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 calculated using \( U = -\vec{p} \cdot \vec{E} \), where \( \vec{p} \) is the dipole moment and \( \vec{E} \) is the electric field.
Step 2: Initially, when the dipole moment \( \vec{p} \) aligns with the electric field \( \vec{E} \), the potential energy is \( U_i = -p E \).
Step 3: After the electric field's direction changes by \( 60^\circ \), the new potential energy \( U_f \) is \( U_f = -p E \cos 60^\circ \), which simplifies to \( U_f = -p E \times \frac{1}{2} \).
Step 4: The change in potential energy \( \Delta U \) is determined by \( \Delta U = U_f - U_i \). Substituting the values, we get \( \Delta U = -p E \times \frac{1}{2} - (-p E) = \frac{p E}{2} \).
Step 5: Given \( p = 10^{-30} \, \text{Cm} \) and \( E = 10^5 \, \text{V/m} \), the change in potential energy is \( \Delta U = \frac{10^{-30} \times 10^5}{2} = 5 \times 10^{-26} \, \text{J} \).
Therefore, the change in potential energy is \( 5 \times 10^{-26} \, \text{J} \).