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

Options

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

Answer

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

### 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 is given by Coulomb's law:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]
Here, \( \epsilon_0 \) is the permittivity of free space.

2. **Force in a Dielectric Medium:**
When the charges are placed in a medium with dielectric constant \( K \), the force \( F' \) is given by:
\[ F' = \frac{1}{4 \pi \epsilon} \frac{q_1 q_2}{r'^2} \]
where \( \epsilon = K \epsilon_0 \) is the permittivity of the medium.

3. **Setting the Forces Equal:**
To find the distance \( r' \) in the dielectric medium where the force is the same as in air, we set \( F = F' \):
\[ \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} = \frac{1}{4 \pi K \epsilon_0} \frac{q_1 q_2}{r'^2} \]

4. **Canceling Common Terms:**
We can cancel \( \frac{1}{4 \pi \epsilon_0} \) and \( q_1 q_2 \) 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 charges will experience the same force in the medium of dielectric constant \( K \) is:
\[ r' = \frac{r}{\sqrt{K}} \]