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.Download Link: