Question

The water (H2O) molecule has an electric dipole moment of 6.3 x 10^-30 C · m. A sample contains 10^21 water molecules, with the dipole moments all oriented in the direction of an electric field of magnitude 2.5 x 10^5 N/C. How much work is required to rotate the dipoles from this orientation (θ = 0°) to one in which all the moments are perpendicular to the field (θ = 90°)?

Name: the water h2o molecule has an electric dipole moment of 63 x 10 30 c m a sample contains 1021 water molecules with the dipole moments all oriented in the direction of an electric field of ma 25687
Uploaded: 2022-07-30T07:24:29-08:00
Duration: 2 min 40 s
Channel: Adi S
Description: the water h2o molecule has an electric dipole moment of 63 x 10 30 c m a sample contains 1021 water molecules with the dipole moments all oriented in the direction of an electric field of ma 25687

          The water (H2O) molecule has an electric dipole moment of 6.3 x 10^-30 C · m. A sample contains 10^21 water molecules, with the dipole moments all oriented in the direction of an electric field of magnitude 2.5 x 10^5 N/C. How much work is required to rotate the dipoles from this orientation (θ = 0°) to one in which all the moments are perpendicular to the field (θ = 90°)?

Added by Jason H.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

07/30/2022

Step 1

Given: Number of water molecules, n = 10^21 Dipole moment of a single water molecule, p = 6.3 x 10^-30 C⋅m Total dipole moment, P = n * p P = (10^21) * (6.3 x 10^-30) P = 6.3 x 10^-9 C⋅m Show more…

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The water (H2O) molecule has an electric dipole moment of 6.3 x 10^-30 C · m. A sample contains 10^21 water molecules, with the dipole moments all oriented in the direction of an electric field of magnitude 2.5 x 10^5 N/C. How much work is required to rotate the dipoles from this orientation (θ = 0°) to one in which all the moments are perpendicular to the field (θ = 90°)?