Question

In a uniform electric field E, the work done in rotating the electric dipole of dipole moment P from $-45^{\circ}$ to $45^{\circ}$ with respect to the field direction is

• A. 0.5 PE
• B. 0.414 PE
• C. 0.866 PE
• D. PE
• E. zero

Hint:

Logic Tip: The potential energy of a dipole depends strictly on $\cos\theta$. Any rotation between two symmetric angles (like $-\theta$ to $+\theta$) results in the dipole returning to an identical potential energy state, meaning the net work done by the external agent over the complete path is exactly zero.

Answer 1

Answer: (E)

Concept:
The work done ($W$) in rotating an electric dipole with dipole moment $p$ in a uniform electric field $E$ from an initial angle $\theta_1$ to a final angle $\theta_2$ is equal to the change in its potential energy: $$W = \Delta U = U_f - U_i$$ Where the potential energy of a dipole in an electric field is $U = -pE \cos\theta$. Thus, $W = pE(\cos\theta_1 - \cos\theta_2)$.
Step 1: Identify the initial and final angles.
Initial angle, $\theta_1 = -45^{\circ}$ Final angle, $\theta_2 = 45^{\circ}$
Step 2: Substitute into the work formula.
$$W = PE(\cos(-45^{\circ}) - \cos(45^{\circ}))$$
Step 3: Evaluate the cosine values.
Recall that the cosine function is an even function, meaning $\cos(-\theta) = \cos(\theta)$. Therefore, $\cos(-45^{\circ}) = \cos(45^{\circ}) = \frac{1}{\sqrt{2}}$.
Step 4: Calculate the final work done.
$$W = PE \left( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \right)$$ $$W = PE (0)$$ $$W = 0$$