Two charges exert a force of 10 N on each other when separated by a distance 0.2 m in air. When they are placed in another medium of dielectric constant `K = 4`, and separated by distance R. they exert same force. The distance R equats to

Options

A. 2m
B. 1m
C. 0.2m
D. 0.1m

Answer

To solve the problem, we will use Coulomb's law, which states that the force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) in a medium with dielectric constant \( K \) is given by:

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

Where \( \epsilon \) is the permittivity of the medium, given by:

\[ \epsilon = K \epsilon_0 \]

Here, \( \epsilon_0 \) is the permittivity of free space.

### Step 1: Write the force equation in air
In the first case, when the charges are in air (where \( K = 1 \)), the force is given as:

\[ F = 10 \, \text{N} \]

The distance \( r_1 = 0.2 \, \text{m} \). Thus, we can write:

\[ 10 = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{(0.2)^2} \]

### Step 2: Write the force equation in the medium with dielectric constant \( K = 4 \)
In the second case, when the charges are in a medium with dielectric constant \( K = 4 \), the force remains the same (10 N) but the distance is \( R \). Thus, we can write:

\[ 10 = \frac{1}{4 \pi (4 \epsilon_0)} \frac{q_1 q_2}{R^2} \]

### Step 3: Set the two equations equal to each other
Since both expressions equal 10 N, we can set them equal to each other:

\[ \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{(0.2)^2} = \frac{1}{4 \pi (4 \epsilon_0)} \frac{q_1 q_2}{R^2} \]

### Step 4: Cancel out common terms
We can cancel \( \frac{1}{4 \pi \epsilon_0} \) and \( q_1 q_2 \) from both sides:

\[ \frac{1}{(0.2)^2} = \frac{1}{4} \cdot \frac{1}{R^2} \]

### Step 5: Rearrange the equation
Rearranging gives:

\[ R^2 = 4 \cdot (0.2)^2 \]

### Step 6: Calculate \( R^2 \)
Calculating \( (0.2)^2 \):

\[ (0.2)^2 = 0.04 \]

Thus,

\[ R^2 = 4 \cdot 0.04 = 0.16 \]

### Step 7: Find \( R \)
Taking the square root of both sides gives:

\[ R = \sqrt{0.16} = 0.4 \, \text{m} \]

### Conclusion
The distance \( R \) at which the charges exert the same force in the medium with dielectric constant \( K = 4 \) is:

\[ R = 0.4 \, \text{m} \]