Percentage Increase in Capacitance of Half Filled Dielectric Capacitor

To find the percentage increase in the capacitance, we analyze the parallel plate capacitor before and after the dielectric slab is inserted.

1. Initial Capacitance

Let the area of each plate be $A$ and the distance between the plates be $d$. The initial capacitance of the air capacitor is given by: $C = \frac{\varepsilon_0 A}{d}$

2. Capacitance with Dielectric

As shown in the figure, a dielectric slab of dielectric constant $K = 5$ and thickness $t = \frac{d}{2}$ is inserted into the capacitor, while the remaining half thickness $\frac{d}{2}$ contains air.

This setup can be represented as two capacitors connected in series:

1. $C_1$: A capacitor filled with dielectric of constant $K = 5$ and thickness $\frac{d}{2}$. $C_1 = \frac{K \varepsilon_0 A}{d/2} = \frac{2 K \varepsilon_0 A}{d} = 2KC = 2(5)C = 10C$

2. $C_2$: A capacitor filled with air of thickness $\frac{d}{2}$. $C_2 = \frac{\varepsilon_0 A}{d/2} = \frac{2 \varepsilon_0 A}{d} = 2C$

The equivalent capacitance $C'$ of the series combination is: $\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2}$

Substitute the values of $C_1$ and $C_2$: $\frac{1}{C'} = \frac{1}{10C} + \frac{1}{2C} = \frac{1 + 5}{10C} = \frac{6}{10C}$

$C' = \frac{10}{6} C = \frac{5}{3} C$

3. Percentage Increase in Capacitance

The percentage increase in capacitance is calculated as: $\text{Percentage Increase} = \left( \frac{C' - C}{C} \right) \times 100\%$

$\text{Percentage Increase} = \left( \frac{\frac{5}{3}C - C}{C} \right) \times 100\% = \frac{2}{3} \times 100\% \approx 66.67\%$

Correct Option: B