1. Initial capacitor

For the original parallel plate capacitor,

$C_0=\frac{\varepsilon_0 A}{d}$

Given:

$C_0=4\,\mu F, \qquad d=2\,\text{m}$

So the geometry satisfies

$\frac{\varepsilon_0 A}{2}=4\,\mu F$

2. Interpret the dielectric filling

"Half of the space between them is filled with dielectric" as shown in the usual figure means the dielectric slab occupies half the separation, i.e. thickness

$d_1=\frac{d}{2}=1\,\text{m}$

with dielectric constant

$K=3$

The remaining half is air of thickness

$d_2=\frac{d}{2}=1\,\text{m}$

Since the layers are along the separation, this is equivalent to two capacitors in series having the same plate area $A$:

• one with dielectric $K=3$, thickness $d/2$
• one with air, thickness $d/2$

3. Capacitance of each part

For the dielectric-filled half:

$C_1=\frac{K\varepsilon_0 A}{d/2}=\frac{2K\varepsilon_0 A}{d}$

Using $\frac{\varepsilon_0 A}{d}=C_0=4\,\mu F$,

$C_1=2K C_0=2\times 3\times 4=24\,\mu F$

For the air-filled half:

$C_2=\frac{\varepsilon_0 A}{d/2}=\frac{2\varepsilon_0 A}{d}=2C_0=8\,\mu F$

4. Series combination

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$

$\frac{1}{C_{eq}}=\frac{1}{24}+\frac{1}{8}=\frac{1+3}{24}=\frac{4}{24}=\frac{1}{6}$

Hence,

$C_{eq}=6\,\mu F$

5. Check options

• A: $2\,\mu F$ ❌
• B: $32\,\mu F$ ❌
• C: $6\,\mu F$ ✅
• D: $8\,\mu F$ ❌

Therefore, the correct answer is:

$\boxed{6\,\mu F}$