1. Initial capacitor

For a parallel plate capacitor with air between plates, $C_0=\frac{\varepsilon_0 A}{d}$ Given: $C_0=5\,\mu\text{F}$

2. After inserting dielectric slab

A dielectric slab of:

• dielectric constant $K=1.5$
• thickness $t=\dfrac{d}{2}$
• area equal to plate area

is inserted between the plates.

So the space between the plates now has two layers in series:

• dielectric of thickness $\dfrac{d}{2}$ and permittivity $K\varepsilon_0$
• air of thickness $d-\dfrac{d}{2}=\dfrac{d}{2}$ and permittivity $\varepsilon_0$

3. Equivalent capacitance using effective separation

For layered media filling the full area, the equivalent capacitance is $C=\frac{\varepsilon_0 A}{\dfrac{d_1}{K_1}+\dfrac{d_2}{K_2}}$

Here, $d_1=\frac{d}{2},\quad K_1=1.5$ $d_2=\frac{d}{2},\quad K_2=1$

Thus, $C=\frac{\varepsilon_0 A}{\dfrac{d/2}{1.5}+\dfrac{d/2}{1}}$

Now simplify: $\frac{d/2}{1.5}=\frac{d}{3}, \qquad \frac{d/2}{1}=\frac{d}{2}$

So, $C=\frac{\varepsilon_0 A}{\frac{d}{3}+\frac{d}{2}}$ $=\frac{\varepsilon_0 A}{\frac{5d}{6}}$ $=\frac{6}{5}\cdot \frac{\varepsilon_0 A}{d}$

But $\frac{\varepsilon_0 A}{d}=C_0=5\,\mu\text{F}$

Therefore, $C=\frac{6}{5}\times 5=6\,\mu\text{F}$

4. Final answer

The capacitance in presence of the slab is $\boxed{6\,\mu\text{F}}$