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: \(66.67\)