Find the amount of work done in rotating an electric dipole of dipole moment `3.2 xx 10^(- 8)`Cm from its position of stable equilibrium to the position of unstable equilibrium in a uniform electric field if intensity `10^(4) N// C`.

Answer

To find the amount of work done in rotating an electric dipole from its position of stable equilibrium to the position of unstable equilibrium in a uniform electric field, we can follow these steps:

### Step 1: Understand the Positions of Equilibrium
- The stable equilibrium position of an electric dipole is when it is aligned with the electric field, which corresponds to an angle \( \theta_1 = 0^\circ \).
- The unstable equilibrium position is when the dipole is aligned opposite to the electric field, corresponding to an angle \( \theta_2 = 180^\circ \).

### Step 2: Identify Given Data
- Dipole moment, \( p = 3.2 \times 10^{-8} \, \text{Cm} \)
- Electric field intensity, \( E = 10^4 \, \text{N/C} \)

### Step 3: Use the Work Done Formula
The work done \( W \) in rotating the dipole in an electric field is given by the formula:
\[ W = pE \left( \cos \theta_1 - \cos \theta_2 \right) \]
Where:
- \( p \) is the dipole moment,
- \( E \) is the electric field intensity,
- \( \theta_1 \) is the initial angle (0 degrees),
- \( \theta_2 \) is the final angle (180 degrees).

### Step 4: Calculate Cosine Values
- \( \cos(0^\circ) = 1 \)
- \( \cos(180^\circ) = -1 \)

### Step 5: Substitute Values into the Formula
Substituting the values into the work done formula:
\[ W = (3.2 \times 10^{-8} \, \text{Cm}) \times (10^4 \, \text{N/C}) \left( \cos(0^\circ) - \cos(180^\circ) \right) \]
\[ W = (3.2 \times 10^{-8}) \times (10^4) \left( 1 - (-1) \right) \]
\[ W = (3.2 \times 10^{-8}) \times (10^4) \times 2 \]

### Step 6: Calculate the Final Value
Calculating the above expression:
\[ W = (3.2 \times 10^{-8}) \times (10^4) \times 2 = 6.4 \times 10^{-4} \, \text{J} \]

### Final Answer
The amount of work done in rotating the electric dipole from stable equilibrium to unstable equilibrium is:
\[ W = 6.4 \times 10^{-4} \, \text{J} \]