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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 $10^6\; Vm^{–1}$. The direction of the field is suddenly changed by an angle of $60^{\circ}$. 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.

1 Answer

Dipole moment of each molecules $= 10^{–29} C m $
As 1 mole of the substance contains $6 \times 10^{23}$ molecules,
Total dipole moment of all the molecules, $p = 6 × 10^{23} × 10^{–29}\; C m = 6 × 10^{–6} Cm$
Initial potential energy, $U_i = –pE \cos \theta = –6 × 10^{–6} \times 10^6 cos 0^{\circ} = –6 J$
Final potential energy (when $\theta = 60^{\circ}), U_f = –6 × 10^{–6} × 10^{6} \cos 60^{\circ} = –3 J$
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.
answered Jun 11, 2014 by meena.p
 

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