Two point charges exert a force `F_((d))`, on each other when placed in vacuum. If the charges are increased to four times, separation between them is doubled and the system is placed in an insulating medium, then also they experience the same force. What should be the dielectric constant of the medium?

Answer

To solve the problem step by step, we will use Coulomb's law and the concept of dielectric constant.

### Step 1: Understand the initial conditions
In vacuum, the force between two point charges \( Q_1 \) and \( Q_2 \) separated by a distance \( r \) is given by Coulomb's law:
\[ F_d = \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2} \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Set up the conditions after changes
When the charges are increased to four times, we have:
\[ Q_1' = 4Q_1, \quad Q_2' = 4Q_2 \]
The distance is doubled, so:
\[ r' = 2r \]
Now, we need to find the force in the insulating medium, which is given by:
\[ F_m = \frac{1}{4 \pi \epsilon} \frac{Q_1' Q_2'}{(r')^2} \]
Substituting the new values:
\[ F_m = \frac{1}{4 \pi \epsilon} \frac{(4Q_1)(4Q_2)}{(2r)^2} \]
This simplifies to:
\[ F_m = \frac{1}{4 \pi \epsilon} \frac{16 Q_1 Q_2}{4r^2} = \frac{4}{4 \pi \epsilon} \frac{Q_1 Q_2}{r^2} \]

### Step 3: Set the forces equal
According to the problem, the forces in vacuum and the insulating medium are equal:
\[ F_d = F_m \]
Substituting the expressions we derived:
\[ \frac{1}{4 \pi \epsilon_0} \frac{Q_1 Q_2}{r^2} = \frac{4}{4 \pi \epsilon} \frac{Q_1 Q_2}{r^2} \]
We can cancel \( \frac{Q_1 Q_2}{r^2} \) from both sides (assuming \( Q_1 \) and \( Q_2 \) are not zero):
\[ \frac{1}{4 \pi \epsilon_0} = \frac{4}{4 \pi \epsilon} \]

### Step 4: Solve for the dielectric constant \( \epsilon \)
Rearranging the equation gives:
\[ \epsilon = 4 \epsilon_0 \]
The dielectric constant \( \epsilon_r \) is given by:
\[ \epsilon_r = \frac{\epsilon}{\epsilon_0} = 4 \]

### Conclusion
Thus, the dielectric constant of the medium is:
\[ \epsilon_r = 4 \]