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

Explanation

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 Force Between Two Charges
The force of attraction \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) in a vacuum is given by Coulomb's law:
\[ F_0 = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Introduce the Dielectric Medium
When a dielectric medium with dielectric constant \( K \) is introduced, the permittivity of the medium \( \epsilon \) becomes:
\[ \epsilon = K \cdot \epsilon_0 \]
Thus, the force of attraction \( F \) in the dielectric medium can be expressed as:
\[ F = \frac{1}{4 \pi \epsilon} \frac{q_1 q_2}{r^2} = \frac{1}{4 \pi K \epsilon_0} \frac{q_1 q_2}{r^2} \]

### Step 3: Relate the Forces
To compare the forces in vacuum and in the dielectric medium, we can write:
\[ F = \frac{F_0}{K} \]
where \( F_0 \) is the force in vacuum.

### Step 4: Analyze the Effect of the Dielectric Constant
From the equation \( F = \frac{F_0}{K} \), we can see that the force \( F \) in the dielectric medium is inversely proportional to the dielectric constant \( K \). Therefore, as \( K \) increases, the force \( F \) decreases.

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

### Final Answer
The correct option is that the force decreases by \( K \) times.

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