An electric dipole of moment `vecp` is placed normal to the lines of force of electric intensity `vecE`, then the work done in deflecting it through an angle of `180^(@)` is

Options

A. `pE`
B. `+2 pE`
C. `-2 pE`
D. zero

Answer

To solve the problem of finding the work done in deflecting an electric dipole through an angle of 180 degrees when it is initially placed normal to the electric field, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Configuration**:
- An electric dipole is represented by a dipole moment vector \( \vec{p} \).
- The electric field vector \( \vec{E} \) is given, and the dipole is initially placed perpendicular (normal) to the electric field.

2. **Initial Position**:
- Since the dipole is normal to the electric field, the angle \( \theta \) between \( \vec{p} \) and \( \vec{E} \) is \( 90^\circ \).
- The potential energy \( U \) of a dipole in an electric field is given by the formula:
\[ U = -\vec{p} \cdot \vec{E} = -pE \cos(\theta) \]
- Substituting \( \theta = 90^\circ \):
\[ U_{\text{initial}} = -pE \cos(90^\circ) = -pE \cdot 0 = 0 \]

3. **Final Position**:
- The dipole is rotated through an angle of \( 180^\circ \). After this rotation, the dipole is still perpendicular to the electric field, but now it points in the opposite direction.
- The angle \( \theta \) remains \( 90^\circ \) even after the rotation.
- Therefore, the potential energy in the final position is:
\[ U_{\text{final}} = -pE \cos(90^\circ) = -pE \cdot 0 = 0 \]

4. **Calculate Work Done**:
- The work done \( W \) in deflecting the dipole is equal to the change in potential energy:
\[ W = U_{\text{final}} - U_{\text{initial}} = 0 - 0 = 0 \]

5. **Conclusion**:
- The work done in deflecting the dipole through an angle of \( 180^\circ \) is \( 0 \).

### Final Answer:
The work done in deflecting the dipole through an angle of \( 180^\circ \) is \( 0 \).

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