A molecule of a substance has a permanent electric moment of magnitude `10^(-29)Cm`. A mole of this substnace is polarised by applying a strong electrostatic field of magnitude `10^(6)Vm^(-1)`. The direction of the field is suddenly changed by a angle of `60^(@)`. estimate the heat released by the substance in alighning its dipoles along the new direction of the field. for simplicity, assume 100% polarisation of the sample.

Explanation

Given, dipoe moment of each molecule=`10^(-29)Cm`
Total dipole moment of 1 mole of the substance `=6.023xx10^(23)xx10^(-29)Cm`
`=6.023xx10^(-6)Cm`
Initial potential energy, `U_(i)=-pE costheta=-6.023xx10^(-6)xx10^(6)xxcos0^(@)=-6.023J`
Final P.E. when, `q=60^(@),U_(f)=-pE cosQ=-6.023xx10^(-6)xx10^(6)xxcos60^(@)`
`=-3.0125J`
Change in potential energy=`U_(f)-U_(i)`
i.e., `DeltaP.E.=[-3.0125-(-6.023)]J=3.0125J`
This energy loss by the substance is in the form of heat.