Problem 2.2
(a) Find the electric field (magnitude and direction) a distance z above the midpoint between
two equal charges, 9, a distance d apart (Fig. 2.4). Check that your result is consistent with
what you'd expect when z >> d.
(b) Repeat part (a), only this time make the right-hand charge - instead of +q.
2P
•Р
Ź
dg
P
di'
(b) Line charge, a
(a) Continuous
distribution
•Р
da'
2 P
di'
9 d/2
d/29
(c) Surface charge, o
(d) Volume charge, p
Figure 2.4
Figure 2.5
Answers
Answer:
Answer:
Answer:
Answer: L=2m,
Answer: L=2m,d=3mm,A=
Answer: L=2m,d=3mm,A= 4
Answer: L=2m,d=3mm,A= 49π
Answer: L=2m,d=3mm,A= 49π
Answer: L=2m,d=3mm,A= 49π ×10
Answer: L=2m,d=3mm,A= 49π ×10 −6
Answer: L=2m,d=3mm,A= 49π ×10 −6 m
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2
Answer: 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= 4
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π
Answer: 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 −6
Answer: 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 11
Answer: 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 30×2
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2
Answer: 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 −5
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 .
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 .