Two point charges placed at a distance r in air exert a force F on each other. The value of distance R at which they experience force 4F when placed in a medium of dielectric constant K = 16 is :

Options

A. r
B. `r/8`
C. `r/4`
D. `r/2`

Answer

To solve the problem, we need to find the distance \( R \) at which two point charges experience a force of \( 4F \) when placed in a medium with a dielectric constant \( K = 16 \).

### 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:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]

2. **Force in a Medium with Dielectric Constant**:
When the charges are placed in a medium with dielectric constant \( K \), the force \( F' \) becomes:
\[ F' = \frac{1}{4 \pi \epsilon} \frac{q_1 q_2}{R^2} \]
where \( \epsilon = K \cdot \epsilon_0 \). For \( K = 16 \):
\[ \epsilon = 16 \epsilon_0 \]

3. **Substituting the Dielectric Constant**:
Therefore, the force in the medium can be expressed as:
\[ F' = \frac{1}{4 \pi (16 \epsilon_0)} \frac{q_1 q_2}{R^2} \]

4. **Setting Up the Equation**:
We know that \( F' = 4F \). Substituting for \( F \) gives:
\[ 4F = 4 \left( \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \right) \]
This means:
\[ 4 \left( \frac{1}{4 \pi (16 \epsilon_0)} \frac{q_1 q_2}{R^2} \right) = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \]

5. **Simplifying the Equation**:
Canceling \( q_1 q_2 \) and \( 4 \pi \epsilon_0 \) from both sides, we get:
\[ \frac{4}{16} \frac{1}{R^2} = \frac{1}{r^2} \]
Simplifying gives:
\[ \frac{1}{4R^2} = \frac{1}{r^2} \]

6. **Cross-Multiplying**:
Cross-multiplying yields:
\[ r^2 = 4R^2 \]

7. **Solving for \( R \)**:
Taking the square root of both sides results in:
\[ R = \frac{r}{2} \]

### Conclusion:
The distance \( R \) at which the charges experience a force of \( 4F \) in a medium with a dielectric constant \( K = 16 \) is:
\[ R = \frac{r}{2} \]