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 step by step, we need to analyze the situation involving two point charges A and B, their initial charges, and the effect of transferring a portion of charge from A to B.

### Step 1: Identify the initial charges and force
Initially, we have:
- Charge of A, \( Q_A = +Q \)
- Charge of B, \( Q_B = -Q \)

The force \( F \) between the two charges can be calculated using Coulomb's law:
\[ F = k \frac{Q_A \cdot Q_B}{r^2} = k \frac{(+Q)(-Q)}{r^2} = -k \frac{Q^2}{r^2} \]
The magnitude of the force is:
\[ |F| = k \frac{Q^2}{r^2} \]

### Step 2: Transfer 25% of charge from A to B
We transfer 25% of charge from A to B. The amount of charge transferred is:
\[ \text{Charge transferred} = 0.25Q \]

After the transfer:
- New charge of A, \( Q_A' = Q - 0.25Q = 0.75Q = \frac{3}{4}Q \)
- New charge of B, \( Q_B' = -Q + 0.25Q = -0.75Q = -\frac{3}{4}Q \)

### Step 3: Calculate the new force between the charges
Now, we can calculate the new force \( F' \) between the charges using the new values:
\[ F' = k \frac{Q_A' \cdot Q_B'}{r^2} = k \frac{\left(\frac{3}{4}Q\right) \cdot \left(-\frac{3}{4}Q\right)}{r^2} \]
This simplifies to:
\[ F' = k \frac{-\frac{9}{16}Q^2}{r^2} = -k \frac{9Q^2}{16r^2} \]

### Step 4: Relate the new force to the initial force
We can express the new force in terms of the initial force \( F \):
\[ F' = \frac{9}{16} \left(-k \frac{Q^2}{r^2}\right) = \frac{9}{16} F \]

### Conclusion
Thus, the new force between the charges after transferring 25% of charge from A to B is:
\[ F' = \frac{9}{16} F \]

### Final Answer
The new force between the charges becomes \( \frac{9}{16} F \).

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