40. The field due to the sheet is E = 2 . The force (in magnitude) on the electron (due to
that field) is F = eE, and assuming its the only force then the acceleration is a= e = slope of the graph ( = 2.0 105 m/s divided by 7.0 1012 s) . 2o m
Thus we obta
39. The charge on the metal plate, which is negative, exerts a force of repulsion on the electron and stops it. First find an expression for the acceleration of the electron, then use kinematics to find the stopping distance. We take the initial direction
38. We use the result of part (c) of Problem 23-35 to obtain the surface charge density. E = / 0 = 0 E = ( 8.85 1012 C2 /N m 2 ) (55 N/C) = 4.9 1010 C/m 2 . Since the area of the plates is A = 1.0 m 2 , the magnitude of the charge on the plate is Q = A =
37. We use Eq. 23-13. (a) To the left of the plates:
E = ( / 2 0 ) ( (from the right plate) + ( / 2 0 )i (from the left one) = 0. i)
(b) To the right of the plates: E = ( / 2 0 ) (from the right plate) + ( / 2 0 ) ( (from the left one) = 0. i i) (c) Betw
36. The charge distribution in this problem is equivalent to that of an infinite sheet of charge with surface charge density = 4.50 1012 C/m2 plus a small circular pad of radius R = 1.80 cm located at the middle of the sheet with charge density . We denot