Question

The dipole moment of a molecule is \( 10^{-30} \, \text{Cm} \). It is placed in an electric field \( \vec{E} \) of \( 10^5 \, \text{V/m} \) such that its axis is along the electric field. The direction of \( \vec{E} \) is suddenly changed by \( 60^\circ \) at an instant. Find the change in the potential energy of the dipole, at that instant.

Hint:

The potential energy of a dipole in an electric field depends on the angle between the dipole moment and the electric field. The energy change is proportional to the cosine of the angle between them.

Answer 1

Step 1: The potential energy \( U \) of an electric dipole in an electric field \( \vec{E} \) is given by:
\[ U = -\vec{p} \cdot \vec{E} \] where \( \vec{p} \) is the dipole moment and \( \vec{E} \) is the electric field.
Step 2: Initially, the dipole moment \( \vec{p} \) is aligned with the electric field \( \vec{E} \), so the potential energy is:
\[ U_i = -p E \] Step 3: When the direction of the electric field is changed by \( 60^\circ \), the new potential energy becomes:
\[ U_f = -p E \cos 60^\circ = -p E \times \frac{1}{2} \] Step 4: The change in potential energy \( \Delta U \) is:
\[ \Delta U = U_f - U_i = -p E \times \frac{1}{2} - (-p E) = \frac{p E}{2} \] Step 5: Substituting \( p = 10^{-30} \, \text{Cm} \) and \( E = 10^5 \, \text{V/m} \):
\[ \Delta U = \frac{10^{-30} \times 10^5}{2} = 5 \times 10^{-26} \, \text{J} \] Thus, the change in potential energy is \( 5 \times 10^{-26} \, \text{J} \).