Two points charges `q_1 and q_2` are placed at a distance of 50 m from each each other in air, and interact with a certain force. The same charges are now put in oil whose relative permittivity is 5. If the interacting force between them is still the same, their separation now is

Options

A. `16.6m`
B. `22.3m`
C. `28.4m`
D. `25.0m`

Explanation

To solve the problem step by step, we will use Coulomb's law and the concept of relative permittivity.

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- Two point charges \( q_1 \) and \( q_2 \) are separated by a distance \( r = 50 \, \text{m} \) in air.
- The force between them in air is given by Coulomb's law:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]

2. **Setting Up the New Scenario**:
- The same charges are placed in oil, which has a relative permittivity \( \epsilon_r = 5 \).
- The new separation between the charges in oil is \( r_1 \).
- The force between the charges in oil is given by:
\[ F_1 = \frac{1}{4 \pi \epsilon_0 \epsilon_r} \frac{q_1 q_2}{r_1^2} \]

3. **Equating the Forces**:
- According to the problem, the force remains the same when the charges are moved from air to oil:
\[ F = F_1 \]
- Therefore, we can write:
\[ \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} = \frac{1}{4 \pi \epsilon_0 \epsilon_r} \frac{q_1 q_2}{r_1^2} \]

4. **Canceling Common Terms**:
- Cancel \( \frac{1}{4 \pi \epsilon_0} \) and \( q_1 q_2 \) from both sides:
\[ \frac{1}{r^2} = \frac{1}{\epsilon_r r_1^2} \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ r_1^2 = \epsilon_r \cdot r^2 \]

6. **Substituting Known Values**:
- Substitute \( r = 50 \, \text{m} \) and \( \epsilon_r = 5 \):
\[ r_1^2 = 5 \cdot (50)^2 \]
\[ r_1^2 = 5 \cdot 2500 = 12500 \]

7. **Finding the New Separation**:
- Taking the square root to find \( r_1 \):
\[ r_1 = \sqrt{12500} = \sqrt{125 \times 100} = 10 \sqrt{125} = 10 \times 11.18 \approx 111.8 \, \text{m} \]

8. **Final Calculation**:
- The new separation \( r_1 \) is approximately \( 111.8 \, \text{m} \).

### Final Answer:
The new separation between the two charges in oil is approximately \( 111.8 \, \text{m} \).