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A molecule of a substance has a permanent electric dipole moment of magnitude 10^(-29) Cm. 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. Assume 100% polarisation of the sample. |
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Answer» Solution :Here, dipole moment of each molecules `= 10^(-29)Cm` 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) Cm = 6 xx 10^(6)Cm` Initial POTENTIAL energy, `U= - pE cos THETA =-6 xx 10^(-6) xx 10^(6) cos 0^@ =- 6 J` Final potential energy `("when "theta= 60^@), U_f= - 6 xx 10^(-6) xx 10^6 cos 60^@= -3 J` Change in potential energy = -3J -(-6J) =3J 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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