Quiz 3 Ch. 23--4, 5, 12; ALSO TRY,  but do not turn in,  3 and 9; 14, 16, 19, 24, 26, 28,  31, 34, 44,  47, 57,  59,  63, 64; ALSO TRY,  but do not turn in,  62;79   
NOTE: You can get the same answers below using my methods in class; see lecture notes. Below may be  different ways of doing problems, but you can use my approaches,   which are more understandable  to some students.  

4. (a) Factoring out the  and substituting numbers gives

:

(b)
Solving for v gives

12. 

       and


14. (a)

(b)  Point a is at higher potential than point b.


(c)  so


16.

 

Setting  gives


24.

(a)  so    






(b) 


(c)

 so the electric field is directed away from the charge.





26.

(a)  so V is zero nowhere except for infinitely far from the charges. The fields can cancel only between the charges, because only there are the fields of the two charges in opposite directions. Consider a point a distance x from Q and  from 2Q, as shown in Figure of class notes.    The other root,  does not lie between the charges.

(b) V can be zero in 2 places, A and B, as shown in Figure of class notes. Point A is a distance x from  and  from 2Q. B is a distance y from  and from 2Q.  



28.

(a) This is outside the sphere, so

(b) This is at the surface of the sphere, so

(c) This is inside the sphere. The potential has the same value as at the surface, 131 V.






34. 

(a)

Substituting numbers gives

(b) We can take any path since the potential is independent of path.

(c) Set Up:   The net force is away from the ring, so the ball will accelerate away. Energy conservation gives

Solving for v gives

 

44.

(a)  No matter what the reference point, we must do work on a positive charge to move it away from the negative sheet. Since we must do work on the positive charge, it gains potential energy, so the potential increases.

(b)Since the electric field is uniform and is equal to  we have

Solving for d gives


64.NOTE: You can get the same answer below using my method in class; see lecture notes. Below may be  a different way of doing the problem, but you can use my approach,   which is more understandable  to some students.  

     

     and