Question

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​

Answer 1

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 .