When the distance between two charged particles is halved, the force between them will become

Options

A. one-third
B. one-half
C. four times
D. five times

Explanation

To solve the problem of how the force between two charged particles changes when the distance between them is halved, we can use Coulomb's Law. Let's break it down step by step.

### Step-by-Step Solution:

1. **Understanding Coulomb's Law**:
Coulomb's Law states that the 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.

2. **Initial Force Calculation**:
Let the initial distance between the two charges be \( r \). The initial force \( F \) can be expressed as:
\[ F = k \frac{q_1 q_2}{r^2} \]

3. **Halving the Distance**:
When the distance between the two charges is halved, the new distance becomes:
\[ r' = \frac{r}{2} \]

4. **New Force Calculation**:
Now, we can calculate the new force \( F' \) when the distance is \( r' \):
\[ F' = k \frac{q_1 q_2}{(r')^2} = k \frac{q_1 q_2}{\left(\frac{r}{2}\right)^2} \]

5. **Simplifying the New Force**:
Simplifying the expression for \( F' \):
\[ F' = k \frac{q_1 q_2}{\frac{r^2}{4}} = k \frac{q_1 q_2 \cdot 4}{r^2} = 4 \left( k \frac{q_1 q_2}{r^2} \right) \]
Therefore, we have:
\[ F' = 4F \]

6. **Conclusion**:
Thus, when the distance between the two charged particles is halved, the force between them becomes four times greater than the initial force.

### Final Answer:
The force between the two charged particles will become **4 times** the original force.

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