Question

Given the dipole moment $ p $ and the electric field $ E $, find the work done to move the dipole from a parallel orientation to an antiparallel orientation with respect to the electric field.

• A. $ pE $
• B. $ -pE $
• C. $ 2pE $
• D. $ -2pE $

Hint:

The work done is negative because the system loses potential energy as the dipole moves from a lower-energy state (parallel) to a higher-energy state (antiparallel).

Answer 1

The Correct Option is D

Step 1: Recall the formula for work done on a dipole
The work done (\( W \)) in rotating a dipole from one orientation to another in an electric field is given by:
\[ W = -\Delta U = - (U_{\text{final}} - U_{\text{initial}}) \] where the potential energy of a dipole in an electric field is:
\[ U = -\vec{p} \cdot \vec{E} = - p E \cos \theta \] Here:
\(\vec{p}\) is the dipole moment,
\(\vec{E}\) is the electric field,
\(\theta\) is the angle between \(\vec{p}\) and \(\vec{E}\).
Step 2: Analyze the orientations
1. Initial orientation: Parallel to the electric field (\(\theta_i = 0^\circ\)).
Potential energy:
\[ U_{\text{initial}} = - p E \cos(0^\circ) = - p E \times 1 = - p E \] 2. Final orientation: Antiparallel to the electric field (\(\theta_f = 180^\circ\)).
Potential energy:
\[ U_{\text{final}} = - p E \cos(180^\circ) = - p E \times (-1) = + p E \] Step 3: Calculate the change in potential energy
\[ \Delta U = U_{\text{final}} - U_{\text{initial}} = (+ p E) - (- p E) = p E + p E = 2 p E \] Step 4: Work done
The work done (\(W\)) is:
\[ W = - \Delta U = - (2 p E) = - 2 p E \] Step 5: Conclusion
The work done to move the dipole from parallel to antiparallel to the electric field is:
\[ {(D) -2 p E} \]