Question

A dielectric of dielectric constant 2.5 is filled in the gap between the plates of a
capacitor. Calculate the factor by which the capacitance is increased, if the
dielectric is only sufficient to fill up one-fourth of the gap

Answer 1

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

We can consider the capacitor to be series connection of two capacitors. 

$$\frac{1}{C'}=\frac{1}{C_1}+\frac{1}{C_2}$$

$$C_1=\frac{\epsilon_0 A}{0.75d}=\frac{4}{3}C$$

$$C_2=\frac{\epsilon\epsilon_0 A}{0.25d}=\frac{2.5\epsilon_0 A}{0.25d}=10C$$

$$\frac{1}{C'}=\frac{3}{4C}+\frac{1}{10C}$$

$$C'=\frac{20}{17}C$$

The capacitance is increased by 

$$\frac{20}{17}\approx 1.18$$