 2.54 × 10 −4 m

and therefore
6 × 8.854 × 10 −12 × 6.45 × 10 −3
C = = 1.349 nF
2.54 × 10 −4
A large plate area is obtained in capacitors of small physical dimensions by
stacking smaller plates in 50- or 100-decker sandwiches, or by rolling up foil plates
separated by a flexible dielectric.
Table C.1 in Appendix C also indicates that materials are available having di-
electric constants greater than 1000.
Finally, the total energy stored in the capacitor is
2 2
S d 2 2 S ρ d
1 2 1 ρ S 1 ρ S 1 S
W E = 2 E dv = 2 2 dz dS = 2 Sd = 2 2
vol 0 0 d
or
Q 2
1 2 1 1 (4)
W E = CV = QV 0 =
2 0 2 2 C
which are all familiar expressions. Equation (4) also indicates that the energy stored
in a capacitor with a fixed potential difference across it increases as the dielectric
constant of the medium increases.