Two charges `+3.2 xx 10^(-19) C` and `-3.2 xx 10^(-9) C` kept `2.4 Å` apart forms a dipole. If it is kept in uniform electric field of intensity `4 xx 10^(5) vol//m` then what will be its electrical energy in equilibrium ?

Options

A. `+3 xx 10^(-23) J`
B. `-3 xx 10^(-23) J`
C. `-6 xx 10^(-23) J`
D. `-2 xx 10^(-23)J`

Explanation

To find the electrical energy of a dipole in a uniform electric field at equilibrium, we can follow these steps:

### Step 1: Identify the given values
- Charge \( q_1 = +3.2 \times 10^{-19} \, C \)
- Charge \( q_2 = -3.2 \times 10^{-9} \, C \)
- Distance between the charges \( d = 2.4 \, \text{Å} = 2.4 \times 10^{-10} \, m \)
- Electric field intensity \( E = 4 \times 10^{5} \, \text{V/m} \)

### Step 2: Calculate the dipole moment \( p \)
The dipole moment \( p \) is given by the formula:
\[ p = q \cdot d \]
Where \( q \) is the magnitude of one of the charges (since they are equal and opposite, we can take either) and \( d \) is the separation distance.

Using \( q = 3.2 \times 10^{-19} \, C \) and \( d = 2.4 \times 10^{-10} \, m \):
\[ p = (3.2 \times 10^{-19} \, C) \cdot (2.4 \times 10^{-10} \, m) = 7.68 \times 10^{-29} \, C \cdot m \]

### Step 3: Calculate the electrical energy \( U \) in the electric field
The electrical energy \( U \) of a dipole in a uniform electric field is given by the formula:
\[ U = -\vec{p} \cdot \vec{E} \]
In equilibrium, the angle \( \theta \) between the dipole moment and the electric field is \( 0^\circ \), thus:
\[ U = -pE \cos(0) = -pE \]
Substituting the values:
\[ U = - (7.68 \times 10^{-29} \, C \cdot m) \cdot (4 \times 10^{5} \, V/m) \]
Calculating this gives:
\[ U = - (7.68 \times 10^{-29} \cdot 4 \times 10^{5}) = -3.072 \times 10^{-23} \, J \]

### Final Answer
The electrical energy of the dipole in equilibrium is approximately:
\[ U \approx -3.072 \times 10^{-23} \, J \]