Example

Question:

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 \(60^\circ\). 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.

Solution:

Dipole moment of each molecule = \(10^{-29}\) C m.
Number of molecules in one mole = \(6 \times 10^{23}\).
Total dipole moment, \(p = 6 \times 10^{23} \times 10^{-29}\) C m = \(6 \times 10^{-6}\) C m.
Initial potential energy, \[ U_i = -pE \cos\theta = -6 \times 10^{-6} \times 10^6 \cos 0^\circ = -6\,\text{J} \] Final potential energy (when \(\theta = 60^\circ\)), \[ U_f = -6 \times 10^{-6} \times 10^6 \cos 60^\circ = -3\,\text{J} \] Change in potential energy, \[ \Delta U = U_f - U_i = -3\,\text{J} - (-6\,\text{J}) = 3\,\text{J} \] Therefore, there is loss in potential energy, and this is the energy released as heat by the substance when aligning its dipoles.