The force between two charges when placed in free space is 5 N.If they are in a medium of relative permittivity 5, the force between them will be

Options

A. `1N`
B. `25 N`
C. `2.5 N`
D. `1/5N`

Explanation

To solve the problem, we need to determine the new force between the two charges when they are placed in a medium with a relative permittivity of 5. We start with the force in free space and apply the relationship between the forces in different media.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Force in free space (F) = 5 N
- Relative permittivity of the medium (ε_r) = 5

2. **Understand the Relationship Between Forces:**
The force between two charges in a medium can be expressed in terms of the force in free space as follows:
\[ F' = \frac{F}{\epsilon_r} \]
where \( F' \) is the force in the medium and \( F \) is the force in free space.

3. **Substitute the Known Values:**
Using the formula derived above, substitute the values:
\[ F' = \frac{5 \, \text{N}}{5} \]

4. **Calculate the New Force:**
\[ F' = 1 \, \text{N} \]

5. **Conclusion:**
The force between the two charges when placed in a medium with a relative permittivity of 5 is **1 N**.