Two point charges A and B, having charges +Q and -Q respectively, are placed at certain distance apart and force acting between them is F, if `25%` charge of A is transferred to B, then force between the charges becomes:

Options

A. `(4F)/(3)`
B. F
C. `(9F)/(16)`
D. `(16F)/(9)`

Explanation

To solve the problem, we need to analyze the forces between two point charges before and after transferring a portion of charge from one to the other.

### Step-by-Step Solution:

1. **Identify the Initial Charges:**
- Charge A: \( +Q \)
- Charge B: \( -Q \)

2. **Calculate the Initial Force (F):**
Using Coulomb's Law, the force \( F \) between two point charges is given by:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{Q \cdot (-Q)}{R^2} = \frac{-Q^2}{4 \pi \epsilon_0 R^2} \]
Since we are interested in the magnitude, we can write:
\[ F = \frac{Q^2}{4 \pi \epsilon_0 R^2} \]

3. **Transfer 25% of Charge from A to B:**
- Amount transferred from A to B:
\[ \text{Charge transferred} = 0.25Q = \frac{Q}{4} \]
- New charge on A:
\[ Q_A' = Q - \frac{Q}{4} = \frac{3Q}{4} \]
- New charge on B:
\[ Q_B' = -Q + \frac{Q}{4} = -\frac{4Q}{4} + \frac{Q}{4} = -\frac{3Q}{4} \]

4. **Calculate the New Force (F'):**
Now, we calculate the new force \( F' \) between the new charges \( Q_A' \) and \( Q_B' \):
\[ F' = \frac{1}{4 \pi \epsilon_0} \frac{\left(\frac{3Q}{4}\right) \cdot \left(-\frac{3Q}{4}\right)}{R^2} \]
Simplifying this:
\[ F' = \frac{1}{4 \pi \epsilon_0} \frac{-\frac{9Q^2}{16}}{R^2} = \frac{9Q^2}{16 \cdot 4 \pi \epsilon_0 R^2} \]

5. **Relate F' to F:**
Now, we relate the new force \( F' \) to the original force \( F \):
\[ F' = \frac{9}{16} F \]

### Final Answer:
The new force between the charges after transferring 25% of charge from A to B becomes:
\[ F' = \frac{9}{16} F \]