Mathematics of Physics and Engineering

Equations in Vacuum 167

is potential: E(P) = - W(P), where U(P) = q/(4ne 0 \OP), P ^ O. By
linearity, the electric field at the point P, produced by n point charges is

E(P) = -VU(P), U(P) = J2Aqk H^1 „, (3-3.13)

where r^ = OOk, k = l,...,n, are the position vectors of the point charges,
rp = OP is the position vector of P, and we assume P ^ Ok for all k (draw
a picture).

EXERCISE 3.3.4.c Verify (3.3.13).

Now assume that there is a continuum of charges in K^3 with density p
so that the charge of a small region GQ around a point Q € M^3 is approxi-
mately P(Q)TTI(GQ), where p is a twice continuously differentiable function
in R^3 and ^.(GQ) is the volume of GQ; the approximation becomes better as
the volume of GQ becomes smaller. We assume that p — 0 outside of some
bounded domain. This continuum of charges produces an electric field E
at every point P e K^3. Using (3.3.13) and the result of Exercise 3.1.24,

E(P) = - W(P), U(P) = JL Jff jW dV, (3.3.14)

R^3

with integration over the points Q £ M^3 where p(Q) is not zero.

EXERCISE S.S.S.*^7 Verify (3.3.14). Hint: by assumption on p, the integral in
(3.3.14) is actually over a bounded domain.
Equalities (3.3.13) and (3.3.14) show that the stationary electric field
E, produced either by finitely many point charges or by a continuum of
distributed charges, is a potential vector field: E = —'VU. The solutions
of the vector differential equation r(t) = E(r(t)) for all possible initial
conditions are called lines of force, and the surfaces {PeR^3 : U(P) =
c}, for all possible values of c, are called equipotential surfaces.

EXERCISE 3.3.6? (a) Show that lines of forces are orthogonal to the equipo-
tential surfaces. Hint: VZ7 gives the normal direction to the surface, and, be-
cause E = — Vf7, also given the tangential direction for the line of force. (b)
Verify that the lines of force for a point charge are straight lines and the
equipotential surfaces are spheres with the charge at the center. Hint: for
a suitable f = f(t) and every unit vector u, the function r(t) = uf{t) satisfies
r(t) = cr(t)/\r(t)f.