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

The potential energy (U) of a dipole in an electric field is given by:

`U = -vecp xx vecE = -pE costheta`

where:

• p is the dipole moment,

• E is the magnitude of the electric field,

• θ is the angle between the dipole moment and the electric field.

Potential Energy Change

Initial potential energy (U₁​):

U₁ ​= −pEcos(0^∘) = −pE ⋅ 1 = −(10⁻³⁰) (10⁵) = −10⁻²⁵ J

`U_2 = -pE cos(60°) = -pE(1/2) = -10^-25/2 = -5xx10^-26 J`

Change in Potential Energy (ΔU)

ΔU = U₂ ​− U₁

​ΔU = −5 × 10⁻²⁶ − (−10⁻²⁵)

ΔU = −5 × 10⁻²⁶ + 10⁻²⁵ = 5 × 10⁻²⁶ J

ΔU = 5 × 10⁻²⁶ J