If two charges `q_1` and `q_2` are separated withdistance `d` and palced in a medium of dielectric constant K. what will be the equivalent distance between charge in air for the same electrostatic force?

Options

A. `ksqrtd`
B. `1.5dsqrtd`
C. `2dsqrtk`
D. `dsqrtk`

Answer

To solve the problem, we need to find the equivalent distance between two charges \( q_1 \) and \( q_2 \) in air that would produce the same electrostatic force as when they are separated by distance \( d \) in a medium with dielectric constant \( K \).

### Step-by-Step Solution:

1. **Understand the Electrostatic Force Formula**:
The electrostatic force \( F \) between two point charges in a medium is given by:
\[ F = \frac{1}{4\pi \epsilon} \cdot \frac{q_1 q_2}{r^2} \]
In a medium with dielectric constant \( K \), the formula modifies to:
\[ F_1 = \frac{1}{4\pi \epsilon K} \cdot \frac{q_1 q_2}{d^2} \]

2. **Write the Force in Air**:
When the charges are in air (or vacuum), the force is given by:
\[ F_2 = \frac{1}{4\pi \epsilon} \cdot \frac{q_1 q_2}{d'^2} \]
where \( d' \) is the distance between the charges in air.

3. **Set the Forces Equal**:
Since we want the forces to be equal, we set \( F_1 \) equal to \( F_2 \):
\[ \frac{1}{4\pi \epsilon K} \cdot \frac{q_1 q_2}{d^2} = \frac{1}{4\pi \epsilon} \cdot \frac{q_1 q_2}{d'^2} \]

4. **Cancel Common Terms**:
We can cancel \( \frac{1}{4\pi \epsilon} \) and \( q_1 q_2 \) from both sides (assuming \( q_1 \) and \( q_2 \) are not zero):
\[ \frac{1}{K} \cdot \frac{1}{d^2} = \frac{1}{d'^2} \]

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

6. **Taking the Square Root**:
Taking the square root of both sides, we find:
\[ d' = d \sqrt{K} \]

### Final Answer:
Thus, the equivalent distance \( d' \) between the charges in air for the same electrostatic force is:
\[ d' = d \sqrt{K} \]