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

Given that,

Dipole moment of a water $p=6.3\times10^{-30}\ C-m$

Molecules of water  = 10¹² molecules

Electric field strength $E=2.5\times10^{5}\ N/C$

We know that,

The torque is

$\tau=Ep\sin\theta$

We need to calculate the work done

Using formula of torque

$W=\int_{0}^{90}{\tau d\theta}$

Put the value of torque

$W=\int_{0}^{90}{(Ep\sin\theta) d\theta}$

$|W|=Ep(1-\cos90)$

$|W|=Ep$

Where, p = dipole moment

E = electric  field of strength

Put the value into the formula

$W=2.5\times10^{5}\times6.3\times10^{-30}$

$W=1.575\times10^{-26}\ N-m$

Hence, The work done is $1.575\times10^{-26}\ N-m$