Page 287 - Schaum's Outline of Theory and Problems of Applied Physics
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272                                   ELECTRICITY                                [CHAP. 23



        SOLVED PROBLEM 23.15
              (a) What potential difference must be applied across two metal plates 15 cm apart if the electric field
              between them is 600 V/m? (b) What is the force on a charge of 10 −10  C in this field? (c) How much
              kinetic energy will the charge have when it has moved through 5 cm in the field, starting from rest?

              (a)                       V = Es = (600 V/m)(0.15 m) = 90 V
              (b)                    F = qE = (10 −10  C)(600 V/m) = 6 × 10 −8  N
              (c)  Since the KE of the charge is equal to the work done on it by the electric field when it travels 0.05 m,
                                      KE = W = Fs = (6 × 10 −8  N)(0.05 m) = 3 × 10 −9  J

        SOLVED PROBLEM 23.16

              What potential difference must be applied to produce an electric field that can accelerate an electron to a
                         7
              velocity of 10 m/s?
                  The kinetic energy of such an electron is
                                         2
                                                             7
                                                                 2
                                             1
                                      1
                                 KE = mv = ( )(9.1 × 10 −31  kg)(10 m/s) = 4.6 × 10 −17  J
                                      2      2
              This KE is equal to the work W that must be done on the electron by the electric field, and so, since W = qV in
              general, we have here
                                      W    KE    4.6 × 10 −17  J   2
                                  V =    =    =            = 2.9 × 10 V = 290 V
                                       q    e   1.6 × 10 −19  C
        SOLVED PROBLEM 23.17
              A 12-V storage battery is being charged at the rate of 15 C/s. (a) How much power is being used to charge
              the battery? (b) How much energy is stored in the battery if it is charged at this rate for 1 h?

              (a) The work done to transfer the charge q from one set of the battery’s electrodes to the other set against the
                  potential difference V is W = Vq. Since power is the rate at which work is being done, here
                                              W   Vq
                                          P =   =    = (12 V)(15 C/s) = 180 W
                                              t    t
              (b) The work done in t = 1h = 3600 s is
                                                                        5
                                         W = Pt = (180 W)(3600 s) = 6.48 × 10 J
                  If the charging process is perfectly efficient, this amount of energy will be stored in the battery as a result.




                                     Multiple-Choice Questions


         23.1. Which one or more of the following statements is true?
               (a) All protons have the same charge.
               (b) Electrons and protons have equal masses.
               (c) Protons and neutrons have equal masses.
               (d) Atomic nuclei contain only protons and neutrons.

         23.2. Electric charge occurs only in separate parcels
               (a)of ±1.6 × 10 −19  C
               (b)of ±1C
               (c) whose value is proportional to the mass of the particle carrying the charge
               (d) that are different for positive and negative particles
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