A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be

66.6%

The correct answer is Option (3) → 66.6%

Capacitance of parallel plate capacitor partially filled with dielectric:

For dielectric filling half the distance, equivalent capacitance is series combination:

$C' = \frac{C_1 C_2}{C_1 + C_2}$

Here, $C_1 = \varepsilon_0 \frac{A}{d/2} = 2C$ (air part)

$C_2 = 5 \varepsilon_0 \frac{A}{d/2} = 10C$ (dielectric part)

Equivalent capacitance:

$\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2C} + \frac{1}{10C} = \frac{5+1}{10C} = \frac{6}{10C}$

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

Percentage increase:

% Increase = $\frac{C' - C}{C} \times 100 = \frac{5C/3 - C}{C} \times 100 = \frac{2}{3} \times 100 \approx 66.67\%$

Percentage increase in capacitance ≈ 66.7%