A parallel plate capacitor with plate area $$A$$ and plate separation $$d$$ is filled with a dielectric material of dielectric constant $$K = 4$$. The thickness of the dielectric material is $$x$$, where $$x < d$$.

Let $$C_1$$ and $$C_2$$ be the capacitance of the system for $$x = \dfrac{1}{3}d$$ and $$x = \dfrac{2d}{3}$$, respectively. If $$C_1 = 2$$ $$\mu$$F, the value of $$C_2$$ is ______ $$\mu$$F.

We have a parallel plate capacitor with plate area $$A$$ and plate separation $$d$$, containing a dielectric of constant $$K = 4$$ and thickness $$x$$.

Treating it as two capacitors in series (one with dielectric and one without), the capacitance is

$$C = \frac{\varepsilon_0 A}{(d - x) + \frac{x}{K}} = \frac{\varepsilon_0 A}{d - x + \frac{x}{4}} = \frac{\varepsilon_0 A}{d - \frac{3x}{4}}$$

Case 1: For $$x = \frac{d}{3}$$,

$$C_1 = \frac{\varepsilon_0 A}{d - \frac{3}{4} \cdot \frac{d}{3}} = \frac{\varepsilon_0 A}{d - \frac{d}{4}} = \frac{\varepsilon_0 A}{\frac{3d}{4}} = \frac{4\varepsilon_0 A}{3d} = 2\;\mu\text{F}$$

Case 2: For $$x = \frac{2d}{3}$$,

$$C_2 = \frac{\varepsilon_0 A}{d - \frac{3}{4} \cdot \frac{2d}{3}} = \frac{\varepsilon_0 A}{d - \frac{d}{2}} = \frac{\varepsilon_0 A}{\frac{d}{2}} = \frac{2\varepsilon_0 A}{d}$$

Now taking the ratio,

$$\frac{C_2}{C_1} = \frac{\frac{2\varepsilon_0 A}{d}}{\frac{4\varepsilon_0 A}{3d}} = \frac{2 \times 3}{4} = \frac{3}{2}$$

So $$C_2 = \frac{3}{2} \times C_1 = \frac{3}{2} \times 2 = 3\;\mu\text{F}$$.

Hence, the answer is $$3$$ $$\mu$$F. So, the answer is $$3$$.