where,

$ C $ is the capacitance,

$ A $ is the area of one of the plates,

$ \varepsilon_0 $ is the permittivity of free space,

$ k $ is the dielectric constant of the material,

$ d $ is the separation between the plates.

For each section with a different dielectric:

First Dielectric:

Dielectric constant $ k_1 = 3 $

Thickness $ d_1 = \frac{d}{3} $

The capacitance $ C_1 $ is:

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

Second Dielectric:

Dielectric constant $ k_2 = 6 $

Thickness $ d_2 = \frac{2d}{3} $

The capacitance $ C_2 $ is:

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

Since these two sections are in series, the total capacitance $ C_{\text{total}} $ is given by:

$ C_{\text{total}} = \frac{C_1 \times C_2}{C_1 + C_2} = \frac{9C \times 9C}{9C + 9C} = \frac{81C^2}{18C} = \frac{9}{2}C $

Given that the original capacitor had a capacitance of $ 9 \, \text{pF} $, substitute this into the total capacitance equation:

$ C_{\text{total}} = \frac{9}{2} \times 9 \times 10^{-12} \text{F} $

Thus,

$ C_{\text{total}} = 40.5 \, \text{pF} $

Therefore, the capacitance of the new capacitor is $ 40.5 \, \text{pF} $.

Capacitor

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