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

Answer

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

\[ F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2} \]

### Step-by-Step Solution:

1. **Initial Setup**:
- Let the initial charges be \( Q_1 \) and \( Q_2 \).
- The initial distance between the charges is \( r \).
- The initial force between the charges is given as \( F \).

\[ F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2} \tag{1} \]

2. **New Conditions**:
- According to the problem, the distance is reduced to half, so the new distance \( r' = \frac{r}{2} \).
- The charges are doubled, so the new charges are \( Q_1' = 2Q_1 \) and \( Q_2' = 2Q_2 \).

3. **Calculating the New Force**:
- The new force \( F' \) can be calculated using the new charges and the new distance:

\[ F' = \frac{1}{4 \pi \epsilon_0} \frac{Q_1' Q_2'}{(r')^2} \]

Substituting the new values:

\[ F' = \frac{1}{4 \pi \epsilon_0} \frac{(2Q_1)(2Q_2)}{(\frac{r}{2})^2} \]

Simplifying the denominator:

\[ F' = \frac{1}{4 \pi \epsilon_0} \frac{4Q_1 Q_2}{\frac{r^2}{4}} = \frac{1}{4 \pi \epsilon_0} \frac{4Q_1 Q_2 \cdot 4}{r^2} \]

This simplifies to:

\[ F' = \frac{16}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2} \]

4. **Relating New Force to Initial Force**:
- From equation (1), we know that:

\[ F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2} \]

Therefore, we can express \( F' \) in terms of \( F \):

\[ F' = 16 \cdot F \]

5. **Conclusion**:
- The new force \( F' \) is \( 16F \).

### Final Answer:
The force between the particles after the changes is \( 16F \).

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