The air capacitor has a capacitance of 2μ F. Now the space between...

Let the area of the plates be $A$ and the distance between them be $d$.

The initial capacitance of the air capacitor is given by:

$C_0 = \dfrac{\epsilon_0 A}{d} = 2 \mu\text{F}$

From the given figure, the terminals are connected to the left and right faces, so they act as the plates of the capacitor. The space between the plates is divided horizontally into two equal halves, each with area $A/2$ and thickness $d$.

This arrangement is equivalent to two capacitors connected in parallel.

Capacitance of the upper half with dielectric $k$ is:

$C_1 = \dfrac{k \epsilon_0 (A/2)}{d} = \dfrac{k}{2} \left( \dfrac{\epsilon_0 A}{d} \right) = \dfrac{k}{2} C_0$

Capacitance of the lower half with dielectric $3k$ is:

$C_2 = \dfrac{3k \epsilon_0 (A/2)}{d} = \dfrac{3k}{2} \left( \dfrac{\epsilon_0 A}{d} \right) = \dfrac{3k}{2} C_0$

The equivalent capacitance of the parallel combination is:

$C_{eq} = C_1 + C_2$

$C_{eq} = \dfrac{k}{2} C_0 + \dfrac{3k}{2} C_0 = \dfrac{4k}{2} C_0 = 2k C_0$

Substituting the value of $C_0 = 2 \mu\text{F}$:

$C_{eq} = 2k \times 2 = 4k \mu\text{F}$

Answer: $4k$