A. Electrostatics

Part I – 100 points

**15.1 Electric Charge**

**1-5, 8, 10**

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**15.2 Electrostatic Charging**

**11, 12, 13, 15 **(What they don’t tell you in the
problem is that when you pull the spheres apart, one should have a positive
charge, and the other a negative charge.
Keep that in mind when you do #15 – obviously they will be charged by
some kind of induction method.)

**Note:** The answer to #11 in the book is INCORRECT. The correct answer is POSITIVE!

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**15.3 Electric Force**

**16, 17, 18, 20, 22, 27, 28, 30, 32, 33, 34**

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18: Anything that is charged electrically won’t stay charged for long. Also, recall the conservation of charge law which states the charge on an isolated system remains constant.

22: Carbon has 6 electrons (thank you chemistry!)

27: Easy, no math required. Just think about it

30: a) find force using columb’s law. b)
find v because F = m(v^{2}/r).
Remember that formula from the circular motion chapter? c)
find a_{c} because a_{c} = v^{2}/r = F/m. Also, for c, since g = 9.8 m/s^{2},
you must divide your answer by 9.8 to find out how many times larger than g it
is.)

33: Find force on q_{2}
caused by q_{1}, q_{3} and q_{4}. Then, add all of these vectors
together. Do you remember how to find
the components of vectors and add them together? You should!

Realize that q_{1}
**PUSHES** q_{2} along the +x axis.
Realize that q_{3} **PULLS** it down along the –y axis. Realize that q_{4} **PULLS** it
at an angle which is 225^{o} from zero (yes, it is a 45^{o}
angle, but if you use the value 225^{o} you will find the solution
gives you the correct sign values for the magnitudes. Do you see where I get the value 225? 180 + 45 = 225)

34: First, find angle q. Then, notice that in the vertical direction, TCosq = mg. Therefore, we can define the tension of the string (T) as being = mg/cosq.

Next, notice
that in the horizontal direction, Tsinq = F_{e}. So, if you substitute mg/cosq for
T in this expression, you get: F_{e}
= mg/cosq
(sinq). Wow, there’s a trig identity in there
somewhere!

Once you have F_{e},
you can use your old friend F_{e} = kq^{2}/r^{2} (and q is squared since we know that the
charges on both objects is the same).
Solve for q.

Answers to Even problems:

2: b

4: no – if you don’t understand why, ask in class.

8: a) -8.0 x 10^{-10} C b)
5.0 x 10^{9} electrons

10: +6.40 x 10^{-19}C

12: b

16: c

18: We are electrically neutral.

20: a) 1/4 as large. b) 9 times as large

22: 1.3 x 10^{-7}N

28: a) nowhere b) nowhere

30: 8.2 x 10^{-8}N b)
2.2 x 10^{6} m/s c)
9.2 x 10^{21} g

32: 3.6 N in the +x direction

34: 3.3 x 10^{-8}C