# 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°.

#### Solution

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

Concept: Electrical Energy of Two Point Charges and of a Dipole in an Electrostatic Field
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#### APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 8 Electrostatics
Exercises | Q 10 | Page 213