The ratio of the forces between two charges placed at a certain distance apart in air at half of the distance apart in medium of dielectric .k. is

Options

A. A) `1: 4k`
B. B) `k: 4`
C. C) `4k: 1`
D. D) `4: k`

Answer

To solve the problem, we need to find the ratio of the forces between two charges placed at a certain distance apart in air and at half of the distance apart in a medium with dielectric constant \( k \).

### Step-by-Step Solution:

1. **Understanding the Force Between Charges**:
The force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( d \) in air (or vacuum) is given by Coulomb's law:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{d^2} \]
where \( \epsilon_0 \) is the permittivity of free space.

2. **Force in Air**:
Let’s denote the force between the charges when they are separated by distance \( d \) in air as \( F_1 \):
\[ F_1 = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{d^2} \]

3. **Force in a Dielectric Medium**:
When the same charges are placed at half the distance apart in a medium with dielectric constant \( k \), the distance becomes \( \frac{d}{2} \). The force \( F_2 \) in the dielectric medium is given by:
\[ F_2 = \frac{1}{4 \pi \epsilon_0 k} \frac{q_1 q_2}{\left(\frac{d}{2}\right)^2} \]
Simplifying this:
\[ F_2 = \frac{1}{4 \pi \epsilon_0 k} \frac{q_1 q_2}{\frac{d^2}{4}} = \frac{1}{4 \pi \epsilon_0 k} \cdot \frac{4 q_1 q_2}{d^2} = \frac{4}{4 \pi \epsilon_0 k} \frac{q_1 q_2}{d^2} \]

4. **Finding the Ratio of Forces**:
Now, we need to find the ratio \( \frac{F_1}{F_2} \):
\[ \frac{F_1}{F_2} = \frac{\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{d^2}}{\frac{4}{4 \pi \epsilon_0 k} \frac{q_1 q_2}{d^2}} \]
The \( \frac{1}{4 \pi \epsilon_0} \), \( q_1 q_2 \), and \( d^2 \) terms cancel out:
\[ \frac{F_1}{F_2} = \frac{1}{\frac{4}{k}} = \frac{k}{4} \]

5. **Final Ratio**:
Thus, the ratio of the forces \( F_1 \) to \( F_2 \) is:
\[ F_1 : F_2 = 1 : 4k \]

### Conclusion:
The correct answer is \( 1 : 4k \).