Two particles having charge `Q_(1)` and `Q_(2)` , when kept at a certain distance exert a force `F` on each other . If the distance between the two particles is reduced to half and the charge on each particle is doubled , the force between the particles would be

Options

A. `2 F`
B. `4 F`
C. `8 F`
D. `16 F`

Explanation

To solve the problem, we will use Coulomb's law, which states that the electrostatic force \( F \) between two point charges \( Q_1 \) and \( Q_2 \) separated by a distance \( r \) is given by:

\[ F = k \frac{Q_1 Q_2}{r^2} \]

where \( k \) is Coulomb's constant.

### Step-by-Step Solution:

1. **Initial Force Calculation**:
- The initial force \( F \) between the two charges \( Q_1 \) and \( Q_2 \) separated by a distance \( r \) is given by:
\[ F = k \frac{Q_1 Q_2}{r^2} \]

2. **Change in Distance and Charge**:
- According to the problem, the distance between the two charges is reduced to half, so the new distance \( r' \) is:
\[ r' = \frac{r}{2} \]
- The charges are doubled, so the new charges \( Q_1' \) and \( Q_2' \) are:
\[ Q_1' = 2Q_1 \quad \text{and} \quad Q_2' = 2Q_2 \]

3. **New Force Calculation**:
- We need to calculate the new force \( F' \) with the new charges and new distance:
\[ F' = k \frac{Q_1' Q_2'}{(r')^2} \]
- Substituting the new values:
\[ F' = k \frac{(2Q_1)(2Q_2)}{(\frac{r}{2})^2} \]

4. **Simplifying the Expression**:
- Simplifying the denominator:
\[ (\frac{r}{2})^2 = \frac{r^2}{4} \]
- Therefore, the new force becomes:
\[ F' = k \frac{4Q_1 Q_2}{\frac{r^2}{4}} = k \frac{4Q_1 Q_2 \cdot 4}{r^2} = k \frac{16Q_1 Q_2}{r^2} \]

5. **Relating New Force to Initial Force**:
- We can relate the new force \( F' \) to the initial force \( F \):
\[ F' = 16 \left( k \frac{Q_1 Q_2}{r^2} \right) = 16F \]

### Conclusion:
Thus, the new force \( F' \) between the two particles after reducing the distance to half and doubling the charges is:

\[ F' = 16F \]