Solve: (a) The electric fields due to the charges at point A are shown in Figure 17.40a. Since the two fields are in opposite directions, we subtract their magnitudes to find the net field. to the right. (b) The electric fields at points B are shown in Figure 17.40b. Since the fields are in the same direction, we add their magnitudes to find the net field. to the right. (c) At A, to the right. The force on a proton placed at this point would be to the right. 17.41. Set Up: For a point charge, is directed toward a negative charge and away from a positive charge. Let the points a, b and c be the locations where the field is calculated in parts (a), (b) and (c). The three points and the electric fields produced at those points by each of the two charges are shown in Figure 17.41. Figure 17.41 (a) The field is 1050 N/C, in the direction. (b) The field is 312 N/C, in the direction. (c) The field is 845 N/C, in the direction. Reflect:
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