Page 107 - Engineering Electromagnetics, 8th Edition

CHAPTER 4 Energy and Potential 89

Figure 4.4 A simple dc-circuit problem that must be
solved by applying E · dL = 0in the form of Kirchhoff’s
voltage law.

(20) states that no work is involved in carrying a unit charge from A through R 2 and
R 3 to B and back to A through R 1 ,or that the sum of the potential differences around
any closed path is zero.
Equation (20) is therefore just a more general form of Kirchhoff’s circuital law
for voltages, more general in that we can apply it to any region where an electric
field exists and we are not restricted to a conventional circuit composed of wires,
resistances, and batteries. Equation (20) must be amended before we can apply it to
time-varying fields.
Any field that satisfies an equation of the form of Eq. (20), (i.e., where the closed
line integral of the field is zero) is said to be a conservative field. The name arises from
the fact that no work is done (or that energy is conserved) around a closed path. The
gravitational field is also conservative, for any energy expended in moving (raising)
an object against the field is recovered exactly when the object is returned (lowered)
to its original position. A nonconservative gravitational field could solve our energy
problems forever.
Given a nonconservative field, it is of course possible that the line integral may
be zero for certain closed paths. For example, consider the force field, F = sin πρ a φ .
Around a circular path of radius ρ = ρ 1 ,wehave dL = ρ dφ a φ , and

2π 2π
F · dL = sin πρ 1 a φ · ρ 1 dφ a φ = ρ 1 sin πρ 1 dφ
0 0
= 2πρ 1 sin πρ 1

The integral is zero if ρ 1 = 1, 2, 3,... , etc., but it is not zero for other values of ρ 1 ,
or for most other closed paths, and the given field is not conservative. A conservative
field must yield a zero value for the line integral around every possible closed path.

D4.6. If we take the zero reference for potential at infinity, find the potential
at (0, 0, 2) caused by this charge configuration in free space (a)12 nC/m on the
line ρ = 2.5m, z = 0; (b) point charge of 18 nC at (1, 2, −1); (c)12 nC/m on
the line y = 2.5, z = 0, −1.0 < x < 1.0.
Ans. 529 V; 43.2 V; 66.3 V