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Problem 2.2 (a) Find the electric field (magnitude and direction) a distance z above the midpoint betweentwo equal charges, 9, a distance d apart (Fig. 2.4). Check that your result is consistent withwhat you'd expect when z >> d.(b) Repeat part (a), only this time make the right-hand charge - instead of +q.2P•РŹdgPdi'(b) Line charge, a(a) Continuousdistribution•Рda'2 Pdi'9 d/2d/29(c) Surface charge, o(d) Volume charge, pFigure 2.4Figure 2.5 |
| Answer» ANSWER: Answer: Answer: L=2m,Answer: L=2m,d=3mm,A= Answer: L=2m,d=3mm,A= 4ANSWER: L=2m,d=3mm,A= 49πAnswer: L=2m,d=3mm,A= 49πAnswer: L=2m,d=3mm,A= 49π ×10 Answer: L=2m,d=3mm,A= 49π ×10 −6Answer: L=2m,d=3mm,A= 49π ×10 −6 m Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2ANSWER: L=2m,d=3mm,A= 49π ×10 −6 m 2 Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 4Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49πAnswer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49πAnswer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mmAnswer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm .Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm . Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm . | |