1. Capacitance with air only

For a parallel plate capacitor of plate area $A$ and separation $d$, with air between the plates:

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

2. Interpretation of the new arrangement

"Half space between the plates is filled with dielectric" means half of the area between the plates is filled with dielectric of relative permittivity $\varepsilon_r$, while the other half remains air.

So the system behaves like two capacitors in parallel, each having:

• plate separation $d$
• area $A/2$

3. Capacitance of each part

• Air-filled half:

$C_1 = \frac{\varepsilon_0 (A/2)}{d} = \frac{C_0}{2}$

• Dielectric-filled half:

$C_2 = \frac{\varepsilon_0 \varepsilon_r (A/2)}{d} = \frac{\varepsilon_r C_0}{2}$

4. Equivalent capacitance

Since these two parts are in parallel:

$C = C_1 + C_2$

$C = \frac{C_0}{2} + \frac{\varepsilon_r C_0}{2}$

$C = \frac{C_0}{2}(1+\varepsilon_r)$

5. Check options

• A: $\dfrac{C_0}{2}(1+\varepsilon_r)$ ✅
• B: $C_0 + \varepsilon_r$ ❌ dimensionally incorrect
• C: $\dfrac{C_0\varepsilon_r}{2}$ ❌ misses air half contribution
• D: $C_0(1+\varepsilon_r)$ ❌ twice the correct value

Hence, the correct answer is:

$\boxed{\frac{C_0}{2}(1+\varepsilon_r)}$