Practice Problems 1 Solutions

xxSolutions Practice Problems I P23.5 (a) (b) The electric force is . (c) If with and , then . P23.7 FIG. P23.7 P23.18 (a) FIG. P23.18 P23.10 Let the third bead have charge Q and be located distance from the left end of the rod. This bead will experience a net force given by . The net force will be zero if , or . This gives an equilibrium position of the third bead of . The equilibrium is . P23.14 : P23.15 The point is designated in the sketch. The magnitudes of the electric fields, , (due to the charge) and (due to the charge) are (1) (2) FIG. P23.15 Equate the right sides of (1) and (2) to get or which yields or . The negative value for d is unsatisfactory because that locates a point between the charges where both fields are in the same direction. Thus, . P23.33 Due to symmetry , and where , so that, where and . FIG. P23.33 Thus, . Solving, . Since the rod has a negative charge, . P23.42 electron: in a direction opposite to the field proton: in the same direction as the field P23.43 (a) (b) (c) (d) P24.2 P24.10 (a) : But Q is negative since E points inward. (b) The charge has a charge distribution. P24.24 (a) (b) (c) (d) The direction for each electric field is . P24.44 (a) (b) (c) (d) P24.48 (a) (b) P24.55 (a) (b) The charge distribution is spherically symmetric and . Thus, the field is directed . (c) for . (d) Since all points within this region are located inside conducting material, for . (e) (f) (g) (radially outward) for . (h) (i) (radially outward) for . (j) From part (d), for . Thus, for a spherical gaussian surface with , where is the charge on the inner surface of the conducting shell. This yields . (k) Since the total charge on the conducting shell is , we have . (l) This is shown in the figure to the right. FIG. P24.55(l)