Page 119 - Engineering Electromagnetics, 8th Edition
P. 119

CHAPTER 4   Energy and Potential          101

                         Wemaystartbyvisualizinganemptyuniverse.Bringingacharge Q 1 frominfinity
                                                                        2
                     to any position requires no work, for there is no field present. The positioning of
                     Q 2 at a point in the field of Q 1 requires an amount of work given by the product of
                     the charge Q 2 and the potential at that point due to Q 1 .We represent this potential
                     as V 2,1 , where the first subscript indicates the location and the second subscript the
                     source. That is, V 2,1 is the potential at the location of Q 2 due to Q 1 . Then

                                          Work to position Q 2 = Q 2 V 2,1
                         Similarly, we may express the work required to position each additional charge
                     in the field of all those already present:

                                  Work to position Q 3 = Q 3 V 3,1 + Q 3 V 3,2
                                  Work to position Q 4 = Q 4 V 4,1 + Q 4 V 4,2 + Q 4 V 4,3
                     and so forth. The total work is obtained by adding each contribution:
                                   Total positioning work = potential energy of field

                                     = W E = Q 2 V 2,1 + Q 3 V 3,1 + Q 3 V 3,2 + Q 4 V 4,1
                                                                                     (39)
                                       +Q 4 V 4,2 + Q 4 V 4,3 +· · ·
                         Noting the form of a representative term in the preceding equation,

                                                     Q 1         Q 3
                                       Q 3 V 3,1 = Q 3    = Q 1
                                                  4π  0 R 13   4π  0 R 31
                     where R 13 and R 31 each represent the scalar distance between Q 1 and Q 3 ,we see that
                     it might equally well have been written as Q 1 V 1,3 .If each term of the total energy
                     expression is replaced by its equal, we have
                                                                                     (40)
                         W E = Q 1 V 1,2 + Q 1 V 1,3 + Q 2 V 2,3 + Q 1 V 1,4 + Q 2 V 2,4 + Q 3 V 3,4 + ···
                     Adding the two energy expressions (39) and (40) gives us a chance to simplify the
                     result a little:
                                       2W E = Q 1 (V 1,2 + V 1,3 + V 1,4 + ···)
                                             + Q 2 (V 2,1 + V 2,3 + V 2,4 +· · ·)
                                             + Q 3 (V 3,1 + V 3,2 + V 3,4 +· · ·)
                                             +· · ·
                     Each sum of potentials in parentheses is the combined potential due to all the charges
                     except for the charge at the point where this combined potential is being found. In
                     other words,

                                           V 1,2 + V 1,3 + V 1,4 +· · · = V 1


                     2  However, somebody in the workshop at infinity had to do an infinite amount of work to create the
                     point charge in the first place! How much energy is required to bring two half-charges into coincidence
                     to make a unit charge?
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