Question

The dipole moment of a molecule is 10⁻³⁰ Cm. It is placed in an electric field `vec E` of 10⁵ V/m such that its axis is along the electric field. The direction of `vec E` is suddenly changed by 60° at an instant. Find the change in the potential energy of the dipole, at that instant.

Answer 1

Given:

P = 10⁻³⁰ cm

E = 10⁵ V/m

To find:

Δu = ?

Formulae:

1. U = −pE cos θ
2. Δu = −PE(cos θ₂ − cos θ₁)

Calculations:

Initial Potential Energy (U₁):

U₁ ​= −pE cos(0°)

= −pE

= −(10⁻³⁰)(10⁵)

= −10⁻²⁵ J

Final Potential Energy (U₂​):

U₁ ​= −pE cos(60°)

= `-pE * 1/2`

= `-1/2(10^-30)(10^5)`

= −5 × 10⁻²⁵ J

Change in Potential Energy (Δu):

Δu = U₂ − U₁

= −5 × 10⁻²⁵ − (−10⁻²⁵)

= 5 × 10⁻²⁵ J