Two charges `q_1` and `q_2` are placed in vacuum at a distance d and the force acting between them is F . If a medium of dielectric constant 4 is introduced around them, the force now will be

Options

A. `4 F `
B. `2F`
C. `F/2`
D. `F/4`

Answer

To solve the problem, we need to understand how the introduction of a dielectric medium affects the electrostatic force between two charges. Here’s a step-by-step solution:

### Step 1: Understand the initial force between the charges
The force \( F \) between two charges \( q_1 \) and \( q_2 \) placed at a distance \( d \) in a vacuum is given by Coulomb's law:
\[ F = \frac{k \cdot q_1 \cdot q_2}{d^2} \]
where \( k \) is the electrostatic constant in vacuum, which can be expressed as:
\[ k = \frac{1}{4 \pi \epsilon_0} \]
Thus, we can rewrite the force as:
\[ F = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{d^2} \]

### Step 2: Introduce the dielectric medium
When a dielectric medium with a dielectric constant \( \epsilon_r = 4 \) is introduced, the new electrostatic constant \( k' \) becomes:
\[ k' = \frac{1}{4 \pi \epsilon_0 \cdot \epsilon_r} = \frac{1}{4 \pi \epsilon_0 \cdot 4} = \frac{1}{16 \pi \epsilon_0} \]

### Step 3: Calculate the new force
The new force \( F' \) between the charges in the dielectric medium is given by:
\[ F' = \frac{k' \cdot q_1 \cdot q_2}{d^2} \]
Substituting for \( k' \):
\[ F' = \frac{1}{16 \pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{d^2} \]

### Step 4: Relate the new force to the original force
Now, we can express the new force \( F' \) in terms of the original force \( F \):
\[ F' = \frac{1}{4} \cdot \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{d^2} = \frac{F}{4} \]

### Conclusion
Thus, the force acting between the charges when the dielectric medium is introduced is:
\[ F' = \frac{F}{4} \]