The dipole moment of a water molecule
is 6.3 × 10–30 Cm. A sample of water
contains 1021 molecules, whose dipole
moments are all oriented in an electric
field of strength 2.5 × 105
N /C. Calculate
the work to be done to rotate the dipoles
from their initial orientation θ1 = 0 to one
in which all the dipoles are perpendicular
to the field, θ2
= 90
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 .