Solution:

Let’s solve the problem step by step:

Given:

• Initial capacitance: $C$
• The separation between the plates is doubled, and a dielectric medium is inserted.
• The new capacitance becomes $2C$.

Step 1: Capacitance formula

The capacitance of a parallel plate capacitor is given by:

$$C = \frac{\varepsilon_0 A}{d}$$

where:

• $C$ is the capacitance,
• $\varepsilon_0$ is the permittivity of free space,
• $A$ is the area of the plates, and
• $d$ is the separation between the plates.

Step 2: Effect of doubling the separation and adding a dielectric

When the separation $d$ is doubled, the capacitance would normally decrease by a factor of 2 (since capacitance is inversely proportional to $d$).

Now, when a dielectric of dielectric constant $K$ is inserted, the capacitance increases by a factor of $K$. So, the new capacitance $C'$ is:

$$C' = K \cdot \frac{\varepsilon_0 A}{2d} = K \cdot \frac{C}{2}$$

We are told that the new capacitance is $2C$, so:

$$K \cdot \frac{C}{2} = 2C$$

Step 3: Solve for $K$

$$K = 4$$

Final Answer:

The dielectric constant of the medium is 4.