Chapter 24 Class Examples

Chapter 24 Class Examples - Chapter 24 Class Examples 7. We...

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1 Chapter 24 Class Examples 7. We connect A to the origin with a line along the y axis, along which there is no change of potential (Eq. 24-18: E ds z 0). Then, we connect the origin to B with a line along the x axis, along which the change in potential is V E ds xdx x       F H G I K J z z 4 00 4 00 4 2 2 0 4 0 4 . . which yields V B – V A = –32.0 V. 18. When the charge q 2 is infinitely far away, the potential at the origin is due only to the charge q 1 : V 1 = 1 0 4 q d  = 5.76 × 10 7 V. Thus, q 1 / d = 6.41 × 10 17 C/m. Next, we note that when q 2 is located at x = 0.080 m, the net potential vanishes ( V 1 + V 2 = 0). Therefore, 21 0 0.08 m kq kq d  Thus, we find q 2 = 1 ( / )(0.08 m) qd = –5.13 × 10 18 C = –32 e . 24. The potential is 9 2 2 12 2 rod rod 0 0 0 1 1 (8.99 10 N m C )(25.6 10 C) 4 4 4 3.71 10 m 6.20 V. P dq Q V dq R R R     We note that the result is exactly what one would expect for a point-charge – Q at a distance R . This
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Chapter 24 Class Examples - Chapter 24 Class Examples 7. We...

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