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A molecule of a substance has a permanent electric dipole moment of magnitude `10^(-29) `C m. A mole of this substance is polarized at low temperature by appling a strong elecrostatic 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 dipole along the new direction of the field. For simplicity, assume `100%` polarisation of sample.

Answer» Here, dipole moment of each molecules` = 10^(-29) C m` As 1 mole of the substance contains `6 xx 10^(23)` molecules, total dipole moment of all the molecules, `p = 6 xx 10^(23) xx 10^(-29) C m`
`= 6 xx 10^(-6) C m`
Initial potential energy,`U_(i) = -pE cos theta = -6 xx 10^(-6) xx 10^(6) cos 0^(@) = -6J`
Final potential energy (when `theta = 60^(@)`), `U_(f) = –6 xx 10^(-6) × 10^(6) cos 60^(@) = -3 J`
Change in potential energy `= -3 J - (-6J) = 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.


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