Solution:

To solve this, we use the formula for the capacitance of a parallel plate capacitor:

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

Where:

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

Given:

• Initial capacitance without dielectric: $C_1 = 10^{-12} \, \text{F}$
• Final capacitance with wax dielectric: $C_2 = 2 \times 10^{-12} \, \text{F}$
• The separation between plates is doubled, so $d_2 = 2d_1$.

Step 1: Relate the initial and final capacitances

The capacitance of a capacitor is directly proportional to the dielectric constant and inversely proportional to the distance between the plates. So when the separation doubles and a dielectric with dielectric constant $k$ is inserted, the capacitance will change as follows:

$$C_2 = k \times \frac{C_1}{2}$$

Step 2: Solve for $k$

$$2 \times 10^{-12} = k \times \frac{10^{-12}}{2}$$ $$2 \times 10^{-12} = \frac{k \times 10^{-12}}{2}$$ $$k = 4$$

Final Answer:

The dielectric constant of wax is 4.