Quiz    17  questions: 1, 2, 3, 5, 6, Problems 1, 2, 3, 4, 10, 11, 14, 15, 18, 19, 20, 21, 22, 31, 32, 34, 35, 36, 37, 38, 39, 46, 47, 48, 49 (ANSWERS)
TURN IN (EC means extra credit) : Questions 2, 3, 5, 6; Problems 2, 10, 18, 20, 22 (EC-Hint: see question 5),  36, 38, 46, 48
Simulation: Click here for the motion of an electron in a uniform electric field.
Click here  and here for a suggested experiments to measure equipotential  lines. 
(The following notes are from a previous term and do not apply to Sp'11.) Note: The  "e/m lab"  on

http://session.masteringphysics.com

has a warm up  problem,

"Repulsive Force between Two Spheres. "

with a new due date:  2/23/09 midnight.

The online assignment from Quiz 16 will be posted soon on masteringphysics.com .
I will email you when it is ready.
Problem discussions are below. You must also turn in the questions; the problems may help you answer the questions. 
2. Work by electric field  = -(charge)*( change in potential) . This formula works for any sign of charge and change in potential. Explanation: (change in potential) =  (change in potential energy)/(charge)  for  any sign of charge and change in potential.  Since Work by electric field =  - (change in potential energy), the very  first equation at the top of this hint  follows.
ANSWER: 2.88x10-17 J
10. Change in kinetic energy = Net Work
Also:
Net Work = (work by external force) + (work by electric field). See Chapter 6 and the Work Energy Theorem,   equations 6-2 and 6-4.

The (change in kinetic energy) is given. The (work by external force) is also given.  Solve for the (work by electric field). Once you find (work by electric field) , find the (change in potential) using the concepts discussed in the previous problem  hint, i.e. problem 2.
ANSWER: -1.20x102 V

18. (a) Equation  17-5. (b) Use info in Example 17-5.  Modify the example so that the two charges have equal values.  Let Q for both charges be the proton charge (1.6x10 - 19 C )  which you can easily look up in the book. Obviously you will use the distance prescribed  by this problem.
ANSWERS:
(a) 5.8x105 V
(b) 9.2x10-14 J
20. Use conservation of energy  as in Chapter 6, section 6-7.  See problem 43 or 50 , Quiz 6.  

KEi + PEi = KEf  + PEf,   where KE stand for kinetic energy,  PE for potential energy; i and f refer to initial and final respectively.  Use the definition that  PE = (charge)*(potential) =q*V. This definition works for any sign charge q. V is the potential at some point in space. See Example 17-2 in section 17.1.   See also  Conceptual Example 17-1 and surrounding passages.  In this case,

KEi + PEi = KEf  + PEf  means

(1/2)*m*vi2   +  q*Vi  =    (1/2)*m*vf2   +  q*Vf   .       

Now, the initial speed vi = 0. Note that small case v means speed and velocity and V means potential.  
Vi = initial potential, found from equation 17-5 using the  value of Q and and initial r given by this problem. Convert r to meters. q = -e = - 1.6x10 - 19 C and m = electron mass.  Vf = final  potential, found from equation 17-5 using the  same Q value  but  the final  r =  infinity. 
ANSWER: 3.49x107 m/s
36. 
Q = C*V

Thus,
(change in Q) = C*(change in V).

Find C.
ANSWER: 7.5x10-7 F
38. C = Aeo/d. = Q/V.   We want A.  Now, V = E*d. Thus,  Aeo/d  = Q/(E*d)
ANSWER: 0.29 m2.
46. Capacitor PE = (1/2)*Q2/C = (1/2)*C*V2 .  These are two definitions of the same PE. Read the book ! (Section 17-9)
ANSWER: 4.6x10 -4 J
48.  Energy in the electric field  = capacitor PE. Thus,  you can evaluate  PE = (1/2)*Q2/C  or PE = (1/2)*C*V2  
The first equation may be easiest to use since Q is given. You can find C from the given information. 
ANSWER: 2.3x103 J.
More hints later!
``