When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance:

Options

A. increases K times
B. remains unchanged
C. decreases K times
D. increases `K^(-1)` times

Answer

To solve the problem of how the maximum force of attraction between two charges changes when air is replaced by a dielectric medium of constant \( K \), we can follow these steps:

### Step 1: Understand the Original Force in Air
The force between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) in air is given by Coulomb's law:
\[ F = \frac{1}{4\pi \epsilon_0} \cdot \frac{q_1 q_2}{r^2} \]
Here, \( \epsilon_0 \) is the permittivity of free space, and in air, we can consider the dielectric constant \( K \) to be approximately 1.

### Step 2: Introduce the Dielectric Medium
When air is replaced by a dielectric medium with a dielectric constant \( K \), the force between the charges changes. The new force \( F' \) in the dielectric medium is given by:
\[ F' = \frac{1}{4\pi \epsilon} \cdot \frac{q_1 q_2}{r^2} \]
where \( \epsilon = K \cdot \epsilon_0 \).

### Step 3: Relate the New Force to the Original Force
Substituting \( \epsilon \) into the equation for \( F' \):
\[ F' = \frac{1}{4\pi K \epsilon_0} \cdot \frac{q_1 q_2}{r^2} \]
This can be rewritten in terms of the original force \( F \):
\[ F' = \frac{F}{K} \]
This shows that the new force \( F' \) is equal to the original force \( F \) divided by the dielectric constant \( K \).

### Step 4: Conclusion
Since \( K \) is greater than 1 for any dielectric medium, it follows that:
\[ F' < F \]
Thus, the maximum force of attraction between the charges decreases by a factor of \( K \).

### Final Answer
The maximum force of attraction between the two charges decreases by a factor of \( K \) when air is replaced by a dielectric medium of constant \( K \).