A parallel plate air capacitor has a capacitance $C$. When it is half filled as show in figure with a dielectric constant $K = 5$, the percentage increase in the capacitance is _______.

Initial capacitance of the air capacitor is given by:

$C = \dfrac{\epsilon_0 A}{d}$

When the capacitor is half-filled with a dielectric of thickness $d/2$, it can be considered as two capacitors connected in series.

Capacitance of the dielectric-filled part:

$C_1 = \dfrac{K \epsilon_0 A}{d/2} = \dfrac{2K \epsilon_0 A}{d} = 2KC$

Capacitance of the air-filled part:

$C_2 = \dfrac{\epsilon_0 A}{d/2} = \dfrac{2 \epsilon_0 A}{d} = 2C$

The equivalent capacitance $C_{eq}$ of the series combination is:

$C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2} = \dfrac{(2KC)(2C)}{2KC + 2C} = \dfrac{2K}{K+1} C$

Substituting $K = 5$:

$C_{eq} = \dfrac{2(5)}{5+1} C = \dfrac{10}{6} C = \dfrac{5}{3} C$

The increase in capacitance is:

$\Delta C = C_{eq} - C = \dfrac{5}{3} C - C = \dfrac{2}{3} C$

The percentage increase in capacitance is:

$\% \text{ increase} = \dfrac{\Delta C}{C} \times 100 = \dfrac{2}{3} \times 100 \approx 66.67\%$

Answer: $66.67$