ANSWERS

Quiz 1 Chapter 21 problems: 11*, 12*, 13, 15*, 16*, 21*, 22*, 24*, 25*, 26*, 27*, 28*, 35*, 36*, 38*, 42*, 61* 
* means discussion provided. 

Big application: photocopy machines-http://en.wikipedia.org/wiki/Photocopier

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

Masteringphysics.com:  COURSE ID = PHYSICSCALI4BFALL10

11. q1 + q2 = QT   and F = k*q1*q2 / d2.   (a) F = (QT - q2)*q2 / d2    is a minimum. Differentiate with respect to q2 ,  set the expression equal to zero and solve for  q2   and  q1   . There will be two solutions for q2 corresponding to the minimum and the maximum.  To tell the difference, evaluate the  second derivative for each solution.  Review Math 1 and 2 !
12. This should be an easy problem. Please email  with comments on this exercise. What gave you trouble? 

Here are some suggestions: Label the charges, starting from the left, as 1, 2 and 3. 
On charge 1: Charge 2 repels it and charge 3 attracts it, so there two forces on opposite directions. 
On charge 2: Charge 1 repels it and charge 3 attracts it, so again there two forces on opposite directions. 
On charge 3: Charge 2 attracts it and charge 1 attracts it, so there are two forces in SAME direction, pointing left. 
Here is an example of how to do charge 1's net force computation.
Net force has magnitude : F1NET   =  k|q1q3|/(0.70)2   -   k|q1q2|/(0.35)2  .  If this difference is negative,  the force points left. Otherwise, it points  right. 
Try the same subtraction method with charge 2. Be careful;  the distances used in the formulae are equal. 
For charge 3, the computation is similar to charge 1  but the force points  left; get the magnitude by adding  force magnitudes due to  charge 1 and 2.
13. Hints later. Use the component method shown in 15 and 16 below but applied to an equilateral  triangle in which the internal angles are 60 degrees.  
15.  Please email  with comments on this exercise. What gave you trouble?
I can say this: Each of the 4 forces points away from the center along a diagonal. 

Let us label the charges counterclockwise starting  from the charge in the upper right corner, which we will call  charge q1.  Without loss of generality, let us find the net force on that charge using the component method .  
F1NETx  =  F12  + F13cos45.   
F1NETy  =  F13sin45 +  F14. 

 
F12 =   k|q1q2|/a2   .
F13 =   k|q1q3|/(2a2).
F14 =   k|q1q4|/a2 .

Note: a = 0.100 m and 4.15x10-3 C =  q1   = q2  =  q3 = q4  .

For charge 2:

F2NETx  =  - F21  -  F24cos45.  
F2NETy =  F24sin45 +  F23.

F21 =   k|q2q1|/a2   .
F24 =   k|q2q4|/(2a2).
F23 =   k|q2q3|/a2 .

Same procedure for charge 3 and charge 4.

16. Please email  with comments on this exercise. What gave you trouble?
I can say this: Each of the 4 forces points along a diagonal. 

Let us label the charges counterclockwise starting  from the charge in the upper right corner, which we will call  charge q1.  Without loss of generality, let us find the net force on that charge using the component method .  
F1NETx  =  -F12  + F13cos45.   
F1NETy  =  F13sin45 -  F14. 

 
F12 =   k|q1q2|/a2   .
F13 =   k|q1q3|/(2a2).
F14 =   k|q1q4|/a2 .

Note: a = 0.100 m and 4.15x10-3 C =  | q1  | = | q2 | =  | q3 |= | q4 |
For charge 2:

F2NETx  =   F21  -  F24cos45.  
F2NETy =  F24sin45  -  F23.

F21 =   k|q2q1|/a2   .
F24 =   k|q2q4|/(2a2).
F23 =   k|q2q3|/a2 .

Same procedure for charge 3 and charge 4.

 
21. Hint:  The force is opposite  the electric field . See Tipers Excecise ET5-CRT1.
22. Hint:  The force is in the same direction as  the electric field . See Tipers Excecise ET5-CRT1 as a contrast. 
24. See the previous two hints. Which one of these hints applies here? 
25. Use unit vector notation; see previous three hints. 
26. Use unit vector notation; see previous four hints. 
27. Hint:  The force is opposite  the electric field . See Tipers Excecise ET5-CRT1. Find the magnitude of the force and use Newton's Second  Law to find the magnitude of the acceleration. How do the directions of the acceleration  and the field compare?  Are they in the same or opposite directions? 
28. Net electric field has magnitude : F1NET   =  k|q1|/(0.040)2   +   k|q2|/(0.040)2 , where -8x10-6 C =  q1  and  7x10-6 C =  q2  . We assume that q2 is on the right end and q1 is on the left end of the horizontal segment between the two charges. 
Note the sum above is positive. What is the direction of the net force,  left or right?
35. For starters, read Section 21-11 carefully and see the derivations leading to equations 21-11 and 21-12. However, Problem 35 evaluates the field on the central axis between the charges, which requires a different solution. More on this later.  The distance between P and + Q is x + a. The distance between P  and -Q is x - a. The net electric field clearly points left since the negative charge is closer. The magnitude of the net electric field is  ENET = E- - E+ , where E- is the magnitude of the electric field due to the negative charge -Q and E+ is the magnitude of the electric field due to the positive charge +Q.    Each one of these magnitudes have the form of equation 21-4b. The denominators are x + a and x - a.  Insert into the formulae  for the magnitude of the electric field and then and subtract. 
36. Find x by setting E- - E+  = 0, where E- is the magnitude of the electric field due to the negative charge Q1 and E+ is the magnitude of the electric field due to the positive charge Q2.  The distance between P and  Q1 is x. The distance between P  and Q2 is x + 0.12 cm. Each one of these magnitudes have the form of equation 21-4b. The denominators contain x   and x + 0.12 cm, each raised to the second power.   Insert into the formulae  for the magnitude of the electric field and then subtract, set expression to zero and solve for x.  After simplifying you may have a quadratic equation  in x in standard form. 
38. At origin O, the direction of the field due to the charge at A is down. The direction of the field due to the charge at B is  thirty degrees with the negative x-axis:

ENETx  =  - EBcos30. 
ENETy  =  -EA -  EB.sin30.  

Note that the magnitudes are equal: EA  =  EB  .
42. See the book's example. Look closely at Figure 21-34 b. The field at point A will point vertically upward along the positive y-axis. The field at point B will be in the first quadrant of a right-handed x-y coordinate system. 
Here is a hint for point  A:  
ENETx  = 0. 
ENETy  = 2*Ecos (theta)
Theta =  tan-1(5/10) . 
Note that E = kQ/(0.052 + 0.102)1/2  .
Here is a hint for point B:
ENETx  is not zero. 
ENETy  is positive.
61. This is  interesting and the most powerful of the entire homework: the simple harmonic oscillation   of   a charged particle. The problem's model contains, however,  one big  departure from the actual oscillation of molecules in solids. Real oscillations are not symmetric about the local origin  where the force momentarily is zero (at equilibrium). Consider equilibrium position  x = 0 in the excitation of a mass attached to a horizontal  spring and thus under the influence of Hooke's force F = -kx leading to symmetric motion about the origin.   In  real solids,  the amplitude would be  double valued in that an asymmetric  spring swings  further to the left, say,  than to the right  before momentarily coming to rest and turning around.  This asymmetry  accounts for the expansion of solids when they are heated as you will discover in 4C. For further info , visit: http://homepage.mac.com/phyzman/phyz/BOP/2-06ADHT/G-Thermal_Expansion.pdf
and read to the bottom  the document. For posterity, you may want to download  the page into your hard drive as I did for future reference. Required reading. The is a lucid application  of the course  to stable,  ubiquitous  electro-static oscillations in solids. 

In any event you want to show the x-component  of force F may be written as 
F = -Cx, where C is a positive constant.

Clearly the charged particle is attracted back the the ring whether the value of x is positive or negative: we would thus write,

F = q*E, where E is the x-component  of electric field given by the  solution to the charged ring but with some qualifications.  We obtained the formula for E, but  be mindful in this case the charge symbol Q would be replaced by
 - Q , producing a force whose component is negative  when x is positive and positive when x is negative - the prerequisite for  simple harmonic motion achieved when you mathematically simplify  using approximation  | x |<<R. 

Finally , to achieve all results, write the left hand side of the above equations using Newton's 2nd law:

F  = m*d^2x/dt^2 = m*acceleration,  where b^2 is the expression b for squared.

From 4A, you can now compute the  desired ratio for the  frequency of oscillation.

One question you might to ask. What if the particle was oscillating  in a vertical gravitational  field? Would  the gravitational  force on the particle influence the problem in a significant way?