The force of attraction between two charges at distance r, when the air is replaced by a medium of dielectric constant k.

Options

A. Increases `k^(2)` times
B. decreases K times
C. remains uncharged
D. increases K times

Explanation

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

### Step 1: Understand the Formula for Force Between Charges
The force of attraction \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) in a vacuum (or air) is given by Coulomb's Law:
\[ F = \frac{k_e |q_1 q_2|}{r^2} \]
where \( k_e \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

### Step 2: Introduce the Dielectric Constant
When the medium is replaced by a dielectric medium with a dielectric constant \( k \), the force of attraction is modified. The presence of the dielectric reduces the force between the charges by a factor of \( k \). Therefore, the modified formula becomes:
\[ F' = \frac{k_e |q_1 q_2|}{k r^2} \]

### Step 3: Write the Final Expression
Thus, the force of attraction between the two charges when the medium is replaced by a dielectric with constant \( k \) is:
\[ F' = \frac{F}{k} \]
where \( F \) is the original force in air.

### Final Answer
The force of attraction between the two charges at distance \( r \) when the air is replaced by a medium of dielectric constant \( k \) is:
\[ F' = \frac{k_e |q_1 q_2|}{k r^2} \]

---