Page 100 - Engineering Electromagnetics, 8th Edition
P. 100

82                 ENGINEERING ELECTROMAGNETICS

                                        One of the pitfalls in evaluating line integrals is a tendency to use too many minus
                                     signs when a charge is moved in the direction of a decreasing coordinate value. This is
                                     taken care of completely by the limits on the integral, and no misguided attempt should
                                     be made to change the sign of dL. Suppose we carry Q from b to a (Figure 4.2b).
                                     We still have dL = dρ a ρ and show the different direction by recognizing ρ = b as
                                     the initial point and ρ = a as the final point,

                                                                  a  ρ L  d ρ  Qρ L  b
                                                       W =−Q               =      ln
                                                                b 2π  0 ρ    2π  0  a
                                     This is the negative of the previous answer and is obviously correct.

                                        D4.2. Calculate the work done in moving a 4-C charge from B(1, 0, 0) to
                                        A(0, 2, 0) along the path y = 2 − 2x, z = 0in the field E = (a)5a x V/m;
                                        (b)5xa x V/m; (c)5xa x + 5ya y V/m.

                                        Ans. 20 J; 10 J; −30 J


                                        D4.3. We will see later that a time-varying E field need not be conservative.
                                        (If it is not conservative, the work expressed by Eq. (3) may be a function of the
                                        path used.) Let E = ya x V/m at a certain instant of time, and calculate the work
                                        required to move a 3-C charge from (1, 3, 5) to (2, 0, 3) along the straight-line
                                        segments joining: (a)(1, 3, 5) to (2, 3, 5) to (2, 0, 5) to (2, 0, 3); (b)(1, 3, 5) to
                                        (1, 3, 3) to (1, 0, 3) to (2, 0, 3).

                                        Ans. −9J;0


                                     4.3 DEFINITION OF POTENTIAL
                                            DIFFERENCE AND POTENTIAL
                                     We are now ready to define a new concept from the expression for the work done
                                     by an external source in moving a charge Q from one point to another in an electric
                                     field E, “Potential difference and work.”

                                                                        final
                                                             W =−Q        E · dL
                                                                      init
                                        In much the same way as we defined the electric field intensity as the force on a
                                     unit test charge, we now define potential difference V as the work done (by an external
                                     source) in moving a unit positive charge from one point to another in an electric field,


                                                                                 final
                                                                                   E · dL             (9)
                                                     Potential difference = V =−
                                                                               init
   95   96   97   98   99   100   101   102   103   104   105