Modern Physics for Scientists and Engineers Solutions

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Modern Physics for Scientists and Engineers
                Solutions

Joseph N. Burchett after consultations with John C. Morrison

                         August 4, 2010




                  Copyright  2010 by Elsevier, Inc. All rights reserved.
Introduction - Solutions

1. We are given a spherical distribution of charge and asked to calculate the
potential energy of a point charge due to this distribution. There are two
situations to consider which deliver quite different results. The problem does
not specify whether the point charge of interest is inside or outside the charge
distribution so we must take both possibilities into consideration. Before we
get started, let's set up an appropriate description of our charge distribution
so that we may proceed with each of the above situations. It is most useful
to express the charge distribution in terms of a charge density :
                                                 Q
                                      =      4
                                             3
                                               R3

This is the charge per unit volume as found by merely dividing the total
charge Q by the total volume of the sphere. We now handle the "outside the
sphere" configuration. All of the charge inside the sphere may be considered
as though it is concentrated at a single point. We may then use the Coulomb's
Law result for the force acting on the point charge q :
                                            1 Qq
                                    F =
                                           4 0 r2
We integrate to find the potential energy:
                                            r
                                  Qq       1        Qq 1
                       V =-                 2
                                              dx =
                                 4 0      r        4 0 r
This result is also identical to the one obtained in the text in the case of the
two-point charge configuration. We see that this should in fact be the case
since we started with the same quantity of force.
    Now, to handle the "inside the sphere" situation, we shall use the above
stated charge density. We easily see that the distance r from the center of the

                                             3




                         Copyright  2010 by Elsevier, Inc. All rights reserved.
4

distribution of charge to the point charge is less than the radius R. Imagine
now a sphere of radius r that is inside the larger spherical charge distribution
but centered at the same point. Only the charge within this inner sphere will
act on the point charge. We may now use our charge density to find the
charge contained in the smaller sphere:
                                             4
                                        Qin = r 3 
                                             3
The force on a charge q due to the inner sphere is then given by Coulomb's
Law:
                                    q 4r 3       qr
                            F =                =
                                  4 0 3r 2        30
Before we integrate to find the potential energy, we must make an important
distinction from the "outside the sphere" procedure above. We were able to
integrate directly from infinity to the location of our point charge because the
amount of source charge taken into account did not change (Q ). However, as
we enter the sphere, the amount of source charge begins to decrease until we
reach our point at distance r from the center. We split the integration into
two parts corresponding to both regions to find the potential energy:
                                    R                         r
                                         Qq 1                     qr
                      V = -[                    dr +                 dr ]
                                        4 0 r 2              R    30
                              Qq 1    q 2
                              =     -    (r - R2 )
                             4 0 R 6 0
Note that the first integral above was identical to the one for the outside
configuration, just evaluated at r = R. Now, we just substitute our charge
density  and simplify:
                                   Qq 1   Qqr2      Qq
                         V =            -      3
                                                 +
                                  4 0 R 8 0 R      8 0 R
                                       3Qq   Qqr2
                                  =        -
                                      8 0 R 8 0 R3


3. Looking at the definition of kinetic energy:
                                            1
                                        KE = mv 2
                                            2




               Copyright  2010 by Elsevier, Inc. All rights reserved.
                                                                                      5

Let's solve for velocity:
                                                2(KE )
                                      v=
                                                  m
Let KE = 4KE .
                            2(KE )            2(4KE )                2(KE )
                  v=               =                  =2
                              m                  m                     m
                                             = 2v
So, the speed will increase by a factor of 2 if the kinetic energy is increased
by a factor of 4. By the definition of momentum:
                                p = mv  m(2v ) = 2p
We should also expect the momentum of the particle above to increase by
a factor of 2. In fact, as long as the mass does not change, the momentum
should increase or decrease by the same factor as the changing speed.

5. The frequency and wavelength of light are related by equation I.23:
                                      c
                                f=
                                      
The constant c refers to the speed of light which we know to be 3.00108 m/s.
Taking care to convert our wavelength, which is given in nanometers, to
meters:
                            3.00  108 m
                       f=           -  9
                                         s
                                           = 6  1014 Hz
                           500  10 m


7. Equation I.25 relates the photon energy to the wavelength:
                                      hc
                                           E=
                                       
Substituting the value hc = 1240 eV nm:
                                   1240 eV nm
                            E=                 = 2.48 eV
                                     500nm
9. The energy of a quantum of light (the photon) must be equal to the
difference in energies of the two states:
                                   Ephoton = E2 - E1




                            Copyright  2010 by Elsevier, Inc. All rights reserved.
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