An air filled capacitor of capacitance C is filled with dielectric (k = 3) of width d/3, where d is separation between plates. The new capacitance is

Let the initial capacitance of the air-filled capacitor be $C$. The distance between the parallel plates is $d$, and let the plate area be $A$. The capacitance of a parallel plate capacitor filled with air (or vacuum) is given by:
$C=\frac{{\epsilon }_{0}A}{d}$

Now, a dielectric slab of dielectric constant $k=3$ and thickness $t=\frac{d}{3}$ is inserted between the plates. The remaining space of thickness $d-t=d-\frac{d}{3}=\frac{2d}{3}$ is still filled with air.

This setup can be modeled as two capacitors connected in series:
1. A capacitor $C_1$ filled with dielectric of thickness $d_1=\frac{d}{3}$ and dielectric constant $k=3$.
2. A capacitor $C_2$ filled with air of thickness $d_2=\frac{2d}{3}$.

Let's calculate the individual capacitances:
$C_1=\frac{k{\epsilon }_{0}A}{d_1}=\frac{3{\epsilon }_{0}A}{d/3}=9·\frac{{\epsilon }_{0}A}{d}=9C$
$C_2=\frac{{\epsilon }_{0}A}{d_2}=\frac{{\epsilon }_{0}A}{2d/3}=\frac{3}{2}·\frac{{\epsilon }_{0}A}{d}=\frac{3}{2}C$

The equivalent capacitance $C^{\text{'}}$ of two capacitors connected in series is given by:
$\frac{1}{C^{\text{'}}}=\frac{1}{C_1}+\frac{1}{C_2}$

Substitute the values of $C_1$ and $C_2$ into the equation:
$\frac{1}{C^{\text{'}}}=\frac{1}{9C}+\frac{1}{\frac{3}{2}C}=\frac{1}{9C}+\frac{2}{3C}$

Find a common denominator to add the fractions:
$\frac{1}{C^{\text{'}}}=\frac{1+6}{9C}=\frac{7}{9C}$

Taking the reciprocal to find $C^{\text{'}}$:
$C^{\text{'}}=\frac{9}{7}C$

Therefore, the new capacitance of the capacitor is $\frac{9}{7}C$.