Solution:

Given:

• Initial capacitance with air as the dielectric: $C = 9 \, \text{pF}$
• Dielectric constants: $k_1 = 3$, $k_2 = 6$
• Thicknesses of the dielectrics: $d_1 = \frac{d}{3}$ and $d_2 = \frac{2d}{3}$

Step-by-step solution:

1. Capacitance formula:

The capacitance of a parallel plate capacitor is:

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

Where:

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

When we insert dielectrics in series, we treat the system as two capacitors in series with different dielectric constants.

2. Capacitance of each section:

• For the dielectric with $k_1 = 3$ and thickness $d_1 = \frac{d}{3}$, the capacitance $C_1$ is:

$$C_1 = \frac{k_1 \varepsilon_0 A}{d_1} = \frac{3 \varepsilon_0 A}{d/3} = \frac{9 \varepsilon_0 A}{d}$$

• For the dielectric with $k_2 = 6$ and thickness $d_2 = \frac{2d}{3}$, the capacitance $C_2$ is:

$$C_2 = \frac{k_2 \varepsilon_0 A}{d_2} = \frac{6 \varepsilon_0 A}{2d/3} = \frac{9 \varepsilon_0 A}{d}$$

3. Total capacitance:

The total capacitance of the system is found by treating the two capacitances in series. For capacitors in series, the total capacitance $C_{\text{total}}$ is given by:

$$\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2}$$

Substitute $C_1 = C_2 = \frac{9 \varepsilon_0 A}{d}$:

$$\frac{1}{C_{\text{total}}} = \frac{1}{\frac{9 \varepsilon_0 A}{d}} + \frac{1}{\frac{9 \varepsilon_0 A}{d}} = \frac{2}{\frac{9 \varepsilon_0 A}{d}}$$

Thus,

$$C_{\text{total}} = \frac{9 \varepsilon_0 A}{2d}$$

4. Relating to the original capacitance:

The original capacitance with air as the dielectric is:

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

Therefore, the total capacitance becomes:

$$C_{\text{total}} = \frac{9}{2} C = 4.5 C = 4.5 \times 9 \, \text{pF} = 40.5 \, \text{pF}$$

Final Answer:

The total capacitance with the two dielectrics is 40.5 pF.

Thus, 40.5 pF is the correct answer.