0195136047.pdf

1.1 ELECTRICAL QUANTITIES 5

R

Q 1

Q 2

a 21

a 12

F 21

F 12

Figure 1.1.1Illustration of Coulomb’s law.

Equation (1.1.2) is the defining equation for the electric field intensity (with units of N/C or V/m),
irrespective of the source of the electric field. One may then conclude:

F ̄ 21 =Q 1 E ̄ 2 (1.1.3a)

F ̄ 12 =Q 2 E ̄ 1 (1.1.3b)

whereE ̄ 2 is the electric field due toQ 2 at the location ofQ 1 , andE ̄ 1 is the electric field due to
Q 1 at the location ofQ 2 , given by

E ̄ 2 = Q^2

4 πε 0 R^2

a ̄ 21 (1.1.4a)

E ̄ 1 = Q^1

4 πε 0 R^2

a ̄ 12 (1.1.4b)

Note that the electric field intensity due to a positive point charge is directed everywhere
radially away from the point charge, and its constant-magnitude surfaces are spherical surfaces
centered at the point charge.

EXAMPLE 1.1.1

(a) A small region of an impure silicon crystal with dimensions 1. 25 × 10 −^6 m× 10 −^3 m

× 10 −^3 m has only the ions (with charge+1.6 10−^19 C) present with a volume density of

1025 /m^3. The rest of the crystal volume contains equal densities of electrons (with charge

− 1. 6 × 10 −^19 C) and positive ions. Find the net total charge of the crystal.

(b) Consider the charge of part (a) as a point chargeQ 1. Determine the force exerted by this

on a chargeQ 2 = 3 μC when the charges are separated by a distance of2minfree space,

as shown in Figure E1.1.1.

Q 3 = −2 × 10 −6 C

F 12

F 32 F^2

2 m Q^2 = 3 ×^10 −6 C

1 m

76 °

y

Q 1 = 2 × 10 −6 C x

−

+ +

Figure E1.1.1