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.