Page 105 - Engineering Electromagnetics, 8th Edition
P. 105

CHAPTER 4   Energy and Potential           87

                     infinity to the field point at r where we are finding the potential. The volume charge
                     density ρ v (r ) and differential volume element dv combine to represent a differential


                     amount of charge ρ ν (r ) dv located at r . The distance |r − r | is that distance from




                     the source point to the field point. The integral is a multiple (volume) integral.
                         If the charge distribution takes the form of a line charge or a surface charge, the
                     integration is along the line or over the surface:
                                                      ρ L (r ) dL


                                            V (r) =                                  (18)

                                                     4π  0 |r − r |
                                                      ρ S (r ) dS


                                            V (r) =                                  (19)

                                                    S 4π  0 |r − r |
                         ThemostgeneralexpressionforpotentialisobtainedbycombiningEqs.(16)–(19).
                         These integral expressions for potential in terms of the charge distribution should
                     be compared with similar expressions for the electric field intensity, such as Eq. (15)
                     in Section 2.3:
                                                    ρ ν (r ) dv     r − r


                                         E(r) =              2

                                                vol 4π  0 |r − r | |r − r |
                         The potential again is inverse distance, and the electric field intensity, inverse-
                     square law. The latter, of course, is also a vector field.
                                                                                            EXAMPLE 4.3
                     To illustrate the use of one of these potential integrals, we will find V on the z axis for
                     a uniform line charge ρ L in the form of a ring, ρ = a,inthe z = 0 plane, as shown
                     in Figure 4.3.

                     Solution. Working with Eq. (18), we have dL = adφ , r = za z , r = aa ρ , |r−r |=
                     √
                       a + z , and
                        2
                            2
                                             2π                  ρ L a
                                                 ρ L adφ
                                      V =         √        =    √
                                                                  2
                                                     2
                                           0  4π  0 a + z 2  2  0 a + z 2
                     Fora zero reference at infinity, then:
                     1. The potential arising from a single point charge is the work done in carrying a
                         unit positive charge from infinity to the point at which we desire the potential,
                         and the work is independent of the path chosen between those two points.
                     2. The potential field in the presence of a number of point charges is the sum of
                         the individual potential fields arising from each charge.
                     3. The potential arising from a number of point charges or any continuous charge
                         distribution may therefore be found by carrying a unit charge from infinity to
                         the point in question along any path we choose.
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