Two charges `+3.2xx10^(-19)` and `-3.2xx10^(-19)C` placed at `2.4Å` apart from an electric dipole. It is placed in a uniform electric field of intensity `4xx10^(5) "volt"//m`. The electric dipole moment is

Options

A. `15.36 xx 10^(-29) "coulomb" xx m`
B. `15.36xx10^(-19) "coulomb" xx m`
C. `7.68 xx 10^(-29) "coulomb" xx m`
D. `7.68xx10^(-19) "coulomb" xx m`

Explanation

To find the electric dipole moment, we can use the formula:

\[ p = Q \times d \]

where:
- \( p \) is the electric dipole moment,
- \( Q \) is the magnitude of one of the charges,
- \( d \) is the distance between the two charges.

### Step-by-Step Solution:

1. **Identify the Charges**:
The charges given are:
- \( Q_1 = +3.2 \times 10^{-19} \, C \)
- \( Q_2 = -3.2 \times 10^{-19} \, C \)

Here, we can take the magnitude of the charge \( Q = 3.2 \times 10^{-19} \, C \).

2. **Identify the Distance**:
The distance between the two charges is given as:
- \( d = 2.4 \, \text{Å} = 2.4 \times 10^{-10} \, m \)

3. **Calculate the Electric Dipole Moment**:
Using the formula for the dipole moment:
\[ p = Q \times d \]
Substitute the values:
\[ p = (3.2 \times 10^{-19} \, C) \times (2.4 \times 10^{-10} \, m) \]

4. **Perform the Multiplication**:
\[ p = 3.2 \times 2.4 \times 10^{-19} \times 10^{-10} \]
Calculate \( 3.2 \times 2.4 = 7.68 \):
\[ p = 7.68 \times 10^{-29} \, C \cdot m \]

5. **Final Result**:
The electric dipole moment is:
\[ p = 7.68 \times 10^{-29} \, C \cdot m \]