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

Data: p = 6.3 × 10^-30 C∙m, N = 10²¹ molecules,
E = 2.5 × 10⁵ N/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