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/4
C. r/8
D. 2r

Answer

To solve the problem, we need to analyze the forces between two point charges in two different scenarios: in air and in a dielectric medium.

### Step-by-Step Solution:

1. **Initial Force 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:
\[ F = \frac{Q_1 Q_2}{4 \pi \epsilon_0 r^2} \]

2. **Force in Dielectric Medium**:
When the charges are placed in a medium with a dielectric constant \( K \), the force \( F' \) between them at a new distance \( R \) is given by:
\[ F' = \frac{Q_1 Q_2}{4 \pi \epsilon_0 K R^2} \]
Given that \( K = 16 \), we can substitute this value into the equation:
\[ F' = \frac{Q_1 Q_2}{4 \pi \epsilon_0 \cdot 16 \cdot R^2} \]

3. **Relating the Forces**:
According to the problem, the force in the dielectric medium is \( 4F \):
\[ F' = 4F \]
Substituting the expression for \( F \) from step 1 into this equation gives:
\[ \frac{Q_1 Q_2}{4 \pi \epsilon_0 \cdot 16 \cdot R^2} = 4 \left( \frac{Q_1 Q_2}{4 \pi \epsilon_0 r^2} \right) \]

4. **Simplifying the Equation**:
We can cancel \( Q_1 Q_2 \) and \( 4 \pi \epsilon_0 \) from both sides:
\[ \frac{1}{16 R^2} = \frac{4}{r^2} \]

5. **Cross-Multiplying**:
Cross-multiplying gives:
\[ r^2 = 64 R^2 \]

6. **Finding R**:
Taking the square root of both sides, we find:
\[ R = \frac{r}{8} \]

Thus, the distance \( R \) at which the charges experience a force of \( 4F \) in a medium with a dielectric constant \( K = 16 \) is:
\[ \boxed{\frac{r}{8}} \]