Quiz 16 Chapter 16 problems: 12, 13,  14, 15, 23, 24, 25, 26, 27, 28, 37, 38 , 41, 42 
TURN IN: 12, 14, 24, 26, 28, 38 [A only; B is extra Credit (EC)], 42 (EC)

Hints provided soon !
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 again there two forces on opposite directions. 
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.

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.

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.