16.4 Ex 66 - 67

16.4 Ex 66 - 67 - 986 C H A P T E R 16 M U LTI P L E I N T...

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986 CHAPTER 16 MULTIPLE INTEGRATION (ET CHAPTER 15) If you had arranged the axes differently, you could have computed the answer as ³ 0 , 1 12 , 37 48 ´ (depending on orientation). 66. According to Coulomb’s Law, the force between two electric charges of magnitude q 1 and q 2 separated by a distance r is Cq 1 q 2 / r 2 acting along the line through the two charges ( C is a constant). Suppose that a circular disk of radius R has a uniformly distributed charge of density k Coulombs per square centimeter (Figure 25). Let F be the net force on a charged particle P of magnitude Q coulombs located at a distance d cm vertically above the center of the disk (by symmetry, F acts in the vertical direction). (a) Let R be a small polar rectangle of size 1 r × 1 θ located at distance r . Show that R exerts a force on P whose vertical component is ± kCQd ( r 2 + d 2 ) 3 / 2 r 1 r 1 (b) Explain why F is equal to the following double integral and evaluate: F = Z 2 π 0 Z R 0 rdrd ( r 2 + d 2 ) 3 / 2 R d Charged plate Δ r Δ P FIGURE 25 SOLUTION (a)
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16.4 Ex 66 - 67 - 986 C H A P T E R 16 M U LTI P L E I N T...

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