QUIZ 17, CHS. 17

EVEN ANSWERS
SELECTED DISCUSSIONS TO exercises/problems BELOW. I've try to include interesting explanatory hyperlinks throughout. YOU ARE RESPONSIBLE FOR ALL  PROBLEMS EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED EXERCISES. Each sublist is separated by a semi-colon ";" and includes a representative example for  exercises. WE WILL REVIEW THEM IN-CLASS THE LAST WEEK BEFORE  EXAMS. 
This SIMULATION helps convey many concepts in this  problem set. 
CH.   17
CH.  17  MULTIPLE CHOICE 3, 4, 5, 7, 10, 13  - Exercise/problems  example 1: 15, Example 2: 17, Example 3:  22, Example 4: 20, Example 5: 33, 70, : example 6: 37,  Example 7: 43, 40,  Example 9: 59 (See also example 10) , Example 10: 57, Example 11: 62, 63.
TURN IN: 15,  17,  22,  20,  70, 37,  43, 40, 59, 57,  62, 63.
* DISCUSSIONS PROVIDED. CHECK BACK SOON !
15. (a) number of electrons = (atomic number)* (number of moles)*(avogadro's number), where number of moles = mass/atomic mass.
(b) force = 10,000 N = k*q2/D2.  Solve for q then divide by electron charge magnitude e.
(c) {compute answer to (b)}/ {answer to (a)}
MORE DISCUSSION TO FOLLOW:  
17. WOW !  With that much excess charge, if sufficiently close to each other, you two would create a "big bang" of sorts as the computation will show  using coulomb's law. You think that little shock hurts when you get in the car at the gas station? This jolt will really make your day!  |force| = 650 N  = k*q2/D2. Find D; do you expect  it to be large or small? Answer this question before you start writing  and you may not be surprised at the outcome.   
22. See example 3. The 5.00 nC charge is 0.500 m to the left of the 4.00 nC charge.  The middle   negative charge is 0.300 m  to the right of the leftmost charge.  Net force is points  RIGHT from Coulomb's Law since the middle charge is closer to the rightmost charge even though the magnitude of that right charge is the smaller than the leftmost charge. See example 3 for guidance though the problem is different.  
20. PLACE THE PROTON AT THE ORIGIN OF X-Y axes.  Clearly the net force points in the first quadrant. Use components:

x-component: Fxnet  = F + F*cos65 .
y-component: Fynet  = F*sin65 .

Note F = k*q2/D2   .     Look up the value of q in your notes or memory bank.   
70. Cool problem taking you back to 2A and the virtual projectile motion lab. Here all you have to do is pretend gravity is up side down, force pointing up, and use the fact the vertical constant force is F = e*E, where E is the electric field magnitude  between the plates. We see a = F/m = e*E/m.
Use : y = (1/2)*a*t2  and x = vo*t = 0.0200 m.  Solve simultaneously for E.  See this SIMULATION. 
37. The nucleus is a complex animal. First of all, in all atoms except hydrogen, the proton's, which are like charges,  repel each other electrically.  So why is the nucleus relatively stable? Because of the powerful attractive nuclear force between protons, much larger in magnitude than the electric force at small  distances characterizing the atom's core, about 10 -15 m typically,  much smaller than the entire  atom, including electron "cloud," in figure 17.3.
(a) E = k*e/d2 .
(b) E = k*e/D2 .
43. We  provided a hint of this exercise in lecture----SEE 2ND SET OF PHOTOCOPIED LECTURE NOTES. 
40. (a) At point A, the field from the charge on the right points right and the field from the charge on the left points left, This is consistent  with the fact  electric fields point toward NEGATIVE charges. Note: 
magnitude Enet = | k*(12.5x10-9 C)/(0.10 m)2 -   k*(6.25x10-9 C)/(0.15 m)2 |. Note how the magnitude is positive. If I considered  rightward  the positive x direction, I could've  written the x-component of electric field as  Ex = 
Ex =  k*(12.5x10-9 C)/(0.10 m)2 -   k*(6.25x10-9 C)/(0.15 m)2 .
(b) At point B, the field from the charge furthest away  points right and the field from the closest charge also is rightward,
again consistent  with the fact  electric fields point toward negatives. Note: 
magnitude Enet = | k*(12.5x10-9 C)/(0.35 m)2  +   k*(6.25x10-9 C)/(0.10 m)2 |. Note how the magnitude is always positive. If I'd considered  rightward  the positive x direction, I would've  written x-component of electric field as:
Ex =  k*(12.5x10-9 C)/(0.35 m)2   +   k*(6.25x10-9 C)/(0.10 m)2 .
(c) A proton at point A would feel a rightward force  and a leftward force. Clearly the force pointing right would be larger in magnitude based on part (a) results and the formula for force magnitude  F = q*E, where q is the  proton's  charge and E is the NET (RESULTANT) electric  field magnitude.  Note q = +e = +1.6x10 -19C. 
59. (a) Certainly example-10 as well as  example-9 would help here as it does on the next discussed  problem. Meanwhile, let us repeat the world famous Gauss's Law  : Total flux = Qenc/eo  . For this problem there is spherical symmetry, so:   Total flux  = -E*(4*pi*r2) = Qenc/eo  for this spherically symmetric system. Note 2 things: (i.) The flux is negative since the field lines go inward through the gaussian surface toward the negative charge and opposite the normal line, considered to point radially outward. (ii.)  k = 1/(4*pi*eo) as I use in my form of  the celebrated equation.  

Thus, -E*(4*pi*r2) = Qenc/eo , where r = 0.30 m. Set E = 1150 N/C and find Q = Qenc. Once you find Q, which will be negative,  divide by -e = -1.6x10 -19 C to find the number of excess electrons, which will be obviously positive. More discussed in class to sort this out !   

(b) -E*(4*pi*r2) = Qenc/eo , where r = 0.40 m. Note:  0.40 = 0.30 + 0.10. Be careful !! We reference all r values to the sphere center.  Use the enclosed Q you got in part (a) to find E's new value  further way from  sphere center.  
57. Some folks might choose doing this problem ahead of  59,  simply because it's easier conceptually and but it does have the minor complication of negative source charges mentioned in previous problem.  
Here we go:  -E*(4*pi*r2) = Qenc/eo ; the negative sign is on left side again because we have flux lines going inward opposite normal line. 

(a) See example 10 and class notes.  What would be the charge enclosed if the Gaussian source was just inside the spherical shell such that the charge enclosed would be ZERO? If the enclosed charge is zero, then what is E from Gauss's Law??
(b) In the above equation set r = 0.12 m and find E. Note: Qenc  = Q = -15.0 x10 -6 C. This is the enclosed charge for  gaussian surface just outside surface. 
(c)  -E*(4*pi*r2) = Qenc/eo , where r = 0.17 m. Note:  0.17 = 0.12 + 0.05. Be careful !! We reference all r values to the sphere center.  Use the given Qenc = -15.0 x10 -6 C, the enclosed charge  for a concentric gaussian surface whose radius r = 0.17 m.  
62. See figure 17.31 (a), which I covered in class and also figure 17.31 (b), which applies directly to this problem. Model the car as in figure 17.31 (b); we see the  charge on the surface of the cavity is ZERO, proven using Gauss's Law and the rule of zero electric  field inside conductors.  Please read the textbook explanation of figure `17.17 
(b) All the excess charge is on the outer surface.
63. See example 11.  If there was a point charge inside the cavity  the charge on the cavity surface  would be opposite in sign and equal in magnitude as that point charge.  If there was not a point charge inside the cavity, the charge on the cavity surface would be zero.     
Let's use logic based on the information  the conductor is NEUTRAL. Since there is charge on the outer surface, the charge on the cavity surface must be equal and opposite to maintain neutrality. Thus, the cavity surface has charge +12 x10 -6 C to balance the  -12 x10 -6 C charge   on outer surface. If the cavity surface charge is  +12 x10 -6 C, then what must be the value of the point charge within the cavity. Re-read example 11 and #62!  
MORE DISCUSSIONS TO COME in class.  DO NOT LET THEIR ABSENCE PREVENT YOU  FROM COMPLETING  QUIZ TO THE BEST OF YOUR ABILITY.