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*q^{2}/D^{2}. 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*q^{2}/D^{2}. 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: F_{xnet} = F + F*cos65 . y-component: F_{ynet} = F*sin65 . Note F = k*q^{2}/D^{2 } . 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*t^{2} 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/d^{2} . (b) E = k*e/D^{2} . |
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 E_{x} = 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: E_{x} = 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 ^{-19}C. |
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 = Q_{enc}/e_{o}
. For this problem there is spherical symmetry, so: Total
flux = -E*(4*pi*r^{2}) = Q_{enc}/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*e_{o}) as
I use in my form of the celebrated equation. Thus, -E*(4*pi*r^{2}) = Q_{enc}/eo , where r = 0.30 m. Set E = 1150 N/C and find Q = Q_{enc.} 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*r^{2}) = Q_{enc}/e_{o} , 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*r^{2}) = Q_{enc}/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: Q_{enc } = Q = -15.0 x10^{ -6 }C. This is the enclosed charge for gaussian surface just outside surface. (c) -E*(4*pi*r^{2}) = Q_{enc}/e_{o} , 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 Q_{enc} = -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. |