If K = 2 then the relation between force in air and force in medium will be:

Options

A. `2 F_(M) = F_(A)`
B. `4 F_(M) = F_(A)`
C. `F_(M) = F_(A)`
D. `2 F_(A) = F_(M)`

Explanation

To solve the problem, we need to establish the relationship between the force in air (F_A) and the force in a medium (F_M) given that the dielectric constant (K) is equal to 2.

### Step-by-Step Solution:

1. **Understanding the Forces**:
- The force between two charges in a vacuum (or air) is given by Coulomb's law:
\[ F_A = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]
- In a medium with a dielectric constant K, the force is modified to:
\[ F_M = \frac{1}{4 \pi \epsilon_0 K} \frac{q_1 q_2}{r^2} \]

2. **Substituting K**:
- Given that K = 2, we can substitute this value into the equation for the force in the medium:
\[ F_M = \frac{1}{4 \pi \epsilon_0 \cdot 2} \frac{q_1 q_2}{r^2} \]

3. **Finding the Ratio of Forces**:
- To find the relationship between the forces in air and in the medium, we take the ratio \( \frac{F_A}{F_M} \):
\[ \frac{F_A}{F_M} = \frac{\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}}{\frac{1}{4 \pi \epsilon_0 \cdot 2} \frac{q_1 q_2}{r^2}} \]

4. **Simplifying the Ratio**:
- The \( \frac{1}{4 \pi \epsilon_0} \) and \( \frac{q_1 q_2}{r^2} \) terms cancel out:
\[ \frac{F_A}{F_M} = \frac{1}{\frac{1}{2}} = 2 \]

5. **Final Relationship**:
- This means that:
\[ F_A = 2 F_M \]
- Therefore, the force in air is twice the force in the medium.

### Conclusion:
The relationship between the force in air (F_A) and the force in the medium (F_M) when K = 2 is:
\[ F_A = 2 F_M \]