The dipole moment of a water molecule is 6.3 x 10-30 c·m. A sample of water contains 1021 molecules, whose dipole moments are all oriented in an electric field of strength 2.5 x 105N/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°.

p = 6.3 x 10-30 C.m. N = 1021 molecules, E = 2.5 x 105N/C, θ0 = θ1 = 0°, θ= θ2 = 90° W = pE(cos θ0 - cos θ) The total work required to orient N dipoles is W = NpE (cos θ1 - cos θ2) = (1021)(6.3 × 10-30)(2.5 × 105) = 15.75 × 10-4 J = 1.575 mJ

The dipole moment of a water molecule is 6.3 x 10-30 c·m. A sample of water contains 1021 molecules, whose dipole moments are all oriented in an electric field of strength 2.5 x 105N/C. - Physics

Sum

The dipole moment of a water molecule is 6.3 x 10^-30 c·m. A sample of water contains 10²¹ molecules, whose dipole moments are all oriented in an electric field of strength 2.5 x 10⁵N/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°.

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