Page 47 - Engineering Electromagnetics, 8th Edition
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CHAPTER 2   Coulomb’s Law and Electric Field Intensity    29

                         The force expressed by Coulomb’s law is a mutual force, for each of the two
                     charges experiences a force of the same magnitude, although of opposite direction.
                     We might equally well have written

                                                 Q 1 Q 2       Q 1 Q 2
                                     F 1 =−F 2 =       a 21 =−        a 12            (5)
                                                      2
                                                4π  0 R 12    4π  0 R 2 12
                         Coulomb’s law is linear, for if we multiply Q 1 by a factor n, the force on Q 2 is
                     also multiplied by the same factor n.Itis also true that the force on a charge in the
                     presence of several other charges is the sum of the forces on that charge due to each
                     of the other charges acting alone.


                        D2.1. A charge Q A =−20 µCis located at A(−6, 4, 7), and a charge Q B =
                        50 µCisat B(5, 8, −2) in free space. If distances are given in meters, find:
                        (a) R AB ;(b) R AB . Determine the vector force exerted on Q A by Q B if   0 =
                             −9
                        (c)10 /(36π) F/m; (d)8.854 × 10 −12  F/m.
                        Ans. 11a x + 4a y − 9a z m; 14.76 m; 30.76a x + 11.184a y − 25.16a z mN; 30.72a x
                        + 11.169a y − 25.13a z mN


                     2.2 ELECTRIC FIELD INTENSITY
                     If we now consider one charge fixed in position, say Q 1 , and move a second charge
                     slowly around, we note that there exists everywhere a force on this second charge;
                     in other words, this second charge is displaying the existence of a force field that is
                     associated with charge, Q 1 . Call this second charge a test charge Q t . The force on it
                     is given by Coulomb’s law,

                                                     Q 1 Q t
                                                F t =      a 1t
                                                          2
                                                    4π  0 R 1t
                     Writing this force as a force per unit charge gives the electric field intensity, E 1 arising
                     from Q 1 :
                                                  F t    Q 1
                                            E 1 =    =       2  a 1t                  (6)
                                                  Q 1  4π  0 R 1t
                     E 1 is interpreted as the vector force, arising from charge Q 1 , that acts on a unit positive
                     test charge. More generally, we write the defining expression:

                                                        F t
                                                   E =                                (7)
                                                        Q t
                     in which E,avector function, is the electric field intensity evaluated at the test charge
                     location that arises from all other charges in the vicinity—meaning the electric field
                     arising from the test charge itself is not included in E.
                         The units of E would be in force per unit charge (newtons per coulomb). Again
                     anticipating a new dimensional quantity, the volt (V), having the label of joules per
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