Two charges `+3.2 xx 10^(-19)C and -3.2 xx 10^(-19)C` placed at `2.4A^(@)` apart form an electric dipole. It is placed in a uniform electric field of intensity `4 xx 10^(5)` volt/m. The electric dipole- moment is

Options

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

Answer

To find the electric dipole moment, we can follow these steps:

### Step 1: Identify the charges and the distance between them
We have two charges:
- \( q_1 = +3.2 \times 10^{-19} \, \text{C} \)
- \( q_2 = -3.2 \times 10^{-19} \, \text{C} \)

The distance between the charges is given as:
- \( d = 2.4 \, \text{Å} \)

### Step 2: Convert the distance from angstroms to meters
1 Å (angstrom) = \( 10^{-10} \, \text{m} \), so:
\[ d = 2.4 \, \text{Å} = 2.4 \times 10^{-10} \, \text{m} \]

### Step 3: Calculate the electric dipole moment
The electric dipole moment \( p \) is given by the formula:
\[ p = q \times d \]
where \( q \) is the magnitude of one of the charges and \( d \) is the distance between the charges.

Since we need to consider the distance between the two charges, we will use \( 2L \) where \( L \) is half the distance between the charges:
\[ L = \frac{d}{2} = \frac{2.4 \times 10^{-10}}{2} = 1.2 \times 10^{-10} \, \text{m} \]

Thus, the total distance \( 2L = 2 \times 1.2 \times 10^{-10} = 2.4 \times 10^{-10} \, \text{m} \).

Now, substituting the values into the dipole moment formula:
\[ p = q \times 2L = (3.2 \times 10^{-19}) \times (2.4 \times 10^{-10}) \]

### Step 4: Perform the multiplication
Calculating the product:
\[ p = 3.2 \times 2.4 \times 10^{-19} \times 10^{-10} = 7.68 \times 10^{-29} \, \text{C m} \]

### Final Answer
The electric dipole moment is:
\[ p = 7.68 \times 10^{-29} \, \text{C m} \]