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
B. `r//K`
C. `r//sqrtK`
D. `r sqrtK`

Answer

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

### Step-by-Step Solution:

1. **Understanding the Force in Air:**
The electrostatic force \( F \) between two point charges \( Q_1 \) and \( Q_2 \) separated by a distance \( r \) in air (where the dielectric constant \( K = 1 \)) is given by Coulomb's law:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2} \]

2. **Understanding the Force in a Medium:**
When the charges are placed in a medium with dielectric constant \( K \), the force \( F' \) between the same charges \( Q_1 \) and \( Q_2 \) separated by a distance \( r' \) is given by:
\[ F' = \frac{1}{4 \pi K \epsilon_0} \frac{Q_1 Q_2}{(r')^2} \]

3. **Setting the Forces Equal:**
According to the problem, we want the forces to be equal:
\[ F = F' \]
Therefore, we can equate the two expressions:
\[ \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2} = \frac{1}{4 \pi K \epsilon_0} \frac{Q_1 Q_2}{(r')^2} \]

4. **Canceling Common Terms:**
We can cancel \( Q_1 Q_2 \) and \( 4 \pi \epsilon_0 \) from both sides:
\[ \frac{1}{r^2} = \frac{1}{K (r')^2} \]

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

6. **Solving for \( r' \):**
To find \( r' \), we take the square root of both sides:
\[ (r')^2 = \frac{r^2}{K} \]
\[ r' = \frac{r}{\sqrt{K}} \]

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