Two point charges placed at a distance `r` in air experience a certain force. Then the distance at which they will experience the same force in a medium of dielectric constant `K` is

Options

A. `r//K`
B. `Kr`
C. `(r )/(sqrt(K))`
D. `r+sqrt(K)`

Explanation

To solve the problem, we need to determine the distance at which two point charges will experience the same force in a medium with a dielectric constant \( K \) as they do in air at a distance \( r \).

### Step-by-Step Solution:

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

2. **Force in a Medium with Dielectric Constant \( K \)**:
When the same charges are placed in a medium with dielectric constant \( K \), the force \( F_m \) is given by:
\[ F_m = \frac{q_1 q_2}{4 \pi \epsilon_0 K r'^2} \]
where \( r' \) is the new distance between the charges in the medium.

3. **Set the Forces Equal**:
According to the problem, we want the force in the medium to be equal to the force in air:
\[ F = F_m \]
Substituting the expressions for \( F \) and \( F_m \):
\[ \frac{q_1 q_2}{4 \pi \epsilon_0 r^2} = \frac{q_1 q_2}{4 \pi \epsilon_0 K r'^2} \]

4. **Cancel Common Terms**:
Since \( q_1 \) and \( q_2 \) are non-zero, we can cancel them from both sides:
\[ \frac{1}{r^2} = \frac{1}{K r'^2} \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ K r'^2 = r^2 \]

6. **Solving for \( r' \)**:
Dividing both sides by \( K \):
\[ r'^2 = \frac{r^2}{K} \]
Taking the square root of both sides:
\[ r' = \frac{r}{\sqrt{K}} \]

### Final Answer:
The distance at which the two point charges will experience the same force in a medium of dielectric constant \( K \) is:
\[ r' = \frac{r}{\sqrt{K}} \]