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QUIZ 1 |
| 12. -150 N, 560 N, -420 N are the forces starting with the charge on the left. Note these are x-components. When negative, the force points to the left. |
| 16. For each point charge, there are two forces that point towards the adjacent corners, and one force that points away from the center of the square. Thus for each charge, the net force will be of magnitude 1.42X107N and will lie along the line from the charge inwards towards the center of the square. |
| 22. 1.36X105 N/C south |
| 24. 9.5X 105 N/C up |
| 26. (2400i - 3100j)( N/C), where i and j represent our standard Cartesian unit vectors. |
| 28. 7.8X107 N/C, toward negative charge. |
| 36. 35cm. Note that when you solve for this value of x, you can simplify the quadratic equation into a linear equation with two roots. Just take the square root of both sides immediately and choose the correct root that gives a positive value for x. |
| 38. (a) 31/2 *k*Q/L2. Angle = 240 degrees with the positive x-axis---the vector points in the third quadrant. (b) k*Q/L2 , 330 degrees with the positive x-axis, pointing in the 4th quadrant. |
| 42. (a) 3.7X106 N/C, vertically up (b) 9.5X106 N/C, angle = 56 degrees, See Figure 21-34b for confirmation. The vector in both cases is tangent to the electric field line at the point of evaluation. Clearly parts (a) and (b) of this problem are consistent with that figure. |