Page 194 - Modern physical chemistry

186 Equilibria in Condensed Phases

In principle, one could map out an electric field using a small test charge. The force
per unit charge acting on the test charge, at rest at a certain point, is the electric inten-
sity E at that point.
Coulomb's law, in its simplest form, states that the electric intensity produced by a
point charge of magnitude q is

E=-q-rl> [8.51 ]
4ner 2
where vector r is drawn from the source charge to the point at which the test charge is
placed, r l is the unit vector giving the direction of r (r = rr l ), and e is the dielectric con-
stant of the inteIVening medium.
One may surround the source with a surface S, as figure 8.1 shows. A differential
element of the surface is represented by vector dS whose magnitude equals the size of
the element and whose direction is perpendicular to the element in the outward direc-
tion. The solid angle dO. subtended by the element equals the projection of dS on vector
r divided by r. So we have
E·dS = Ecos(}dS = Er2 dO., [8.52]
where () is the angle between E and dS.
Combining equations (8.51) and (8.52), and integrating yields

fE.dS= f Er2 <in = f~r2 dn=~f <in=.[. [8.53]
s s s 4ner 4ne s e
The integral of the solid angle over the closed surface equals 4Jr; since the area of a sphere
of unit radius is 4Jr in magnitude.

s

FIGURE 8.1 Surface element dS subtending the solid angle d.{l at point charge q.