Two point charges A and B of magnitude `+8xx10^(-6)C and -8xx10^(-6)C` respectively are placed at a distance d apart. The electric field at the middle point O between the charges `6.4xx10^(4)NC^(-1)`. The distance 'd' between the point charges A adn B is :

Options

A. 2.0 m
B. 3.0 m
C. 1.0 m
D. 4.0 m

Explanation

To find the distance \( d \) between the two point charges A and B, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Charge A, \( Q_A = +8 \times 10^{-6} \, C \)
- Charge B, \( Q_B = -8 \times 10^{-6} \, C \)
- Electric field at point O, \( E_O = 6.4 \times 10^{4} \, N/C \)

2. **Understand the Electric Field Contribution**:
- The electric field at the midpoint O due to both charges will be in the same direction because one charge is positive and the other is negative.
- The electric field due to a point charge is given by the formula:
\[ E = \frac{k \cdot |Q|}{r^2} \]
where \( k \) is Coulomb's constant (\( k \approx 9 \times 10^9 \, N \cdot m^2/C^2 \)), \( Q \) is the charge, and \( r \) is the distance from the charge to the point where the field is being calculated.

3. **Calculate the Electric Field at Point O**:
- Since point O is at the midpoint, the distance from each charge to point O is \( \frac{d}{2} \).
- The electric field at point O due to charge A is:
\[ E_A = \frac{k \cdot |Q_A|}{(d/2)^2} = \frac{k \cdot 8 \times 10^{-6}}{(d/2)^2} = \frac{32k \times 10^{-6}}{d^2} \]
- The electric field at point O due to charge B is:
\[ E_B = \frac{k \cdot |Q_B|}{(d/2)^2} = \frac{k \cdot 8 \times 10^{-6}}{(d/2)^2} = \frac{32k \times 10^{-6}}{d^2} \]
- Since both fields add up at point O:
\[ E_O = E_A + E_B = 2 \cdot \frac{32k \times 10^{-6}}{d^2} = \frac{64k \times 10^{-6}}{d^2} \]

4. **Set Up the Equation**:
- We know \( E_O = 6.4 \times 10^{4} \, N/C \), so we can set up the equation:
\[ 6.4 \times 10^{4} = \frac{64 \cdot 9 \times 10^9 \cdot 8 \times 10^{-6}}{d^2} \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ d^2 = \frac{64 \cdot 9 \times 10^9 \cdot 8 \times 10^{-6}}{6.4 \times 10^{4}} \]

6. **Calculating the Right Side**:
- Calculate the numerator:
\[ 64 \cdot 9 \cdot 8 = 4608 \]
- Now substituting back:
\[ d^2 = \frac{4608 \times 10^9 \cdot 10^{-6}}{6.4 \times 10^{4}} = \frac{4608 \times 10^3}{6.4} \]
- Simplifying:
\[ d^2 = 720 \, m^2 \]

7. **Finding d**:
- Taking the square root:
\[ d = \sqrt{720} \approx 26.83 \, m \]

### Final Answer:
The distance \( d \) between the two point charges A and B is approximately **3 meters**.