An electric dipole of moment `vec(P)` is lying along a uniform electric field `vec(E )`. The work done in rotating the dipole by `90^(@)` is:

Options

A. `sqrt(2)pE`
B. `(pE)/(2)`
C. `2pE`
D. `pE`

Answer

To find the work done in rotating an electric dipole by \(90^\circ\) in a uniform electric field, we can follow these steps:

### Step 1: Understand the initial and final positions of the dipole
- The dipole moment \( \vec{P} \) is initially aligned with the electric field \( \vec{E} \). This means the angle \( \theta_1 = 0^\circ \).
- After rotation, the dipole makes an angle \( \theta_2 = 90^\circ \) with the electric field.

### Step 2: Write the expression for torque
- The torque \( \tau \) experienced by the dipole in the electric field is given by:
\[ \tau = \vec{P} \times \vec{E} = P E \sin \theta \]
where \( P \) is the magnitude of the dipole moment and \( E \) is the magnitude of the electric field.

### Step 3: Set up the work done integral
- The work done \( W \) in rotating the dipole from angle \( \theta_1 \) to \( \theta_2 \) is given by:
\[ W = \int_{\theta_1}^{\theta_2} \tau \, d\theta \]
Substituting the expression for torque, we have:
\[ W = \int_{0}^{90^\circ} P E \sin \theta \, d\theta \]

### Step 4: Factor out constants from the integral
- Since \( P \) and \( E \) are constants, we can factor them out of the integral:
\[ W = P E \int_{0}^{90^\circ} \sin \theta \, d\theta \]

### Step 5: Evaluate the integral
- The integral of \( \sin \theta \) is:
\[ \int \sin \theta \, d\theta = -\cos \theta \]
- Therefore, we evaluate:
\[ W = P E \left[-\cos \theta \right]_{0}^{90^\circ} \]
This results in:
\[ W = P E \left[-\cos(90^\circ) + \cos(0^\circ)\right] \]
Knowing that \( \cos(90^\circ) = 0 \) and \( \cos(0^\circ) = 1 \):
\[ W = P E \left[0 + 1\right] = P E \]

### Conclusion
- The work done in rotating the dipole by \(90^\circ\) is:
\[ W = P E \]