| Quiz 16 Chapter 16 problems: 12,
13, 14, 15, 23, 24, 25, 26, 27, 28, 37, 38 , 41, 42 (ANSWERS) |
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TURN IN: 12, 14, 24, 26, 28, 38 [A only; B is extra Credit (EC)],
42 (EC) |
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LAB:
LIGHT AS A LENS TO COULOMB'S LAW EXP 32 VERNIER: SEE PAGE
154, FIG. 9.5, of Conceptual Physics (11ed) by Paul
Hewitt or this u-tube video: http://www.youtube.com/watch?v=JW3tT0L2gpc
if you do not have that book. In the figure, the
THICKNESS OF THE LAYER OF PAINT IS like the intensity.
When you double the distance, the thickness of the layer of paint
or butter thickness on the toast (see video above) goes down
by a factor of 4, not 2, since you are squaring the
denominator. The further out you go, the larger the radius r, and
the smaller the intensity, just like quiz 5 problem , #11:
http://www.nvaphysics.com/p11/au10/quizzes/QUIZ4AU10SAMPLECH9.htm
for Physics 11.
Note the formula for the surface area of a sphere is 4*pi*r2,
where r2 means the radius squared and pi = 3.14..... The
intensity = Power/(area), where Power is the energy per unit time
emitted by the source and area is the surface area of the sphere:
surface area = 4*pi*r2. The power does not
depend on the radius r, only the intensity does. Thus, Power is a
constant for any distance r from the source in the above formula.
Power does not change as r changes.
2. If the light source is not a true point source , then the
radiation is NOT emitted with wave fronts in perfect spheres as
shown in FIGURE 2 OF THE LAB HANDOUT. The picture may not
look like figure 9.5 in all directions for the same radius r and
the u-tube video would not apply. Figure 9.5 and the video
assume the inverse square law is true in all directions. Now
assume the source is not a true point source, but is
extended and/or may not have a spherical shape. Would light be
emitted equally in all directions? Would the intensity obey the
inverse square law?
ANALYSIS QUESTIONS
1. BASED ON THE SHAPE OF THE GRAPHS, WHICH FORMULA SHOULD
YOU PICK FROM THE CHOICES GIVEN?
2. From the curve fit menu, which curve did you pick
from the choices suggested to you on the white board? Which
function gave the best match to your curve? What was the formula
you used? Include the actual numbers you recorded from the
curve fit menu.
3. See preliminary questions and figure 2. If the model of
your experiment obeys the inverse square law, then the
concentric spherical model discussed in the preliminary questions
is true.
4. Explain experimentally why your source may not be
considered a true point source as assumed in the preliminary
questions and figure 2. What were the limitations of the
experimental set up that caused your results not to match the
relationship predicted in the preliminary questions? List these
limitations. Think about measurement errors and reflections.
What did you measure directly? What could be the cause of
reflections. Was there light from other sources in the room when
you made your measurements?
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| DISCUSSIONS BELOW. |
| 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, staring 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 there two forces
in the same direction.
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 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. |
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14. 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 6x10-3 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.
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| 24. A proton is positive, so the force points
in the same direction as the electric field vector. Note: E = F/Q.
Look up the proton charge using Google or the textbook |
| 26. Use equation 16-4a or 16-4b. You should be able
to find the field direction given that the lines leave positive
charges. |
28. See the hints for the computation of the net
electric force on charge 3, problem 12. See also example
16-8.
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, but the force points
left. |
| More hints later. |
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