Electricity (电学答案).pdfVIP

Electricity 13-1 We obtain that the field evaluated at the center of curvature due to a charged distribution on a circular arc is given by   E sin 4 r  0 along the symmetry axis, with  = q/r with  in radians (弧度）. In this problem, each charged quarter-circle produces a field of magnitude |q | 1  / 4 1 2 2 |q | |E | sin 2 . r/ 2 4 r  / 4 4 r 0 0 That produced by the positive quarter-circle points at – 45°, and that of the negative quarter-circle points at +45°. (a) The magnitude of the net field is  1 2 2 |q | 1 4|q | Enet,x 2  2 cos 45 2 4 r  4 r  0  0 9 2 2 12 (8.9910 N m C )4(4.5010 C) 20.6 N/C. (5.00102 m)2 ˆ (b) By symmetry, the net field points vertically downward in the j direction, or 90 counterclockwise from the +x axis. 13-2 We assume q 0. Using the notat