QUIZ 4 pt 1 and 2
 
NOTE: MOST OF QUIZ 4 PART 1 DISCUSSIONS WERE EMAILED INSTEAD OF POSTED ON THE WEB SITE.
4. -53.8 (V)
12. Positive work by an external force on a negative particle means you, the external force agent, are moving that charge toward other negative charges; thus you are  moving the negative particle  from a high to a low potential region of space.
Thus the change in potential  is negative: -53.8 (V) 
14. (a) 3.8 x10-8 C/m2  (b) 4.4 (m)
22. This is a fairly complex problem, though do-able with a certain level of mental discipline and rule following.
(a) out side the sphere: E = (3/2)*kQ/r2.  Within the conductor, E = 0  Within the cavity a distance r from the central   charge, E =  (1/2)*kQ/r2.
(b)  out side the sphere: V = (3/2)*kQ/r.  (c) Within the conductor, V = (3/2)*kQ/r2 = potential of conducting shell. 
(d) Within the cavity a distance r from the central   charge,
V = potential of the conducting shell - integral (E vector)*(dl vector) from r1 to r =
(3/2)*kQ/r2  +  (1/2)*kQ[ 1/r  - 1/r1). Using  r2 = 2*r1   , we get (kQ/2)*[ 1/r2  +  1/r] 
26. Assume negative Q2 is on the left end of the segment and positive Q1 is on the right end.
 (a) 16 cm left of the negative Q2 (b) 1.9 cm to the right of the negative chargeQ2  towards the positive charge Q1 and 7.1 cm to the left of the negative charge Q2  away from the positive charge Q1.
28.  (a) 3.6x104 (V)  (b)  |Vector Eb - Vector Ea | = 570,000 V/m 
34. (k*1.41*Q)/(2L)*( 1 + 1.41), where 1.41 = square root of 2.
36. Q/(4*epsilon*L)
38. (1/2)*(kQ/L)*ln[ {sqrt(L2 + y2) + L}/{ sqrt(L2 + y2) - L } ]
44.( a)  0.444 m (b) 0.49 m  (c) 21 m
48. Component of vector E in the r direction = E because  the field is radial. E = -dV/dr =
2400000000000000000000 V/m = 2.1x10 21V/m,  very large because the radius is so small.
50. vector E = (y2 - a2)*b/(a2 + y2)2, in the positive or negative   y-direction depending on the value of y.   See ICQ 10-15, with a flexible deadline TBA (To be announced) .
54. 6.9x10-18 J.
60.
(a) (k*Q2/b)*(4 + 1.41) , where 1.41 = square root of 2.
(b) U5th charge = (k*Q2/b)*(4*1.41), where 1.41 = square root of 2.
(c) Giancoli is shrewd in that he recognizes, as all professional physicists do,  there is NO location of stable electrostatic equilibrium in a three dimensional (3-D) approach. So he adds the proviso that the charge is constrained to remain on the plane.  In this artificial 2-D situation, the location of a positive charge at the center would be that of stable equilibrium and (d)  the location of a negative charge at the center would be that of unstable equilibrium.
76. 6.5 degrees.