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

Let the plates have area $$A$$ and separation $$d$$.
For air between the plates the capacitance is

$$C=\frac{\varepsilon_0 A}{d} \quad -(1)$$

The dielectric slab of constant $$K=5$$ completely occupies one-half of the gap (thickness $$d/2$$).
Hence the space between the plates now behaves like two capacitors in series:

Case 1: Region filled with dielectric
Thickness $$=\frac{d}{2}$$, permittivity $$=\varepsilon_0 K$$

$$C_1=\frac{\varepsilon_0 K A}{d/2}=2K\,\frac{\varepsilon_0 A}{d}=2K\,C \quad -(2)$$

Case 2: Region still filled with air
Thickness $$=\frac{d}{2}$$, permittivity $$=\varepsilon_0$$

$$C_2=\frac{\varepsilon_0 A}{d/2}=2\,\frac{\varepsilon_0 A}{d}=2\,C \quad -(3)$$

For capacitors in series:

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

Substituting from $$(2)$$ and $$(3)$$:

$$\frac{1}{C'}=\frac{1}{2K\,C}+\frac{1}{2\,C}=\frac{1+K}{2K\,C}$$

$$\Rightarrow\; C'=\frac{2K\,C}{1+K} \quad -(4)$$

For $$K=5$$:

$$C'=\frac{2\times5\,C}{1+5}=\frac{10\,C}{6}=\frac{5}{3}C\approx1.6667\,C$$

Percentage increase in capacitance:

$$\%\text{ increase}= \frac{C'-C}{C}\times100 =\left(\frac{5}{3}-1\right)\times100=\frac{2}{3}\times100=66.67\%$$

Therefore, the capacitance increases by $$66.67\%$$.

Option B which is: $$66.67$$