717_Physics ProblemsTechnical Physics

717_Physics ProblemsTechnical Physics - Chapter 25 P25.18...

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Chapter 25 57 P25.18 (a) E kq x x x ee =+ = 1 2 2 2 200 0 . af becomes Ek q x q x xe = + + F H G I K J = 22 2 0 . . Dividing by k e ,2 2 0 0 2 2 qx q x =− . xx 2 400 400 0 +− = .. . Therefore E = 0 when x = −± + 160 160 2 483 . . m . (Note that the positive root does not correspond to a physically valid situation.) (b) V x x = 12 0 . or Vk q x q x e = + F H G I K J = 2 0 . . Again solving for x 2 0 0 qx q x . . For 02 0 0 ≤≤ x . V = 0 when x = 0667 m and q x q x = 2 2 . For x < 0 x m . P25.19 q r V V i i i = × −+ L N M O Q P × 8 99 10 7 00 10 1 00100 1 1 00387 110 10 110 96 7 ... ej e j V M V FIG. P25.19 P25.20 (a) U qQ r = = ×− × × × −− 4 5 00 10 3 00 10 8 99 10 0350 386 10 0 99 9 7 π . . . C C V m C m J e j e j The minus sign means it takes 3 86 10 7 . ×
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