Two point charges placed in a medium of dielectric constant 5 are at a distance r between them, experience an electrostatic force ‘F’. The electrostatic force between them in vacuum at the same distance r will be-

Options

A. 5F
B. F
C. F/2
D. F/5

Answer

To solve the problem, we need to understand the relationship between the electrostatic force in a dielectric medium and in a vacuum. Here’s a step-by-step solution:

### Step 1: Understand the Force in a Dielectric Medium
The electrostatic force \( F \) between two point charges \( Q_1 \) and \( Q_2 \) in a dielectric medium is given by the formula:
\[ F = \frac{1}{4 \pi \epsilon_r} \cdot \frac{Q_1 Q_2}{r^2} \]
where \( \epsilon_r \) is the dielectric constant of the medium. In this case, \( \epsilon_r = 5 \).

### Step 2: Write the Force in the Dielectric Medium
Substituting the value of the dielectric constant into the formula, we have:
\[ F = \frac{1}{4 \pi (5 \epsilon_0)} \cdot \frac{Q_1 Q_2}{r^2} \]
This can be simplified to:
\[ F = \frac{Q_1 Q_2}{20 \pi \epsilon_0 r^2} \]

### Step 3: Calculate the Force in Vacuum
In a vacuum, the electrostatic force \( F' \) is given by:
\[ F' = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_1 Q_2}{r^2} \]

### Step 4: Relate the Forces
We can relate the force in vacuum \( F' \) to the force in the dielectric medium \( F \):
\[ F' = 5F \]
This is because the dielectric constant reduces the force by a factor equal to its value. Therefore, the force in vacuum is 5 times the force in the dielectric medium.

### Step 5: Conclusion
Thus, the electrostatic force between the charges in vacuum at the same distance \( r \) will be:
\[ F' = 5F \]

### Final Answer
The electrostatic force between the charges in vacuum at the same distance \( r \) will be \( 5F \).

---