Question

A molecule of a substance has a permanent electric dipole moment of magnitude 10^-29 C m. A mole of thus substance is polarised by applying a strong electrostatic field of magnitude 10⁶ V/m. 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. For simplicity, assume 100% polarisation of the sample.​

Answer 1

AnswEr:

• $\bf{3\:joules}$

ExplanaTion:

• Here , dipole moment of each molecules = $\sf{10^{-29}C\:m}$

As 1 mole of the substance contains $\sf{6\times 10^{23}}$ molecules.

Total dipole moment of all the molecules :

$\longrightarrow{\sf{ p=6\times 10^{23}\times 10^{-29} C\:m}}$

$\longrightarrow{\sf{6\times 10^{-6} C\:m}}$

Initial potential energy,

U1= -pEcos∅

→ -6×10-⁶ ×10⁶ cos0°

→ -6 J

Final potential energy (when ∅= 60°)

Uf = -6×10-⁶ ×10 cos 60°

→ -3 J

Change in potential energy:

→-3 J-(-6 J)

→ 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.

Answer 2

Dipole moment of molecule of substance = 10^–29 C-m

| E | applied = 106 Vm^–1

Change of angle of electric field = 60º

No of molecules in one mole = n = 6.023 × 10^23

Amount of heat released in aligning the dipoles along new direction.

= + ΔU = ΔWext

= + (Uf - Ui)

= n [ -PEcos60° - (-PE)]

= n [ -PE/2 + PE] = n PE/2

= ( 6.023 *10^23 * 10^-29 * 10^6) / 2

= 6.023 / 2 ≈ 3.0115 J