A. Electrostatics

Part I – 50 points

15.1 Electric Charge

1-5, 8, 10


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!


15.3 Electric Force

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


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(v2/r).  Remember that formula from the circular motion chapter?  c)  find ac because ac = v2/r = F/m.  Also, for c, since g = 9.8 m/s2, you must divide your answer by 9.8 to find out how many times larger than g it is.)

33:  Find force on q2 caused by q1, q3 and q4.  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 q1 PUSHES q2 along the +x axis.  Realize that q3 PULLS it down along the –y axis.  Realize that q4 PULLS it at an angle which is 225o from zero (yes, it is a 45o angle, but if you use the value 225o 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 = Fe.  So, if you substitute mg/cosq for T in this expression, you get:  Fe = mg/cosq (sinq).  Wow, there’s a trig identity in there somewhere!

     Once you have Fe, you can use your old friend Fe = kq2/r2   (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 109 electrons

10:  +6.40 x 10-19C

12:  b

16:  c

18:  We are electrically neutral.

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

22:  1.3 x 10-7N

28:  a) nowhere    b) nowhere

30:  8.2 x 10-8N    b)  2.2 x 106 m/s      c) 9.2 x 1021 g

32:  3.6 N in the +x direction

34:  3.3 x 10-8C