A molecule of a substance has a permanent electric dipole moment of magnitude `10^(-29)` C m. A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude `10^(6) V m^(-1)` . The direction of the field is suddenly changed by an angle of `60^(@)` . Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100% polarisation of the sample.

Explanation

Here dipole moment of each molecules,
`= 10^(-29) cm `
Total dipole moment P of one mole molecules
`= 10^(-29) xx N_(A)`
`= 10^(-29) xx6 xx10^(23)`
`= 6xx10^(-6) C m`

Electrostatic field E `=10^(6) V//m theta_(2) = 60^(@) theta_(1)=0^(@)`
Intial potential energy ,
`U_(i) =-6 J [ because cos 0^(@) =1]`
Final potential energy
`U_(f) = -pE cos theta_(2)`
`:. U_(f) =-6xx10^(-6) xx10^(6) xx cos 60^(@)`
= - 3 J
`:.` Change in potential energy
`Delta U = U_(f) - U_(i)`
`= -3 -(-6) `
= 3 J
So, there is loss in potential energy. This must be the energy released by the substance in the form of heat in aligning its dipoles.