ANSWERS

Quiz 4 part 1  Chapter  23 problems: 1, 2, 4, 5, 11, 12, 14, 9, 19,  22, 26, 28, 34, 35, 36, 37
Quiz 4 part 2  Chapter  23 problems: 38*, 39*, 43, 44, 45*, 47*, 48*, 49*, 50*, 54*, 55*, 57*, 60*, 61*, 75*, 76*

Simulation: Click here for the motion of an electron in a uniform electric field.

DISCUSSIONS TO PROBLEMS  WILL BE POSTED OR COVERED IN CLASS
38. THIS IS A NICE EXAMPLE OF HOW A SCALAR CAN LEAD TO EASIER INTEGRALS FOR DISTRIBUTED CHARGED SYSTEMS.

dV = k*(lambda)*dx/r, where r2 = y2 + x2  .
Here is a guide to the trig substitutions you need to solve this problem:

First off, let the angle theta  be between the y axis and the line segment of length r between the point of evaluation on the y-axis and the charge element dq on the x axis .
x = y*tan theta.

Thus ,

dx = y*sec2theta*d(theta)   .
Note also that happily r =  y/cos theta = y*sec theta. Thus the good news is,
 k*(lambda)*dx/r =  k*(lambda)*y*sec2theta*d(theta)/r and
k*(lambda)*y*sec2theta*d(theta)/ (y*sec theta) = k*(lambda*sec theta*d(theta).
Integrate     k*(lambda*sec theta*d(theta) between theta = - tan-1 (L/y)
and theta = + tan-1 (L/y)   using an integral table or you can revert back to  rectangular coordinates x and y and use Appendix B4 . The choice is yours.  

39. This can be done without  consulting an integral table and I would expect you to know how to perform the calculation from memory with no notes whatsoever.  Place your potential evaluation point   on the x axis say a distance D to the right of the right end of the rod. Thus the distance between that location and an element of charge on the rod is D - x , where x is between - L and  + L. Now, dV =   k*(lambda)*dx/(D + L  - x). This logarithmic integral can easily be done  using a substitution like  u =  D + L - x , so that dx = -du, and the integrand becomes  -k*(lambda)*du/u with new limits of integration.
43. delta V = E*L = 100 (V), where L is the distance between equi-potential surfaces.  Find  L using the definition of  field magnitude  E in terms of  surface charge density.
44. There are only a finite number of equi-potential surfaces separated by 100 (V).  If  n is the number of surfaces from the first  one at ro   , then  n*100 < kQ/ro . 

Now, n*100 (V)  = kQ*(1/ro - 1/rn), where n = 0, 1, 2, 3, 4 ....<  kQ/(100*ro).
Note: rn  is the  nth radius from ro.   Set n = 1, 10 and 100 and solve for r1, r10 and  r100.  Verify  the maximum
n = 100 < kQ/(100*ro).
45. See page 617.
47. See equation 23-8.
48.  Find the electric field by differentiating as you did in #47.
49.  See section 23-7.
50. See section 23-7.
54.  See section 23-8. The problem of three charges is illustrated.
Note: Even though the electrons are negative, the potential energy is positive. The interactions are repulsive. The three protons would produce the same qualitative result.  
55. See class notes. Net work = change in kinetic energy.
Net work = field work = - change in potential energy. Change in potential  energy = Q*change in voltage. 
Thus:

 - Q*change in voltage = change in kinetic energy.  Solve for change in voltage; your answer will be negative because the positive particle moves from higher to lower potential. 

Note that 1 KeV = 1000 eV and 1 eV = 1.6x10 -19 J. 
57. The magnitude of the change in potential energy = the magnitude of the chemical energy released. The potential energy for this problem has the form U =  k*Q1*Q2/r, where r is the distance between the proton and electron. Now Q1 = e and Q2 = -e if  1 and 2 represent the proton and electron respectively. Now initially 
U =  Ui = k*Q1*Q2/ri, and finally U = Ui = k*Q1*Q2/ri  . Evaluate the difference in these potential energies to get the potential energy lost in this reaction. The two , initial and final , distances are given.
60. 
(a) There are 6 interactions, the 4 sides and the 2 diagonals.
(b) Here we are talking about the potential energy of the added charge alone. Assume we place it dead center.  In this case, the answer is Q*Vp. , where P is a point  at center. Vp = sum of potentials due to the 4  corner charges. Each charge produces a potential  kQ/r, where r is the length of *half* the square's  diagonal .  See for example  Test 1's problem #6 . 
(c) This is sort of obvious, but I'll let you make that determination. There are two types of motions possible, perpendicular or within the square's plane:

(1) Clearly all the charges are positive, so if you raised or lowered the added center charge Q *above or below* the square's plane and released it, it would keep on going higher or lower; it would not return . It is stable or unstable? 
(2) If you moved the center charge through a displacement *along the square's plane*, the repulsion would tend to push it back when you released it. Is that stable or unstable? 
(d) Clearly changing center charge's sign would "reverse" the previous  two results.
61. See class notes; see #55 discussion, which applies to negative or positive charges.  

Net work = field work = - change in potential energy. Change in potential  energy = Q*change in voltage. 
Thus:

 - Q*change in voltage = change in kinetic energy.  

(a) The electron moves from A to B. For the electron, Q = -e. Solve for change in voltage, given the positive 1.33 KeV change in  energy. For energy unit conversions, see #55. You will find the change in voltage is positive, since negative particles move from low to high potential. 

Now use - Q*change in voltage = change in kinetic energy. But now Q = +e.   Since positive particles move from high to low potential, you must use -(change in voltage), where (change in voltage) was found above.  This is the same as saying the proton moves "from B to A"  instead  of "A to B" above. Solve for the the proton's change in kinetic energy. Are you surprised the result is the same as above's 1.33 KeV? I hope not. 
(b) Now (1/2)*mp*Vp2 = (1/2)*me*Ve2,   where indexes refer to proton and electron. Find the ratio of  speeds.
75. Class notes. This a "green" problem, illustrating alternative energy sources. My PhD thesis dealt with thin films related to this  technology; go here:   http://adsabs.harvard.edu/abs/1986PhDT........34A

Obviously I will have to break my research down to you some day. But back the  problem. This is nothing more than the photoelectric effect. First let us talk about the electron escaping from the Barium:
Think of a baseball at the bottom of a well of height h. The ball is rolling back and forth. It can  escape  if  I do work on it to lift it to the top. KEi + Ui + (my work) = KEf + Uf . Now Ui = 0 and  Uf = mgh.
Thus at the top, 
KEf = KEi + Ui - Uf   +   (my work) 
KEf = KEi   -  mgh  +   (my work).  
You might be able to approximate  KEi = 0.
Thus, KEf = (my work) -  mgh .
Thus (my work) must  be large enough to overcome the -mgh  term so that KEf >0. Under this condition, the electron has "escaped" the Barium. Here,  mgh  represents the chemical binding potential  energy needed for the electron to escape the barium's "well." 

 Once the electron has escaped barium,  we will slow it down with the negative potential difference of -3.02 (V); after the electron escapes, we write:
 - Q*change in voltage = change in kinetic energy. The change in voltage = -3.02 (V) and Q = -e. 
The change in kinetic energy = 0 - KEi.  = - KEi.  Solve for KEi and solve for the initial speed. 
76. See class notes and the e/m discussion here for the motion of an electron in a uniform electric field.