733_Physics ProblemsTechnical Physics

733_Physics ProblemsTechnical Physics - Chapter 25 *P25.69...

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Chapter 25 73 *P25.69 (a) V kq r r rr ee e =−= 12 1 2 21 bg From the figure, for ra >> , rr a 2 −≅ cos θ . Then V a kp r ≅≅ 2 2c o s cos . (b) E V r r r e =− = 2 3 cos In spherical coordinates, the component of the gradient is 1 r F H G I K J . FIG. P25.69 Therefore, E r V r e F H G I K J = 1 3 sin . For >> E r r e 0 2 3 °= af and E r 90 0 , E 00 and E r e 90 3 . These results are reasonable for >> . Their directions are as shown in Figure 25.13 (c). However, for rE →→ ,. This is unreasonable,
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