Two charges `+3.2xx10^(-19)C and -3.2xx10^(-19)C` placed 2.4`A^(0)` apart form an electric dipole. It is placed in a uniform electric field of intensity `4xx10^(5)V//m` the work done to rotate the electric dipole form the equilibrium position by `180^(@)` is

Options

A. `3xx10^(-23)J`
B. `6xx10^(-23)J`
C. `12xx10^(-23)J`
D. zero

Explanation

To solve the problem of calculating the work done to rotate an electric dipole from its equilibrium position by 180 degrees in a uniform electric field, we can follow these steps:

### Step 1: Understand the electric dipole moment
The electric dipole moment \( \mathbf{p} \) is defined as:
\[ \mathbf{p} = q \cdot \mathbf{d} \]
where \( q \) is the magnitude of one of the charges and \( \mathbf{d} \) is the separation vector between the charges. In this case, the charges are \( +3.2 \times 10^{-19} \, \text{C} \) and \( -3.2 \times 10^{-19} \, \text{C} \), and the distance \( d = 2.4 \, \text{Å} = 2.4 \times 10^{-10} \, \text{m} \).

### Step 2: Calculate the dipole moment
Using the formula for the dipole moment:
\[ \mathbf{p} = q \cdot d = (3.2 \times 10^{-19} \, \text{C}) \cdot (2.4 \times 10^{-10} \, \text{m}) \]
Calculating this gives:
\[ \mathbf{p} = 7.68 \times 10^{-29} \, \text{C m} \]

### Step 3: Determine the work done to rotate the dipole
The work done \( W \) to rotate the dipole from an angle \( \theta_1 \) to \( \theta_2 \) in an electric field \( \mathbf{E} \) is given by:
\[ W = -\mathbf{p} \cdot \mathbf{E} \cdot (\cos \theta_2 - \cos \theta_1) \]
In this case, \( \theta_1 = 0^\circ \) (equilibrium position) and \( \theta_2 = 180^\circ \). Therefore:
\[ \cos(0^\circ) = 1 \quad \text{and} \quad \cos(180^\circ) = -1 \]
Substituting these values into the work done formula:
\[ W = -\mathbf{p} \cdot \mathbf{E} \cdot (-1 - 1) = 2 \mathbf{p} \cdot \mathbf{E} \]

### Step 4: Substitute the values
Given that the electric field \( E = 4 \times 10^5 \, \text{V/m} \), we can substitute the values:
\[ W = 2 \cdot (7.68 \times 10^{-29} \, \text{C m}) \cdot (4 \times 10^5 \, \text{V/m}) \]
Calculating this gives:
\[ W = 2 \cdot 7.68 \times 10^{-29} \cdot 4 \times 10^5 = 61.44 \times 10^{-24} \, \text{J} \]
or
\[ W = 6.144 \times 10^{-23} \, \text{J} \]

### Conclusion
The work done to rotate the electric dipole from the equilibrium position by 180 degrees is:
\[ W = 6.144 \times 10^{-23} \, \text{J} \]