Fundamentals of Materials Science and Engineering: An Integrated Approach, 3e

GTBL042-12 GTBL042-Callister-v2 August 13, 2007 18:22

12.19 Field Vectors and Polarization • 503

Table 12.6 Primary and Derived Units for Various Electrical Parameters and

Field Vectors

SI Units

Quantity Symbol Derived Primary

Electric potential V volt kg-m^2 /s^2 -C

Electric current I ampere C/s

Electric field strength e volt/meter kg-m/s^2 -C

Resistance R ohm kg-m^2 /s-C^2

Resistivity ρ ohm-meter kg-m^3 /s-C^2

Conductivity σ (ohm-meter)−^1 s-C^2 /kg-m^3

Electric charge Q coulomb C

Capacitance C farad s^2 -C^2 /kg-m^2

Permittivity  farad/meter s^2 -C^2 /kg-m^3

Dielectric constant r dimensionless dimensionless

Dielectric displacement D farad-volt/m^2 C/m^2

Electric polarization P farad-volt/m^2 C/m^2

EXAMPLE PROBLEM 12.5

Computations of Capacitor Properties

Consider a parallel-plate capacitor having an area of 6.45× 10 −^4 m^2 (1 in.^2 )

and a plate separation of 2× 10 −^3 m (0.08 in.) across which a potential of 10 V

is applied. If a material having a dielectric constant of 6.0 is positioned within

the region between the plates, compute

(a)The capacitance.

(b)The magnitude of the charge stored on each plate.

(c)The dielectric displacementD.

(d)The polarization.

Solution

(a)Capacitance is calculated using Equation 12.26; however, the permittivity

of the dielectric medium must first be determined from Equation 12.27

as follows:

=r 0 =(6.0)(8. 85 × 10 −^12 F/m)

= 5. 31 × 10 −^11 F/m

Thus, the capacitance is

C=

A

l

=(5. 31 × 10 −^11 F/m)

(

6. 45 × 10 −^4 m−^2

20 × 10 −^3 m

)

= 1. 71 × 10 −^11 F

(b)Since the capacitance has been determined, the charge stored may be com-

puted using Equation 12.24, according to

Q=CV=(1. 71 × 10 −^11 F)(10 V)= 1. 71 × 10 −^10 C