Question

An electric dipole of dipole moment \(12\mu \text{ C m}\) is suspended in a uniform electric field of intensity \(10^6 \text{Vm}^{-1}\). The work done in rotating it from \(0^\circ\) to \(60^\circ\) with respect to the field direction is

• A. 3 J
• B. 6 J
• C. 1.5 J
• D. 12 J
• E. 2 J

Hint:

Always check the powers of 10. Here, \(\mu (10^{-6})\) and the field (\(10^6\)) cancel each other out perfectly, simplifying the calculation to just the dipole moment and the cosine difference.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When an electric dipole is placed in a uniform electric field, it experiences a torque. To rotate the dipole from one angular position to another against this field, work must be done. This work is stored as potential energy.

Step 2: Key Formula or Approach:
The work done in rotating an electric dipole from angle \(\theta_1\) to \(\theta_2\) is given by: \[ W = pE(\cos\theta_1 - \cos\theta_2) \] where \(p\) is the dipole moment and \(E\) is the electric field intensity.

Step 3: Detailed Explanation:
Given values:
\(p = 12 \mu\text{C m} = 12 \times 10^{-6} \text{ C m}\)
\(E = 10^6 \text{ Vm}^{-1}\)
Initial angle \(\theta_1 = 0^\circ\)
Final angle \(\theta_2 = 60^\circ\)
Substituting into the formula:
\[ W = (12 \times 10^{-6}) \times (10^6) \times (\cos 0^\circ - \cos 60^\circ) \]
\[ W = 12 \times (1 - 0.5) \]
\[ W = 12 \times 0.5 = 6 \text{ J} \]

Step 4: Final Answer:
The work done is 6 J.